Method of orbital launch trajectories

Abstract

A method is provided for efficient orbital launch trajectories. A payload (e.g., satellite) is launched as high as a first radius with respect to the center of the Earth. The method decreases the payload altitude in response to a gravitational pull of the Earth, and ultimately the payload attains a stable orbit around the Earth at a second radius with respect to the center of the Earth. If the first radius is twice the second radius, the payload acquires a gravitational first potential energy at the first radius, and in the stable orbit the payload has a second (potential and kinetic) energy equal to the gravitational first potential energy, with the second kinetic energy being equal to the energy required to maintain a stable orbital velocity. Advantageously, the stable orbit can potentially be at any orbital inclination angle in the range between 0 and 360 degrees.

Description

METHOD FOR ORBITAL LAUNCH TRAJECTORIESRELATED APPLICATIONS

[0001] Any and all applications, if any, for which a foreign or domestic priority claim is identified in the Application Data Sheet of the present application are hereby incorporated by reference under 37 CFR 1.57.

[0002] BACKGROUND OF THE INVENTION

[0003] Field of the Invention

[0004] Fig. 1 A depicts the orbital speed in [m / s] (meter per second) versus radius measured with respect to the stationary coordinate system where its origin is in the center of Earth and altitude measured from sea level. Fig. IB depicts the orbital speed in Mach number, versus radius measured with respect to the stationary coordinate system where its origin is in the center ofEarth and altitude measured from sea level, where 1 Mach is speed of sound at sea level (1 Mach= 340 [m / s]) and the numbers 1, 5, and 10 correspond to the row numbers in Table 2. Since the first satellite launch of Sputnik, on 4thof October 1957, all satellites, of which 7,702 of them still in orbit and 6,718 of them operational, have been launched with the same launch strategy. The final goal for any orbital launch is to put the spacecraft or satellite to its desired orbit. For the spacecraft or satellite to stay in that orbit with no thrust, the magnitude of its velocity must satisfy the orbital speed related to its desired orbital height or radius as shown in Figs. 1 A andIB. As can be seen these orbital speeds are highly hypersonic, many times higher than the speed of sound and are obtainable only by employing rocket propulsion given by the “Tsiolkovsky’sRocket Equation” [1-3], Besides this, the direction of the velocity vector should also be tangent to the desired orbital trajectory.

[0005] Fig. 14 depicts a rocket being launched into orbit using a conventional pitchover or gravity turn maneuver (prior art). To accomplish the goals of obtaining orbit the rocket carrying the satellite or spacecraft is launched from the launch pad with trust-to-weight ratio (T2W) greater than 1, typically 1.5-2 during lift-off, allowing the rocket to accelerate vertically [6-8],After a safe altitude is reached, the rocket rotates around its vertical axes aligning its directional reference point towards to the true north and aligns its azimuth to match its desired inclination.As part of a “gravity turn” or “zero-lift turn”, a “roll” or “pitchover" maneuver is executed to minimize the exertion of energy. However, the rocket still expends a great deal of energy after the gravity turn to obtain the orbital velocity needed to maintain a stable orbit. The roll maneuver can be difficult to observe for cylindrical launches such as the Titan II, Falcon, orDelta, but are more easily observable for the space shuttle, Falcon Heavy, and Ariane 5 due to their asymmetrical geometry. As can be seen, to put anything in orbit successfully, a choreographed sequence of events must be done in a timely manner, which puts time pressure on every event and mission planners take many factors into consideration in this critical planning stage [8-10], An excellent explanation of each maneuver along with very impressive threedimensional graphical material related to this sequence of events is given in [6, 8, 10], Material presented in the podcast [8] by Alfanso Gonzales also gives excellent computer-generated visuals based on the solutions of equation of motion using publicly available Python

[0034] codes related to the problems. The inclination angle i of a satellite orbit is one of the five orbital parameters defining its orbit. It is the angle between the orbital plane and a reference plane, typically Earth’s equatorial plane [8], As an example, the International Space Station (ISS) has an inclination of 51.6°. Due to certain safety limitations, all the launch sites are only permitted a limited range of azimuth angle (angle from the true north) launches. There are several reasons for launch azimuth limitations for a launch site, mainly it is to avoid damage from falling debris in case of an accident or avoiding damage from the falling initial booster stages. Generally preferred permitted launch azimuths are towards to the ocean or uninhabited areas. Inclination angle i, is related to the latitude of the launch site 6 and the launch azimuth A with a simple formula given as,

[0007] This relation puts limits on the inclination angle for the satellite orbit launched from any site. As an example, all launches from Cape Canaveral in central Florida (latitude 6 =28.4°North) are only permitted the launch azimuth angle limits of 35° - 120°, giving the corresponding inclination angle limits of 57° - 39°. Similarly, all launches from Vandenberg Air Force Base in California (latitude 6 =34.7° North) are only permitted the launch azimuth angle limits of 140° - 201°, giving the corresponding inclination angle limits of 56° - 104°.

[0008] Geosynchronous orbit, (GSO) is an orbit which the satellite orbital period matches theEarth’s rotational period and can be at any orbital inclination and its orbital altitude is 35,786[km]. Most communication satellites are in a geostationary orbit, abbreviated as GEO, which is a special case of the geosynchronous orbit having zero inclination and eccentricity, in other words their orbits are circular equatorial orbits. A satellite at geosynchronous orbit remains at a fixed location for an observer on the Earth giving a great advantage in communication and any kind of satellite broadcasting.

[0009] Another orbital inclination limitation arising from the orbital inclination angle formula above is that orbital inclination angles lower than the latitude of the launch site are not possible and require inclination angle change also known as orbital plane change. This maneuver requires a change in the orbital velocity vector at the orbital nodes at the expense of additional propellant. Therefore, a “direct’ ’.launch to an orbital inclination of zero degrees or in other words launching a communication satellite “directly” from any site listed in Table 1 is not possible. There are 580 GEO satellites in orbit and all 580 of them needed energy hungry orbital inclination angle change to put them in geostationary orbit.

[0010] It would be advantageous if rockets could be launched to any inclination angle independent of launch site.

[0011] It would be advantageous if the strain on rocket engines needed to attain stable orbits could be reduced.

[0012] It would be advantageous if rocket engines required less propellant to put their payloads into stable orbits.

[0013] It would be advantageous if rockets could be launched having no inclination angle restrictions at any launch site latitude.

[0014] SUMMARY OF THE INVENTION

[0015] This invention is related to launching satellites into orbit with less propellant, as compared to the standard launch methods, permitting the use of rocket engines with a smaller power output, leading to less demand on rocket engines, and achieving higher reliability with less wear for equivalent mass and orbits. This improved safety due to a lesser stress imposed on the rocket naturally results in a longer life to the launch vehicle, especially if it is intended for multiple launches. In addition to this advantage, the method described herein permits the use of any launch site on Earth to achieve any orbital inclination angle.

[0016] The efficient orbital launch method described herein is based on converting inverse square law gravitational potential energy

[0021] to kinetic energy through the conservation of energy principle [9, 23], Let the desired circular orbital radius of the satellite or spacecraft be RORBIT- A payload (satellite or spacecraft) is launched to a radius r0, which is greater than"•"ORBIT-. the orbital speeds necessary to stay in that orbit are generated by converting its total energy at T0 to kinetic energy during the descent back to rORRIT. The radiuses mentioned here are measured from the center of the Earth and gravitational potential energy is defined as a function of radius. On the other hand, altitude from sea level is the terminology used in satellite or spacecraft business like orbital altitude. Any given altitude h can be simply calculated from a given radius T as h = 7 — TEARTHI radius r can be calculated from altitude h as; r =rEARTH + h-

[0017] It is shown that the total energy of a stationary object with a mass m at a distance of r0away from the center of a stationary spherical body with a mass M is equal to the total energy ofthe same object with the mass 771 when rotating in a circular orbit around the mass M at a radius°f TQRBIT satisfying the relation,

[0018]

[0019] This proof is the key to the launch strategy of the disclosed method and has very interesting computational applications.

[0020] Fig. 3A depicts the gravitational potential energy as a function of radius T0RBIT with respect to Earth’s center, along with the kinetic energy when in orbit at the radius T0RBIT, and its total energy. Also shown is the gravitational potential energy at 2 ■ T0RB1Twith the calculated total energy when in orbit at the radius T0RBITalong with markings of the orbital altitudes given in Table 2. Fig. 3 A appears to show three curves, but there are four curves, two of which are on top of each other. The four curves plotted in Fig. 3 A are the gravitational potential energies at radiuses at T0RBITand 2TORBITmarked as PE(r0RBIT^ and PE(2 ■ rOfiBjT), kinetic energy KE(r0RBIT) and total energy TE(T0RBIT) of the mass 771 at a circular orbit at a radius of TQRBIT from the center of the earth. The marked points on the total energy curve T E (T0RB1T3 are the total energies of a mass of TH at a circular orbit, at orbital altitudes given in Table 2 following the same numbering on the plots just to show thatT E (TORBIT)and PE ^2T0RB1T^ exactly match, which is the numerical verification of the proof given in Section 3.0 and 3.1 in detail.

[0021] Fig. 3B depicts the same plots in Fig. 3 A, where X axes plotted in logarithmic scale. This view provides a clearer view of the tight spacings between the markings for the Table 2 orbital altitudes.

[0022] Relation (1.1) means that if an object at rest is allowed to fall from a radius of FQ towards the stationary body with a mass M , it will acquire a velocity equal to the orbital velocity when rotating in a circular orbit around the mass M at the radius of T0RB1T, when it reaches r0RBrr, as shown in Fig. 3C.

[0023] Fig. 3C depicts the orbital velocity as a function of the orbital altitude VoBB / rC^OBB / T*) as shown in Fig. 1 A with the velocities gained by dropping from To= 2 • T0RB1Tto T0RB1Tfor the orbital altitudes given in Table 2. As can be seen the markers are exactly on theVORBIT(h ORBIT) curve. The orbital velocity is shown in Mach numbers as a function of altitude with the matching markings showing the vertical velocity gained by falling from

[0024] 2 ■ T0RBITto T0RB1Tas a result of the energy conservation principle.

[0025] The only problem is that the velocity vector of the falling object with the mass 771 is towards the center of the stationary spherical body with the mass M, not in the tangent direction to the desired orbital trajectory. Therefore, during the descent a rocket thrust may be needed to steer / divert the descending spacecraft or the satellite towards the desired orbital trajectory. But even with a need for this additional thrust during descent, the overall energy used is less than the thrust / energy / power of a conventional launch strategy. The descent stage rocket thrust is basically used to steer the spacecraft or the satellite towards the desired orbital trajectory, as described below in the explanation of Fig. 19, Fig. 20A, Fig. 20B, and Fig. 20C. Speed gain is achieved by converting the gravitational potential energy to kinetic energy. In fact, due to tire descent stage rocket assist, the rocket typically gains more speed than needed when it descends from Toto T0RB1Tand needs a retro thrust to attain in a permanent circular orbit with the desired radius of T0RBIT. To optimize the rocket assisted descent the relation (1.1) is modified by,

[0026]

[0027] Where, sy is a scaling factor with the bounds as given in (1.2). Ideally, a satellite or spacecraft is launched to the calculated radius having no time pressure, or speed profile that needs to be maintained. In one aspect the satellite or spacecraft reaches the radius rhas given in(1.2) and becomes stationary, or in other words, at that instant it has zero velocity, resulting in zero kinetic energy. That is, its total energy is equal to its gravitational potential energy. Then, the spacecraft or satellite goes into rocket assisted descent maneuver and starts its rocket assistedfall to its desired orbital radius T0RBIT. If Sy is greater than optimum, the satellite needs to be slowed down by retro thrust, and if it is smaller than optimum, it needs to increase its speed with some additional thrust in the direction of travel. Finding an optimal Sy, which minimizes the total propellant mass to launch a given payload to T0RBIT, requires an iterative process usingNewton’s method [27, 29-30] for solving many non-linear trajectory equations.

[0028] Accordingly, a method is provided for efficient orbital launch trajectories. The method launches a payload as high as a first radius (or first radial distance) To, with respect to the center of the Earth, which is associated with a first altitude h0, defined with respect to sea level.Ideally, the payload can be launched at a zero angle with respect to the Earth’s normal surface(vertical). Due to safety related restrictions at the launch site, the launch is performed with a minimum angle with respect to the Earth’s normal surface and towards any allowed launch azimuth to reach the first radius altitude. As used herein, the term “payload” defines the object that eventually attains orbit, which may be spacecraft, a satellite, a rocket, or combinations thereof. The method then decreases the payload altitude in response to a gravitational pull of theEarth, and ultimately the payload attains a stable orbit around the Earth at a second radius (r0RB / r)i with respect to the center of the Earth, which is associated with a second altitude (AORBZT)> defined with respect to sea level. If the first radius is equal to 2 times the second radius (r0= 2 • TORBIJ), then the payload in the stable orbit has a total energy that is the sum of its gravitational second potential energy and kinetic energy, with the payload total energy also being equal to the gravitational first potential energy. The second radius can also be expressed as (h0RB1T+ TEARTH) giving:

[0030] where Ao is the altitude associated with the first radius, as defined with respect to the sea level of Earth;

[0031] where TEARTHt^eEarth’s radius; and,

[0032] where is altitude associated with the second radius, defined with respect tosea level, see Fig. 3D. The relation

[0033] One advantage of the disclosed method is that the stable orbit can potentially be at any orbital inclination angle in the range between 0 and 360 degrees. Further, this advantage can be accrued by launching the payload from any latitude on a surface of the Earth in the range between 90 and -90 degrees.

[0034] More explicitly, the gravitational first potential energy can be expressed as:

[0035]

[0036] where is Newton’s gravitational constant;

[0037] where is the mass of the payload; and,

[0038] where M is the mass of the Earth.

[0039] The second potential energy can be expressed as:

[0040]

[0041] The total energy at the second radius can be expressed as follows:100421

[0043] where VQRBIT payload velocity in the stable circular orbit.

[0044] Once the payload attains the first radius, it needs to be redirected during its fall towards the stationary mass M. So, subsequent to the payload attaining the first radius, the method initiates rocket or payload assisted descent maneuvers, which decreases the payload altitude. In a conventional gravity turn the rocket applies force, in the form of thrust, against the force of gravity (see Fig. 14). In contrast, in a rocket assisted descent maneuver, as defined herein, the payload applies no force to counter the force of gravity, but rather, relies upon the force ofgravity to send the payload into a stable lower altitude orbit. The force is only used to steer the payload towards to the desired orbit at the second altitude. In one aspect the payload is launched to a third radius (F^), which less than the first radius, but greater than the second radius, where:

[0045] 7 ^ — Sf1TQRBIT where 1.1 < Sy < 2;

[0046] Then, subsequent to the rocket assisted descent maneuver a velocity adjustment force may need to be supplied to the payload. The payload has a total energy equal to its gravitational potential energy at the third radius, and the decrease of the payload altitude is responsive to the applied velocity adjustment as well as the gravitation potential energy at the third radius.

[0047] Fig. 3G depicts the velocity that the payload gains, which was at rest at a radius of F^ in terms of the orbital velocity VQRBIT f°r acircular orbit at the radius of T0RBITas a function of Sy when falling from F^ to T0RB1Twhich can be represented with the relation,

[0049] As can be seen when Sy = 2, which corresponds to a radius of (FQ — 2 • TQRB / T)# the vertical velocity of the payload reaches exactly VORBITCJORBIT)asproven with 2 different methods in later sections. If Sy = 3, the VRATIO °f ^e payload hardly reaches 1.16* VORBIT when falling from 3 • r0RBITto T0RB1T. At Sy = 1, since rh=r0RBiT there will no velocity gain and VRATIO becomes 0. When Sy = 1.5, the VRATIO becomes 0.8b VQRBIT- When Sy = 1.2, the VRATIO becomes 0.571VQRBIT- Since it takes more propellant to go higher, finding the right Sy value and deciding the optimal decent trajectory is one of the challenging mathematical parts of the invention. The explanation above assumes a “vertical descent”, from the third radius or altitude along the radial direction towards to the center of the earth and assumes that the mass of the falling object remains the same. Since real maneuvers are three dimensional and since rocket propulsion by its definition is based on ejecting its mass, the real calculation can’t be completely done analytically, it requires solutionof equation of motion. In section 5.2 a more detailed analytical analysis is given for a vertical descent by modifying the rocket equation for constant gravitational acceleration.

[0050] Addition details of the method are provided below.

[0051] BRIEF DESCRIPTION OF THE DRAWINGS

[0052] Fig. 1 A depicts the orbital speed in [m / s] (meter per second) versus radius measured with respect to the stationary coordinate system where its origin is in the center of Earth and altitude measured from sea level.

[0053] Fig. IB depicts the orbital speed in Mach number, versus radius measured with respect to the stationary coordinate system where its origin is in the center of Earth and altitude measured from sea level, where 1 Mach is speed of sound at sea level, 1 Mach = 340 [m / s] and the numbers 1, 5, and 10 correspond to the row numbers in Table 2.

[0054] Fig. 1C depicts the orbital period in hours versus radius measured with respect to the stationary coordinate system where its origin is in the center of the Earth and altitude measured from sea level.

[0055] Fig. ID depicts the escape velocity as a function of radius for the Earth along with markings of the orbital altitudes given in Table 2.

[0056] Fig. IE depicts the density of air as a function of altitude.

[0057] Fig. IF depicts orbital inclination as a function of launch azimuth from 6 locations, representing launch from the Equator, along with the numberings corresponding with the launch latitudes shown in the first column of Table 2.

[0058] Fig. 2A depicts the gravitational acceleration g(Jl) as function of altitude tl from sea level and distance from the earth’s centered radius g(r), where r = h 4- TEARTH.

[0059] Fig. 2B depicts the gravitational acceleration normalized to its sea level value as a function of altitude g (71) and the distance from the Earth’s centered radius r, g(r), with the radius of the Earth taken at the equatorial radius of TEARTH= 6,378 km.

[0060] Fig. 2C depicts the gravitational acceleration normalized to its sea level value as a function of altitude y( / i) with the maximum bum altitude hB(jif = 0) for T2W = 1.1, 2, 4 and, 8 marked on the curve.

[0061] Fig. 3A depicts the gravitational potential energy as a function of radius T0RB1Twith respect to Earth’s center, along with the kinetic energy when in orbit at the radius T0RBJT, and its total energy.

[0062] Fig. 3B depicts the same plots in Fig. 3 A, where x axes plotted in logarithmic scale.

[0063] Fig. 3C depicts the orbital velocity as a function of orbital altitude with the velocity gained when dropping from 2 . TQRBITTO TORBIT along with markings of the orbital altitudes given in Table 2 which matches the orbital velocity as a function of orbital altitude curve.

[0064] Fig. 3D depicts the launch altitude as a function of the orbital altitude along with maridngs of the orbital altitudes given in Table 2.

[0065] Fig. 3E depicts the launch altitude to the orbital altitude ratio along with markings of the orbital altitudes given in Table 2 where the X axis is plotted in logarithmic scale.

[0066] Fig. 3F depicts the fall times from (2 . r ORBIT to r0RB / T) , (2 . T0RB1Tto TEARTH)'“d (joRBITt0 TEARTH)-

[0067] Fig. 3G depicts the velocity that the payload gains, which was at rest at a radius of in terms of the orbital velocity VQRBIT f°r acircular orbit at the radius of T0RBITas a function of Sr when falling from'

[0068] Fig. 3H depicts the rocket thrust assisted velocity reached VB, having rocket thrust being in the same direction of the constant gravitational acceleration shown as VB9(in rocket powered vertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as VB9(in rocket powered vertical ascent) using the solution of modified Tsiolkovsky’s rocket equation for T2W = 1.1 and 4 versus (if.

[0069] Fig. 31 depicts the rocket thrust assisted distance covered having rocket thrust being in the same direction of the constant gravitational acceleration shown as hR9(in rocket poweredvertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as (in rocket powered vertical ascent) using the solution of modifiedTsiolkovsky’s rocket equation for T21V = 1.1 and 4 versus gy.

[0070] Fig. 3J is same data depicted in Fig. 31 in logarithmic X axes to show the covered distances clearer for larger gy values.

[0071] Fig. 3K depicts the ratios between the velocities reached 7 / yB(gy) = Vg9f VB8and ratios of the distances covered1.1.

[0072] Fig. 3L depicts the ratios between the velocities reachedratios of the distances covered4.

[0073] Fig. 4A depicts Vg(jlp T2W) for thrust-to-weight ratio T2W = 1.1 and 8 with a constant acceleration g

[0074] Fig. 4B depicts Vg(jlp T2W) for thrust-to-weight ratio T2W = 1.1 and 8 at a constant acceleration g, where the horizontal axis is in logarithmic scale.

[0075] Fig. 5A depicts h.B(jJ.p T2W) for thrust-to-weight ratio T2W = 1.1, 2, 4, and 8 with a constant acceleration g.

[0076] Fig. 5B depicts hg (gy, T2W) for thrust-to-weight ratio T2W = 1.1 and 8 with a constant acceleration g, where the horizontal axis is in logarithmic scale.

[0077] Fig. 6A depicts the climb altitude h(jlp 7'21V) for the constant gravitational acceleration of g = 9.8 [m / s2] with thrust-to-weight ratio T2W = 1.1 and 8 where the vertical axis is in logarithmic scale.

[0078] Fig. 6B depicts bum height hB^Jlp T2W} and coast height he (.Pp T2W") for the constant gravitational acceleration of g = 9.8 [m / s2] with thrust-to-weight ratio T2W = 1.1 and 8 where the vertical axis is in logarithmic scale.

[0079] Fig. 7 A depicts asymptote X intercept gx$yM(T2W) VS. T2W and gM / w(T2W) by solving the non-linear equation (6.7).

[0080] Fig. 7B depicts climb altitude h.1Ny2^Jlp T2W^ for the inverse square gravitational field with asymptotes where vertical axis is in logarithmic scale for thrust-to-weight ratio T2W = 1.1 and 8.

[0081] Fig. 7C depicts bum height hBINV2T2W) and coast height hCINV2(gy , T2W) at T2W for the inverse square gravitational field with asymptotes where the vertical axis is in logarithmic scale.

[0082] Fig. 7D depicts climb altitude h.iNV2(KiM-f> T2W) for the inverse square gravitational field with asymptotes and h(jip T2W} for the constant gravitational acceleration of g =9.8 [m / s2] where the vertical axis is in logarithmic scale for thrust-to-weight ratio T2W = 1.1 and 8.

[0083] Fig. 7E depicts climb altitude h-iNV2^l1f, 7*2 W ) for the inverse square gravitational field with asymptotes and T2W^ for the constant gravitational acceleration of g =9.8 [m / s2] where the vertical and horizontal axes are in logarithmic scale for thrust-to-weight ratio T2W = 1.1 and 8.

[0084] Fig. 7F depicts climb altitude / l / jvy2( / * / ■» 7*21^) for the inverse square gravitational field with asymptotes for g < g145yjvf(T2W) with thrust-to-weight ratio T2W = 1.1, 2, 4, and 8.

[0085] Fig. 8 depicts climb altitude h.INV2(jlp T2W) for the inverse square gravitational field and climb altitude h T2IV) with the constant gravitational acceleration of g =9.8 [m / s2] vs. T2W, where = 0.1, 0.2, 0.3, and. 0.4.

[0086] Fig. 9 A depicts climb altitudeinverse square gravitational field T2W} functions for T2W = 1.1 and 4 constructed over 400 uniformly spaced discrete sampling points.

[0087] Fig. 9B depicts the functions plotted in Fig. 9A constructed with 40 discrete sampling points marking the curves to show a numerical algorithm for implementing the geometrical solution method in a computer program based on interpolation between the discrete sampling points.

[0088] Fig. 9C depicts a detailed view of the IR 2 region showing the discrete sampling points in the neighborhood of intersection with hGIVEN= 40 km of the

[0089] h1NV2(jip T2W = 1.1) and h(jipT2W = 1.1) functions.

[0090] Fig. 10A depicts a detailed view of the IR 1 and IR 2 regions showing discrete sampling points in the neighborhood of intersection between hGiVEN= 40 km and 400km for the h1NV2(lipT2W = 1.1) and h(gp 7’2lV = 1.1) functions.

[0091] Fig. 10B depicts a detailed view showing discrete sampling points in the neighborhood of the intersection region IR 1 between hGIVEN= 20,200km with thefunctions.

[0093] Fig. 10C shows the convergence properties of Newton’s method employed in solving rocket equations.

[0094] Fig. 11 depicts the skin of a propellant tank.

[0095] Fig. 12 depicts the mass of the propellant tank 771 = / (r, tgfciTl) f°rthe skin thickness of tsklTl=6, and 10 cm carrying 20 tons of propellant mass.

[0096] Fig. 13 A depicts k = / (r, tsklTl) f°r 811aluminum skin thickness of tskin=3, 6, and 10 cm when carrying 20 tons of propellant mass.

[0097] Fig. 13B depicts the plot of relation (8.12), which gives the minimum value of k = kMAXthat needs to be satisfied to give a desired value of flp

[0098] Fig. 13C depicts the height hlNV2(KMAXI T2W^ function, an important relationship between the height that the rocket can reach and the kMAXvalue that the rocket design must satisfy, graphically permitting the determination of kMAXf°rh-GiVEN=400km , 20,200km, and 400,000 km.

[0099] Fig. 13D depicts the SpaceX Starship Super Heavy (loaded with 3,400 tons of propellant) and the ULA Atlas first stage (loaded with 284 tons of propellant) booster heights with a tSKINthickness of 3 [cm], with respective diameters of 9 [m] and 3.81 [m].

[0100] Fig. 13E depicts hINV2(T, 72 W), where F is the rocket diameter of the SpaceX Starship Super Heavy loaded with 3,400 Tons of propellant having tSKIN= 3 [cm] .

[0101] Fig. 13F depicts altitude as a function of time for rocket launching 1,000 [kg] of payload to 40, 100 and 400 [km] altitudes for T2W=1.1 and 4, k = 0.01, NOT an orbital launch.

[0102] Fig. 13G depicts propellant and total mass as a function of time for launching1,000 [kg] of payload to 40 [km] altitude for T2W=1.1 and 4, k = 0.01.

[0103] Fig. 13H depicts altitude as a function of time for rocket launching of 1,000 [kg] of pay load to 20,200 and 35,786 [km] (GPS and GEO) altitudes for T2W=1.1 and 4, k = 0.01,NOT an orbital launch

[0104] Fig. 131 depicts altitude as a function of time for rocket launching of 1,000 [kg] of payload to a circular orbital altitude of 400 [km] (ISS) with Sf = 1.5 and 2 for T2W=1.1 and 4, k = 0.01.

[0105] Fig. 13 J depicts propellant and total mass as a function of time in logarithmic horizontal axes for launching 1,000 [kg] of payload to a circular orbit at 400 [km] (ISS) altitude with Sf= 2 for T2W=1.1 and 4, k - 0.01.

[0106] Fig. 13K depicts propellant and total mass as a function of time in logarithmic horizontal axes for launching 1,000 [kg] of payload to a circular orbit at 400 [km] (ISS) altitude with Sf = 1.5 and 2 for T2W=1.1 and 4, k = 0.01.

[0107] Fig. 13L depicts altitude as a function of time for rocket launching of 1,000 [kg] of payload to a circular orbital altitude of 20,200 [km] (GPS) with Sf = 2 for T2W=1.1 and 4, k = 0.01.

[0108] Fig. 13M depicts altitude as a function of time for rocket launching of 1 ,000 [kg] of payload to a circular orbital altitude of 20,200 [km] (GPS) with Sf= 1.5 and. 2 for T2W==1.1 and 4, k = 0.01.

[0109] Fig. 13N depicts propellant and total mass as a function of time in logarithmic horizontal axes for launching 1,000 [kg] of payload to a circular orbital altitude of 20,200 [km](GPS) with Sf= 2 for T2W-1.1 and 4, k = 0.01.

[0110] Fig. 130 depicts propellant and total mass as a function of time in logarithmic horizontal axes for launching 1,000 [kg] of payload to a circular orbital altitude of 20,200 [km] (GPS) with Sf= 1.5 and 2 for T2W=1.1 and 4, k = 0.01.

[0111] Fig. 14 depicts a rocket being launched into orbit using a conventional pitchover or gravity turn maneuver (prior art).

[0112] Fig. 15 is a flowchart illustrating a method for efficient orbital launch trajectories.

[0113] Fig. 16 is a drawing depicting the relationship between the first radius (ro), second radius ^TQRBIT)' the payload, and the Earth.

[0114] Fig. 17 is a diagram depicting the relationships shown in Fig. 16, as referenced toEarth’s sea level.

[0115] Fig. 18 is a plan view of the Earth centered on the polar axis.

[0116] Fig. 19 is a diagram the relationship between (r0= 2 • rORBIT), second radius ( rORBIT) the payload, the Earth, and (rh) 1900.

[0117] Figs. 20A through 20C are drawings depicting a velocity adjustment applied to the payload during or after the rocket assisted descent maneuver.

[0118] Fig. 21 is a flowchart illustrating a method for minimizing the energy required for an orbital launch.

[0119] Fig. 22 is a diagram depicting a variety of exemplary stable orbits that can be obtained using the disclosed method.

[0120] DETAILED DESCRIPTION

[0121] The Problem to be Solved.

[0122] This invention is related to putting satellites into orbit with smaller rockets employing less propellant compared to the standard rockets with less demanding rocket engines for launching the same orbital mass into orbit. Therefore, an introduction to some very basic orbital mechanics is presented in Section 2 for a satellite having a circular orbit around the earth.

[0123] From the German rocket work, which goes back to 1930’s, the importance of having a low mass rocket to reach a given altitude is a well-known fact and it became a key factor in any rocket design [7], To reach the speeds and altitudes that are desired in rocket applications, the rocket equation enforces that most of the initial rocket mass must be propellant mass Tnprop[1-3], Propellant is contained in the propellant tank and let is the mass ofthe propellant tank and it also includes the mass of the overall volume of the rocket housing the propellant. Let the parameter k is the ratio of the propellant to the propellant tank given as,

[0124]

[0125] Using the parameter k, mass of the rocket engine and the payload massmPAY> the initial mass of the rocket m0, can be given as,

[0126]

[0127] For the rocket to be able to lift-off vertically from the pad its initial thrust-to- weight ratio T2W must be greater than 1 , giving the following thrust equation,

[0128]

[0129] is the thrust generated by the rocket engine and g is the gravitationalacceleration at the launch pad. Thrust, obtained from the solution of the equation (1.5) mustbe a positive value. It is shown that once TnpTOp satisfies the rocket equation, which enables itto reach the desired height, positive thrust value is only possible if the k parameter is less than a specific value defined as kMAX. In its simplest form kMAXis given as,

[0131] Where gy is the ratio between the final mass value of the rocket and its initial rocket mass 7HQ given as,

[0133] Which gy satisfies the non-linear rocket equation formulated including gravitational potential acting upon the rocket to reach a given height above sea level, hGiVEN. Therefore, it is a “solved” quantity and determines the maximum value of k, noted as kMAXgiven in (1.7) in its simplest form. This mathematically relates the k value giving the maximum altitude that the rocket can reach and must be satisfied. If the physical construction of the rocket gives a k larger than k^AX, for the given value of gp the rocket cannot reach the altitude corresponding to

[0134] It is also shown that the opposite is also true. A given k value which can be measured or calculated from some simple rocket geometry, propellant physical and chemical properties define the maximum altitude that the rocket can reach for a given payload, thrust-to- weight ratio at the launch, and rocket engine thrust-to-weight ratio. In its simplest form, the minimum value of gy that can be achieved with this given k value becomes,

[0135]

[0136] Most of the rocket propellant tank shapes determine the shape of the rocket and typically they are cylindrical. For cylindrical geometry, the k value is also a function of rocketradius, therefore rocket diameter can be related to the maximum altitude that it can reach for a given thrust-to-weight ratio at the launch, and rocket engine thrust-to-weight ratio. The detailed derivations on this subject are given in Sections 7 and 8 below.

[0137] Another advantage of the disclosed method is that it allows any launch site on earth to achieve any orbital inclination angle from any launch site [8], The inclination angle of a satellite orbit is one of the five orbital parameters defining its orbit. It is the angle between the satellite orbital plane and the Earth’s equatorial plane. As an example, the International SpaceStation (ISS) has an inclination of 51.6°. Due to certain safety limitations all the launch sites give only a limited range of azimuth angle (angle from the true north) launches. Inclination angle i, is related to the latitude of the launch site 6 and the launch azimuth A with a simple formula given as,

[0139] This relation puts limits on the inclination angle for the satellite orbit launched from any site. As an example, for all launches from Cape Canaveral in central Florida (6 =28.4°North), the launch azimuth angle limits are 35° - 120° giving the corresponding inclination angle limits of 57° - 39°. Similarly, for all launches from Vandenberg Air Force Base in California (9=34.7° North), the launch azimuth angle limits are 140° - 201° giving the corresponding inclination angle limits of 56° - 104°. Therefore, the most popular Polar orbital launches cannot be done from Cape Canaveral, but rather, must be done from Vandenberg Air Force. It is obvious that having no inclination angle restriction is a great advantage and has great commercial value by itself.

[0140] Table 1 gives the most popular launch site latitudes and dates on which they became operational. Fig. IE gives the plot of relation (1.9) for all most popular launch sites with the Azimuthal Limitations shown on the plot for Cape Kennedy and Vandenberg.

[0141] TABLE 1 Most popular launch site latitudes and dates when they became operational.

[0142] In the disclosed method the satellite is launched to a radius of rhwhich could be in any permitted azimuth from the launch site. From there a rocket assisted descent can be made to any direction to get to get to any inclination desired. It is obvious to see that by employing this launch strategy, satellites can be launched into most popular Polar orbits from even CapeCanaveral, French Guiana, or anywhere on Earth.

[0143] Anything related to space has always been a very popular subject in the public and the orbital launch of satellites is becoming a big and competitive industry where profitability is becoming an important part of it. Many published technical materials related to the subject exist in public domain, but unfortunately publications in this area, like in many other areas, take shortcuts in mathematics and derivations. This hides many of the facts and causes one to miss the key important details. As an example, without defining niprop in (1.4) the key results of (1.5) to (1.7) cannot be generated.

[0144] Another goal of this work is to present some of the orbital launch and rocket design related mathematics and mechanics by taking no short cuts. Therefore, in this work there are some chapters which are dedicated to the derivation of the basic principles.

[0145] Simple Orbital Mechanics

[0146] English mathematician and the founder of mechanics among many other earthshaking discoveries, Sir Isaac Newton’s (1642-1726) second law of mechanics allows anyone to calculate the tangential speed required to put a satellite into a circular or elliptical orbit around the earth, at an altitude h from sea level. All the information needed to perform thesecalculations is in “Philosophic Naturalis Principia Mathematica”, which is the foundation of mechanics, integral and differential calculus. It was first published in 1687 with the financial help of Edmond Halley (1656-1742), another famous English astronomer and mathematician named with a comet named after him.

[0147] One cannot mention “differential calculus” without mentioning German mathematician Gottfried Wilhelm von Leibnitz (1646-1716). Leibnitz invented differential calculus independently from Newton with a more convenient notation than used by Newton.Everything mathematically presented in this work is entirely based on the inventions of these two giants of science 334 years after.

[0148] Newton’s gravitational attraction force Fcbetween 2 spherical masses m and M , having a distant d between their centers is given as,

[0150] Where G is Newton’s gravitational constant. This law is also known as the inverse square law of gravitational attraction force or field. This attraction force on the surface of the earth can be defined with the expression,

[0151] FG= rng0(2-2)

[0152] Where go is the gravitational acceleration constant at the surface of the earth and it can be formulated as,

[0154] Where in (2.3) now M and TEARTHrepresent the mass and radius of the earth.For a satellite to stay in a circular orbit, the centrifugal force of the satellite at a radius of T0RB1Tmeasured from the center of the earth, must be equal to the Earth’s gravitational attraction forceas given by (2.3) as,

[0156] Where m and VQRBITare the mass of the satellite and the tangential velocity of the satellite in the circular orbit around the earth at a radius of T"ORBIT measured from the center of the earth, respectively.

[0157] Simplifying (2.4) gives the necessary tangential velocity to keep the satellite in orbit at a radius of T0RBIT, measured from the center of the earth as,

[0158] 0RBIT

[0159] Getting the square root of both sides gives,

[0160]

[0161] In space terminology one usually specifies the orbital altitude h referenced from the sea level giving the relation between T0RBITand h as,

[0162]

[0163] Fig. 1 A is the orbital velocity as a function of radius and altitude as given in (2.6) in [m / s] along with the escape velocity from the Earth. Fig. IB is the orbital velocity as a function of radius and altitude as given in (2.6) in Mach numbers, which is a speed defined in terms of speed of sound at sea level which is 340 [m / s]. The markings on the curves are the orbital velocities of satellites in circular orbit at orbital altitudes given in Table 2. As can be seen, the speed of any satellite in any of the orbits given in Table 2 is hypersonic.

[0164] The numerical values of the constants in (2.3) are,

[0165] Universal Gravitational Constant G

[0166]

[0167] The Earth’s Polar and Equatorial radiuses are slightly different and are,

[0168] Since the radius of the orbit and the constant speed of the satellite is known, one can easily calculate the time T that it takes for the satellite to complete one revolution at the circular orbit as,

[0170] This is nothing more than Johannes Kepler’s (1571-1630) third law applied for a circular orbit. Kepler, a German mathematician, used Tyco Brahe (1546-1601), a Danish astronomer known for the most accurate astronomical data to date collected from his observatory at Benatky nad Jizerou, near Prague of today’s Czech Republic, for validating his three laws of planetary motion while he was Tyco’s assistant.

[0171] The orbital period as a function of radius as given in (2.8) and altitude is plotted in Fig. 1C. The markings on the curves are the orbital periods of the satellites in circular orbit at orbital altitudes given in Table 2.

[0172] Fig. ID depicts the escape velocity as a function of radius for the Earth along with markings of the orbital altitudes given in Table 2.

[0173] Fig. IE depicts the density of air as a function of altitude.

[0174] Fig. IF depicts orbital inclination as a function of launch azimuth from 6 locations, representing launch from the Equator, along with the numberings corresponding with the launch latitudes shown in the first column of Table 1.

[0175] Regarding expressions (2.6) and (2.8), Aristarchus of Samos (310 BC-230 BC), an ancient Greek mathematician and astronomer, was the first to propose that the sun is the center of the Solar system, and the Earth is rotating around sun. Before him, Aristotle (384 BC-322 BC) another ancient Greek philosopher postulated that earth was round and was at the center of the universe, and everything rotated around it. Claudius Ptolemy (100-170), another ancientGreek mathematician supported the “geocentric earth” model. Copernicus (1473-1543), thePolish mathematician and astronomer, brought back the idea of the earth revolving around the sun.

[0176] All the debates related to the subject were settled after Halley asked Newton what the trajectory of the Halley Comet was, which changed the flow of history. To this question,Newton simply said, “(i)t is an Eclipse.” and when Halley asked how he knew that Newton simply replied , “I calculated it”, which was an unheard-of task then. Halley asked how it was calculated and after Newton showed how he calculated it, a completely surprised and impressedHalley told him that he must publish all this, which Newton said it is too expensive to publish and he couldn’t afford it. Totally impressed, Halley decided to finance Newton’s monumental work in “Philosophic Naturalis Principia Mathematica” which undoubtedly became the most important scientific publication in human history, which is the mathematics for explaining the entire mechanics of objects.

[0177] Another useful parameter related to the subject is escape velocity, which gives a good idea of the speeds required in the space launch business. Escape velocity V ESCAPE ™ simplest form is derived between two spherical objects with masses m and M and radiuses for each are noted as Tmand rMrespectively. In the formulation it is also assumed that M » m and » Tmand the larger mass, M will also assumed to be stationary and both masses alsonon-rotating. If the initial center-to-center distances between these spherical masses is d, the escape velocity VESCAPEis the initial velocity that needs to be provided to the smaller mass m in the opposite direction to the gravitational attraction force between these two masses, which will allow it to escape the gravitational attraction force of M and reaching center-to-center distance d -> co. For this condition escape velocity VESCAPEis given as,

[0179] As an example, for an object like a spaceship launched straight up from the surface of the earth, the escape velocity VESCAPE relation given in (2.9) will become,

[0180]

[0181] Where rEARTHand m-EARTH are radius and the mass of the earth. The escape velocity VESCAPEas given in (2.9) can be calculated by applying conservation of energy principle, and by equating the kinetic energy to the potential energy for the mass m for the two- mass system with properties explained as m and M above. Since the method disclosed herein is derived using the energy conservation law, it is useful to derive the escape velocity using this very important concept in detail.

[0182] Potential energy PEof the object with the mass m at a radius of rA, measured from the center of the reference mass M, can be calculated with the work integral W applied to the inverse square law gravitational attraction force field from infinite to rAas [21,23],

[0184] In other words, relation (2.11) is the work done for bringing the mass 771 with zero initial speed from infinite to a distance away from the stationary mass M. During thisfree-fall from infinite to the mass 771 will gain speed and will have a value ofAs aresult, its kinetic energy K , becomes,

[0185]

[0186] The energy conservation law states that the total energies at initial and final states must be equal, which can be expressed with an equation,

[0187]

[0188] Energy conservation equation (2.13) explicitly becomes,

[0190] Solving (2.14) gives,

[0191]

[0192] Due to reciprocity, this is also the magnitude of the escape velocity , inthe opposite direction. Escape velocity from earth then can be formulated by substituting instead ofgiving the same expression as given in (2.10) as,

[0193]

[0194] Fig. 1 A shows the escape velocity from earth along with the orbital velocity as a function of radius and altitude. As can be seen, escape velocity is greater than any of the orbital velocities. Since escape velocity has the radius of the Earth in its formulation it should also be a function of altitude as well.

[0195] Fig. ID depicts the escape velocity as a function of radius for Earth along with markings of the orbital altitudes given in Table 2.

[0196] At this point it is very interesting to note that if VESCAPEIasgiven by relation(2.10), is equated to speed of light C, it becomes the non-relativistic definition of a “Black Hole” which was first conceived by an English scientist John Mitchell (1724-1793) in 1783. The mass / radius calculations of his “Dark Star”, derived from (2.15) is not very different than the relativistic calculations based on the solution of Einstein’s field equations describing his general relativity work of 1915. Einstein’s field equation describing general relativity, which was first solved by Karl Schwarzschild in 1915 gives the “Schwarzschild Radius” of the event horizon as,

[0197]

[0198] Where, C is the speed of light. Although John Mitchell’s result is 122 years prespecial relativity, with no restrictions on the maximum velocity of anything, gives identical numerical value to (2.16) which is another remarkable coincidence in science.

[0199] Fig. 1 A depicts the orbital speed in [m / s] (meter per second) versus radius measured with respect to the stationary coordinate system where its origin is in the center ofEarth and altitude measured from sea level.

[0200] Fig. IB depicts the orbital speed in Mach number, versus radius measured with respect to the stationary coordinate system where its origin is in the center of Earth and altitude measured from sea level, where 1 Mach is speed of sound at sea level (1 Mach = 340 [m / s]) and the numbers 1, 5, and 10 correspond to the row numbers in Table 2. As is conventional, speed isrepresented herein as a Mach number. One of the early questions to be explored in rocketry was to determine if rocket propulsion could generate velocities sufficient to put an object into orbit, or sufficient to create an escape velocity V ESCAPE from the Earth. To illustrate, escape velocity^ESCAPE isshown superimposed in Figs. 1 A.

[0201] Fig. 1C depicts the orbital period in hours versus radius measured with respect to the stationary coordinate system where its origin is in the center of the Earth and altitude measured from sea level. This is a way of presenting the extreme hypersonic speeds needed for putting a satellite into orbit in terms of orbital periods. For low Earth orbits (LEOs) the orbital periods are on the order of 90 minutes, meaning that satellites in LEOs rotate around the Earth every hour or two. The orbital periods of some well-known satellites are given in Table 2, with related data.

[0202] A list of some orbital radiuses of well-known satellites and Van Allen RadiationBelts are presented in Table 2. These orbital radiuses between the Earth and the Moon, along with orbital periods, velocities, their orbital masses, and free fall times to the Earth surface, are calculated by the analytical formula given in [4-5] from the orbits that are presented in table.These orbitals are marked on the curves in Fig. IB, Fig. 1C, and Fig. 2B.

[0203] Fig. 2 A depicts the gravitational acceleration as function of altitude h. from sea level and distance from the earth’s centered radius S'(r), where r = h + rEARTH.

[0204] Fig. 2B depicts the gravitational acceleration normalized to its sea level value as a function of altitude g(h) and the distance from the Earth’s centered radius r, y(r), with the radius of the Earth taken at the equatorial radius of rEARTH= 6,378 km.

[0205] Fig. 2C depicts the gravitational acceleration normalized to its sea level value as a function of altitude g(h) with the maximum bum altitude hB1.1, 2, 4 and 8 marked on the curve.

[0206] TABLE 2. Some Satellite altitudes and critical distances between the Earth and the Moon along with orbital periods, velocities, typical orbital masses, calculated fall times toEarth, and the ratio of the gravitational acceleration ratio to the go at the Earth’s surface.Included in the table with corresponding numbering in column 1 are:

[0207] GOCE 1,000 [kg], ESA Gravity Field and Steady-State Ocean CirculationExplorer,

[0208] ISS International Space Station

[0209] Hubble 11,110 [kg],

[0210] Iridium 689 [kg]

[0211] GPS Block 1 IF 1,630 [kg]

[0212] Latest Geo-Synchronous satellite 4,276 [kg]

[0213] Inner Van Allen (Min): Lower altitude of the Van Allen inner radiation belt, due to interaction between Earth’s magnetic field and incoming high speed charged particles, mainly from the sun.

[0214] Inner Van Allen (Max): Higher altitude of the Van Allen inner radiation belt

[0215] Outer Van Allen (Min): Lower altitude of the Van Allen outer radiation belt

[0216] Outer Van Allen (Max): Higher altitude of the Van Allen outer radiation belt

[0217] Applying Energy Conservation Principle to the New Satellite LaunchMethod

[0218] All the space launch systems today are very large systems and have rocket motors with very large thrust values giving them large thrust to weight ratios, always much greater than1 to achieve the orbital velocities only achievable with very large bum rates as can be seen inTables 4A and 4B. The potential energy of a satellite with a mass m at a circular orbit with a radius of rORBIT, measured from the center of the earth can be calculated with the integral as done in (2.10) is [21-23],

[0220] The kinetic energy of the satellite at a circular orbit with a radius measured from the center of the earth r0RBITis,

[0221]

[0222] Since satellite is orbiting the earth at a circular orbit with a radius measured from the center of the earth rORBITit must satisfy the VQRBIT relation derived earlier in (2.5).Substituting (2.5) in (3.2) gives,

[0223]

[0224] The total energy of the satellite T E ( rORBIT) is the sum of its kinetic (3.3) and potential (3.1) energies giving,

[0226] Now, the question is to find what radius Fo, measured from the center of theEarth, will give the same total energy (3.4) to the satellite with the same mass m while it is stationary. Since we assume the satellite is stationary at radius FQ, its kinetic energy is zero, meaning that its total energy is equal to its potential energy,

[0228] Equating (3.4) to (3.5) Focan be calculated solving it from,

[0230] Solving FQ from (3.6) gives surprisingly simple result of,

[0231]

[0232] Equation (3.7) means that the satellite in a circular orbit at a radius of T0RBrr, measured from the center of the earth, will have the same total energy while it is stationary at a radius of 2 ■ 1"ORBIT-

[0233] Fig. 3 A shows “apparent” three curves, but there are four curves, which two of them are on top of each other. The four curves plotted in Fig. 3 A are the gravitational potential energies for a mass m equal to 1 [kg] at radiuses at T"QRBIT and 2TQRBIT marked asPE(TORBIT)811(1^(2r0RB / r), kinetic energy KE(r0RBIT) and total energy TE(r0RBIT) of the mass m at a circular orbit at a radius of T0RBITfrom the center of the earth. For convenience the mass m is taken as 1 [kg] in energy calculations. The marked points on the total energy curve TE(TQRB]T')316dietota* energies of the satellite with a mass of m at a circular orbit at orbital altitudes given in Table 2 following the same numbering on the plots just to show that T E (TORBIT) and PE (Zr0RBIT) exactly match, which is the numerical verification of the proof given in Sections 3.0 and 3.1 in detail. To have a clearer view of thetight spacings between the markings for the Table 2 orbital altitudes, the Fig. 3 A data is presented by having its X axes presented in the logarithmic scale in Fig. 3B.

[0234] Fig. 3C depicts the orbital velocity as a function of the orbital altitudeVORB / T(^ORBZT)88shown in Fig. 1A with the velocities gained by dropping from r0= ZTORBIT to r0RBrrfor the orbital altitudes given in Table 2. As can be seen the markers are exactly on the VQRBITCHORBIT') curve.

[0235] The only problem is that the velocity vector of the falling object with a mass m is towards the center of the stationary spherical body with a mass M, not in the tangent direction to the desired orbital trajectory. Therefore, during the descent a rocket thrust may be needed to steer / divert the descending spacecraft or satellite into the desired orbital trajectory.

[0236] Substituting (2.7), which gives the relation between the orbital altitude h of the satellite and T0RB1Tinto (3.7) gives,

[0238] Typically, satellite orbits are given as their altitude from sea level instead of radius from the center of the earth. Therefore writing (3.8) as a function of the required altitude above sea level h0corresponding to r0gives,

[0239]

[0240] Close examination of (3.9) shows that for small orbital altitudes compared to the radius of earth, like Low Earth Orbit (LEO) launches, the initial launch altitude is much larger than the orbital altitude h. In other words, for LEO launches the rocket must travel far longer distances compared to the conventional method of launching a satellite into orbit (h0» ll).

[0241] Fig. 3D depicts the launch altitude as a function of the orbital altitude along with markings of the orbital altitudes given in Table 2. Fig. 3D displays the relation (3.9), but not very clearly for LEO launches, as the marked orbital launch altitudes are very tightly spaced anddifficult to distinguish from each other. Proposed first launch altitude h0to the desired orbital altitude h ratio T) displays this fact much clearly giving,

[0243] Fig. 3E depicts the launch altitude to the orbital altitude ratio along with markings of the orbital altitudes given in Table 2 where the X axis is plotted in logarithmic scale. Fig. 3E depicts 7 / ( / l) function where the X axis is drawn logarithmic to show the LEO ratios much clearly. As can be seen for LEO the ratios are above 25, even for large values of desired orbital altitudes h, the ratio Tj approaches 2.

[0244] Fig. 3F depicts the fall times from (2 ■ TQRBIT to T0RB1T^ , (2 • T0RBITtoTEARTH) and (TORBITTOTEARTHY

[0245] Fig. 3G depicts the velocity that the payload gains, which was at rest at a radius ofTh in terms of the orbital velocity VQRBIT f°r acircular orbit at the radius of TORBITas a function of Sf-when falling from to TORBIT»J

[0246] To demonstrate the derived mechanics for an orbital altitude h =400 [km] with a real numerical example will be useful. a. Launch the satellite to an altitude / LQ from sea level, same altitude of International SpaceStation (ISS) as given by relation (3.9),

[0247]

[0248] The rocket is launched from the surface of the earth such that it climbs to 7, 178[km] of altitude from sea level straight up, in radial direction opposite to the gravitational attraction force of the earth and when it reaches this altitude, it will become stationary, which means it has zero kinetic energy. At this point its potential energy is equal to the total energy, and it is equal to the total energy if it was in an orbit at 400 [km]. It can reach this altitude withany speed profile, and as long as it becomes stationary at the altitude h0, it will have the same total energy as if it was in a circular orbit around the earth at an altitude 400 [km] above sea level.

[0249] ii) In this step the satellite falls back freely towards earth. During the descent, the satellite’s speed will only have a radial component, towards to the center of the earth and when it falls to the desired orbital altitude of 400 [km], the magnitude of its speed will be equal to the speed needed to put it in the orbit at 400 [km] altitude from sea level, but its speed will be in a radial direction, not in a direction tangential to the desired orbital trajectory, and it will continue to fall towards the Earth.

[0250] Hi) Since during the free fall explained above, the descent velocity vector direction is typically never tangential to the orbital trajectory to put the satellite into orbit.Therefore, during the descent some energy must be supplied to the satellite through a proper rocket thrust in the proper direction to provide the change in the descent trajectory.

[0251] This concept, even the math seems to be correct, is counter intuitive to conventional launch strategy. As opposed to conventional logic, a satellite can be put to an orbit of 400 [km] above the sea level (for example), without climbing to the same orbital altitude with a roll and a gravity turn as has been done since 1957, with an orbital velocity tangent to the orbit, by first climbing to an altitude of 7,178 [km] above sea level, which is 17.945 times higher than400 [km], and attaining an energy equal to the total orbiting energy.

[0252] Fig. 131 demonstrates the methodology employed for two different values of sf = 1.5 and. 2 along with the related mass versus time plots of Fig. 13J and Fig. 13K.

[0253] Similar graphical demonstration is given for 20,200 [km] (GPS) orbital launch inFig. 13M with the related mass versus time plots Fig. 13N and Fig. 130.

[0254] At first it may seem counter intuitive and very unnecessary launch strategy of going up for 7,178 [km] to put a satellite to an orbit at 400 [km], as opposed to reaching 400[km] of altitude directly with a curved trajectory which becomes tangential to the orbit with thevelocity vector tangential to the orbital trajectory at the calculated orbital speed. So, it is a good practice to verify this claim by substituting T0RB1Tin the total energy equation (3.4) for the orbital radius T0RBITas,

[0257] Giving the same orbital velocity derived and given in (2.6) as,

[0258]

[0259] This is not exactly a proof, because the formula was derived using the same energy conservation law. While the energy conservation law is well understood, this exercise shows that there is no arithmetic error in the calculations. The result is counter intuitive, and additional proof can be derived using a different approach, like Newton’s equation of motion for a free-falling body in an inverse square law gravitational field to have additional confidence in the derivation. a. Analytical Solution of Newton’s Equation of Motion to Derive Velocity as Function of Radius of a Free-Falling mass in an Inverse Square Law Gravitational Field

[0260] Newton’s equation of motion for a free-falling object from a radius of r0in an inverse square law gravitational field given as,

[0261]

[0262] Where,

[0266] then the proof is certain.

[0267] Where F, G, m, M, T, V and t are the gravitational force between mases m and M, the universal gravitational constant, mass of the object in motion being modeled with the inverse square gravitational field of object with a mass of M, which is assumed to be stationary due to its very large mass with respect to m, radius, which is the center-to-center distance between the two spherical objects with masses m and M, velocity, and time respectively. Since a space launch system from the Earth is the interest, M is the mass of the Earth and m is the mass of the satellite or the spacecraft.

[0268] Although the ordinary differential equation (3.13) is simple, its analytical solution is not very straightforward. Applying the Leibnitz’s chain rule [22-26], which was shown to be an especially useful trick derived in

[0022] for the analytical solution of the non-linear Poisson’s equation in the MOS (Metal-Oxide-Semiconductor), gives,

[0269]

[0270] Substituting (3.16) in (3.18) gives,

[0271]

[0272] Substituting (3.19) in (3.15) gives,

[0276] Rearranging (3.21) gives,

[0278] Equation (3.22) becomes a quite simple differential equation that can be solved either with definite integrals or by solving the integration constant with the boundary condition.The rocket equation derivation is done employing many definite integrals, just to add a different flavor to the work, the solving the integration constant gives the following equation,

[0280] Where C is an arbitrary integration constant which needs to satisfy the initial conditions of the problem. Since at t = 0 the radius is T = r0and the initial velocity is v0, the integration constant C must satisfy the equation,

[0282] Giving,

[0283]

[0284] Substituting (3.25) into (3.23) gives the complete solution as,

[0286] Rearranging terms in (3.26) gives,

[0287]

[0288] Finally, the v(r) relation that we are looking for becomes,

[0289]

[0290] Writing (3.28) in more compact form gives,

[0291]

[0292] For a falling mass to the earth the bounds of r and r0can be given as,

[0293]

[0294] For Vo= 0 (3.29) gives,

[0295]

[0296]

[0297]

[0298] (3.31) becomes,

[0299]

[0300] Finally, the counter intuitive same answer that is being sought appears as,

[0301]

[0302] To write the result in text, it can be said that if an object in an inverse square law gravitational field falls from any radius Towith respect to a larger mass M, with a zero initial velocity to half of its initial radius Towith respect to the larger mass M, the speed that it will gain is equal to the speed obtained when it is in a circular stable orbital rotational speed around the larger mass M at an orbital radius of rORBIT. In the process of proving the relation (3.15) beyond any doubt, a very general and handy v(r) relation is also given as (3.27) which is also used later in this work. Since more confidence exists in the (3.7) it is also useful to know the fall times between the first and second radiuses. Analytical formulas are available [4, 5] and Fig. 3F depicts the fall times from (2 . rORBITto rORBIT) , (2 . rORB1Tto rEARTH) and(rORBITto rEARTH)as afunction of altitude up to 60,000 [km]. As can be seen, all of the fall times related to LEO satellites are very tightly spaced and difficult to distinguish from each other, which are shown on the left-hand side of Fig. 3F. Table 3A lists the calculated fall times from 2 • rORBITto rORBITfor the satellites listed in Table 2, with the corresponding numbering.

[0303] TABLE 3 A Calculated Fall times from 2 • r0RBITto r0RBITfor the satellites listed in Table 2, with the corresponding numbering. As can be seen if the time to orbit is an issue, the method presented in this disclosure is not necessarily the best choice. a. Ideal Gas Law, gas density

[0304] The gas density calculations will be useful in Section 7 where some basic rocket fuel and oxidizer calculations are given. Modifying the well-known ideal gas law,

[0305]

[0306] Or more conveniently written as,

[0307]

[0308] Where P, V, n, R, T, m and. M are pressure, volume, mol, ideal gas constant, temperature, mass of the gas, and molecular mass respectively. Dividing both sides of (3.34) with V gives the ideal gas law as,

[0310] Where p is the density of the gas which leads to the gas density formula,

[0312] Gases related to this work are Hz, N2, O2, C02,and He. Molecular masses M for these gases can be calculated based on their atomic masses giving approximately 2, 28, 32,44, and 4 [gr / mol] respectively. The gas density relation as given in (3.38) and its function of altitude as shown in Fig. IE gets into the numerical solution of the equation of motion for including the air drag during the rocket’s travel in the atmosphere. Air drag force FDis expressed by the well-known relation,

[0313]

[0314] Where p, kD, S, and V are the air density, drag coefficient, surface area perpendicular to the air flow and air speed respectively.

[0315] If the satellite in orbit faces air drag, as formulated in (3.39), it will either bum out due to frictional heat generated or it will lose speed and energy and eventually spiral down toEarth and will crash to the Earth’s surface. As can be seen in Fig. IE density of air goes very close to zero after altitudes of 100 [km], which gives practically zero drag force, and the satellite can stay in orbit forever after reaching the orbital velocity with the right direction. This is the only reason of putting satellites into orbit higher than 100 [km], or in other words above the atmosphere. Sixty-seven years after the launch of the first satellite in 1957, still many news and social media news sources very incorrectly use “zero Q", “no gravity”, or “zero gravity” terminology at the satellite or space craft launch news. As can be seen in Fig. 2A and 2B the gravitational pull of earth in all LEOs is not less than 0.75 of its value at sea level, not even close to “zero gravity” as often reported. Zero gravity is only possible at an infinite distance from a mass.

[0316] Rocket Thrust and Mass.

[0317] Rockets create thrust by ejecting parts of its mass with high velocity, which can be mathematically formulated using the conservation of momentum principle, and can be considered a straight-forward calculus exercise [1-3], There are many scientists who need to be credited for this derivation going back to 1810 [3], The first record of the derivation of the rocket equation is known to have been done by the British mathematician William Moore in his work “Theory on the motion of Rockets” and “Treatise on the Motion of Rockets and an assay on Naval Gunnery”, which was published in 1813. The minister William Leitch, another British scientist also independently derived the fundamentals of rocketry in 1861. Robert Goddard inthe USA also independently derived the rocket equation in 1912. Hermann Oberth in Germany derived the same equation studying the feasibility of space travel in 1920’s.

[0318] The Russian scientist Konstantin Tsiolkovsky’s (1857-1935) derivation of the rocket equation in 1897 is accepted as being the first to consider whether rockets could achieve the speeds necessary for space travel and therefore it is named as Tsiolkovsky Rocket Equation, which he called “formula of aviation”. He is also the inventor of multi-stage rockets based on the very interesting mathematical properties of the rocket equation. He is also the inventor of the“space elevator” and many other things related to rocket science and is therefore accepted as being the father of rocket science. Hermann Julius Oberth (1894-1989) a German rocket scientist also derived the rocket equation while working in Peenemunde during Second WorldWar for the V-2 ballistic rocket with his student Werner von Braun who ended up running theApollo program at NASA.

[0319] Tsiolkovsky’s Rocket Equation.

[0320] Several forms of derivation are possible [1-3], The equation of motion for the rocket in vector notation can be written as,

[0321]

[0322] are the sum of external forces, the mass of the rocket, which is a function of time, velocity of the rocket, exhaust gas velocity relative to the rocket, and the unit exhaust gas velocity vector with respect to the rocket, respectively. The time dependent mass of the rocket for constant fuel bum rate bTcan be given as,

[0323]

[0325] where 7rt0and. bTare the initial total mass of the rocket and fuel bum rate in[kflf / s]. With no external forces and in scalar notation (5.1) can be written as,

[0327] where Veis the effective exhaust velocity in [m / s], which has a range of 2,500-4,500 [m / s] based on the propellant used, the rocket engine design, its convergent-divergent nozzle geometry, and injection-mix efficiency of the propellant into the thrust chamber for liquid propellants. The range of these some values of Vefor some popular rockets are presented inTable 4A.

[0328] Table 4A. List of exhaust velocity and propellant types of some well-known rockets at lift-off. RP-1 is a rocket fuel based on kerosene and LOX is Liquid OXygen.

[0329] Multiplying both sides of (5.3) with dt gives,

[0330]

[0331] On the other hand, differentiating (5.2) the generated thrust, FT in scalar form in[Newtons] becomes,

[0332]

[0333] The negative sign indicates that the generated thrust is in the opposite direction of the exhaust gas flow relative to the rocket. Multiplying both sides of (5.4) with dt eliminates the time dependency and gives the very simple differential equation where its analytical solution is trivial as,

[0334]

[0335] Integration of both sides of (5.6) using the proper limits gives,

[0336]

[0337] where m0, TTlj, Vo, and Vj- are the initial and final total mass of the rocket in[kg], and initial and final velocity of the rocket, respectively in [m / s]. Finally, the solution of(5.7) leads rocket equation to its most common form as,

[0338] ( )

[0339] where,

[0340]

[0341] where are the difference in velocity, specific impulse inseconds [s], and standard gravity in [m / s2] respectively. Since in a rocketalways larger than the mass ejection velocity or exhaust gas exit velocity from the rocket engine nozzle, which is in the range of 2,500 - 4,500 [m / s] as can be seen in Table 4A. As can be seen relations (5.9) and (5.11) enable the rocket to achieve very large orbital velocities or even escape velocity, which is 11,000 [m / s] on Earth’s surface, and theoretically propel the rocket to an infinite distance away from the Earth ignoring other gravitational effects present.

[0342] For constant fuel bum rate bTthe fuel bum timerelates the final mass tothe bum rateTas,

[0343]

[0344] where 771propis the mass of the (fuel + oxidizer), which is consumed until the end of fuel bum time TB. One of the most important and useful applications of the rocket equation is in relating the initial and final mass of the rocket as a function of desired speed difference Taking the first part of (5.8), it can be written as,

[0345]

[0346] Dividing both sides with Vegives,

[0347]

[0348] (5.12) can also be written as,

[0349]

[0350]

[0351]

[0352]

[0353]

[0354]

[0355]

[0356]

[0357]

[0358] The equation (5.17) is very useful because by entering 2 numbers into it, it is possible to calculate 77lprop> the fuel needed as a percentage of the initial mass mo to gain a given speed and rocket exhaust velocity for the case of no external forces.

[0359] 5.1 Derivation of Tsiolkovsky’s Rocket Equation for ConstantGravitational Acceleration.

[0360] For a rocket moving straight up vertically against Earth’s gravitational force the most important force to consider is the Earth’s gravitational force. The resulting gravitational acceleration g acting upon is formulated as [4-5, 15, 18-21],

[0361]

[0362] where are object distance to the center of the Earth,Newton’s constant of gravitation, mass of the Earth, and altitude measured from the surface of the Earth. Following are the numerical values in (5.18) as,

[0363]

[0364]

[0365] The Earth’s Polar and Equatorial radiuses are slightly different and are,

[0366]

[0367] As shown in Figs. 2A, 2B, and 2C, for all practical purposes the gravitational acceleration g for the range of altitudes involved in this analysis can be assumed constant with a numerical value of 9.81 [m / s2] at sea level. Since rocket propulsion can deliver the very high velocities required to put a satellite into orbit, it became the only practical means of doing so. In this mode the Earth’s gravitational acceleration g must be included in the rocket equation [1-3],

[0368] For this case the rocket equation of motion becomes,

[0369]

[0370] The second term on the right-hand side is the thrust generated in vector form where is the unit vector in the direction of the flight path relative to the rocket, which is opposite to the rocket velocity vector Rearranging (5.19) gives,

[0371] U# L dt

[0372] The constant force generated by a constant acceleration g opposing thedirection of the thrust can be introduced into (5.20), and ignoring air drag gives the onedimensional scalar rocket equation of motion as,

[0373]

[0374]

[0375]

[0376]

[0377]

[0378]

[0379]

[0380]

[0381]

[0382] On the other hand, (ty — to) in (5.25) is the bum time TBfor constant bum rate bTcalculated as,

[0383]

[0384] Arranging (5.25) and substituting (5.26) in it gives,

[0385]

[0386] Equation (5.27) is the corresponding equation (5.11) for the case of no external forces. In this work the rocket in the launch stage moves in the opposite direction of gravitational force, but during the descent or capture stage the rocket moves in the same directionas gravitational force. If the gravitational acceleration is in the direction of the thrust, the sign of the last term in (5.27) changes to a positive sign (+). Covering both cases (5.27) can be written as,

[0387] y

[0388] where the + sign corresponds to gravitational acceleration if it is in the same direction as thrust, the case where it is used in the descent stage of the system described herein. The velocity difference which is defined as the rocket velocities going opposite andin the same direction of gravitational acceleration, has the same mass parameters,

[0389]

[0390] Equation (5.29) shows that the same velocity can be gained with a significantly smaller mass of propellant. Rocket assisted descent is important and requires a more detailed analysis as is given in Section 5.2. Coming back to lift-off case, since the rocket should be able to lift-off the ground with full initial mass 77l0, the thrust Fj must satisfy,

[0391] W g W g

[0392] The parameter T2W is the thrust-to-weight ratio of the rocket at the launch pad, which must be greater than 1 for a successful launch. Due to safety of the launch the typicallyT2W at the launch is set to a number greater than 1.5.

[0393] Relation (5.27) and its more general form (5.28) can be represented better by introducing a parameter / z(t) as,

[0394]

[0395] The minimum value of #z(t) is reached when all the propellant is consumed. Since there is always a payload involved with any launch my > 0, the final value of jif is greater than zero as given in (5.31). The inverse of g(t) can be written as,

[0396]

[0397]

[0398]

[0399] Time dependency of / z(t) in terms of bum rate can be written explicitly as,

[0400]

[0401] Substituting t = 0 and t = TBin (5.34), the upper (final) and lower (initial) limits of (IJin powered flight can be written as,

[0402]

[0403] Some arithmetic performed on the second part of (5.27) and (5.28) gives,

[0404]

[0405] Multiplying dominator and the denominator of (5.36) with VEand T2W gives,

[0406]

[0407] On the other hand, writing thrust in terms of thrust-to-weight ratio T21V gives,

[0408]

[0409]

[0410]

[0411] The equation (5.29) which gives the velocity of the rocket accelerating in the opposite direction of the uniform gravitational acceleration g after burning all its propellantmprop becomes expressed in a very compact form with very simple rocket related variables g.f and. T2W as,

[0412]

[0413] Since the initial velocity of the rocket when it is standing on the launch pad is zero, using vB, where the subscript “B” representing “Burnt”, instead of Av( / zy, T2W), is a more meaningful and more convenient in the following math. Close examination of (5.40) gives,

[0414]

[0415] This result is clearly non-physical and needs correction which is explained inSection 6 below. And,

[0416]

[0417] Fig. 4A depicts VB( / Zp T2W) for thrust-to-weight ratio T2W = 1.1 and 8 with a constant acceleration g. The asymptote at which makes Vy — > oo at Hj- = 0 is marked with dotted vertical line is clearly shown. One of the early questions to be explored in rocketry was todetermine if rocket propulsion could generate velocities sufficient to put an object into orbit, or sufficient to create an escape velocity VESCAPE from the Earth. To illustrate, escape velocity VESCAPE « superimposed on the Vfl( / Zp T2W} curves. As can be seen, for μfvalues very close to zero, Vg^M-p T2W} > VESCAPE » where the rocket escapes the Earth’s gravitational pull and reaches an infinite altitude h -> 4-oo. The linear scale of flj in Fig. 4A does not clearly show how close the μfvalue must become fhypersonic velocities to put a satellite into a desired orbit as shown in Fig. 1 A and IB. This could be achieved by presenting the same data in Fig. 4A where the horizontal axis is drawn in logarithmic scale.

[0418] Fig. 4B depicts Vg^flf, T2W) for thrust-to-weight ratio T2W = 1.1 and 8 with a constant acceleration g, where the horizontal axis is in logarithmic scale. The required values for achievinghypersonic velocities needed to put a satellite into a desired orbit have to be below 0.1, meaning that majority of the rocket mass has to be propellant.

[0419] Writing everything in scalar form for simplicity, if Z represents the distance traveled up, or in the opposite direction of the gravitational acceleration, the velocity along the same direction becomes the derivative of Z traveled expressed as,

[0420]

[0421] Integrating (5.43) as,

[0422]

[0423] The integral (5.44) can be evaluated with a variable transformation applyingLeibnitz’s chain rule [20-24] given as,

[0424]

[0425]

[0426]

[0427] Substituting (5.46) in (5.45) becomes,

[0428]

[0429] Giving the integral with the help of the lower and upper limits of p given in(5.35),

[0430]

[0431] The upper limit ZlBin the integral on the left-hand side of the integral equation(5.48) is the altitude that the rocket reaches after ejecting all its propellant, or at t = TB.

[0432] To integrate the first term in the right hand-side of the integral equation (5.48),

[0433]

[0434] With the variable transformation,

[0435] P

[0436] resulting in the integral which has an open form integral expression [22-24] as,

[0437]

[0438] Substituting y^ in y (5.51) for calculating the integral value at the upper integration limit 12 of (5.49) becomes,

[0439]

[0440] Substituting the lower integration limit g0= 1 in (5.51) for calculating the integral value at the lower limit of (5.49) gives,

[0441]

[0442] The resulting integral value I2 — K of the first part in (5.49) becomes,

[0443]

[0444] Integration of the second term in right hand-side of (5.48) is straightforward giving [22-24],

[0445]

[0446] Applying the integration limits at (5.48) to (5.55) gives

[0447]

[0448] Substituting (5.56) in (5.48) the integral (5.48) finally becomes, 2

[0449]

[0450] The multiplier in front of (5.57) can be simplified further by multiplying denominator and the dominator with (fl. Vg) giving,

[0451]

[0452] Giving the height that the rocket reaches in powered flight at the time of TB, inother words when it runs out of propellant, with a vertical velocity in opposing direction of constant acceleration g as,

[0453]

[0454] At that point the rocket has a velocity VBas given in (5.40) and it keeps gaining altitude. In other words, it coasts until it reaches its final altitude h. also known as “apogee” in the rocket literature where its velocity becomes zero.

[0455] The limits of are worth mentioninggiving,

[0456]

[0457]

[0458] Fig. 5A depicts for thrust-to-weight ratio and 8with a constant acceleration g .

[0459] Fig. 5B depicts ( ) for thrust-to-weight ratio and S wherethe horizontal axis is in logarithmic scale.

[0460] What is interesting to observe is that as given in (5.60) and shownin Figs. 5A and 5B, has a finite value that is less than 600 [km] for any practical thrust to weightratio of T2W. As shown in Figs. 2A, 2B, and 2C, the fl(h) variation for altitudes of 600 [km], which can be considered to be the maximum bum height hBMAX, is negligible, meaning that the constant acceleration g assumption for any condition is an excellent approximation, at least for the bum stage of any rocket launched from Earth. Fig. 2C depicts the gravitational acceleration normalized to its sea level value as a function of altitude g(K) with the maximum bum altitude hB(jif = 0) for T2W = 1.1, 2, 4 and 8 marked on the curve, which graphically displays this important fact for a wide range of T2W values. As can be seen at the highest bum altitude the normalized gravitational acceleration does not go below 0.85 meaning that constant acceleration g assumption is a good approximation for any rocket launched from the Earth for its hB(0, T2W) calculation. This property is used for the derivation of the inverse square law gravitational case in as described in Section 6.

[0461] The horizontal lines superimposed on Fig. 5A and 5B show hG1VEN= 40 and 400 [km]. Their intersection points with the hB(μf, T2W) curves give the corresponding μfvalues which is the graphical solution to the non-linear problem given as the equation hB0. Fig. 5 A shows that for h.B(jlp T2W) = 40 [ / cm], μfshould be roughly in the interval of 0.3 < < 0.6 for all the ranges of T2W, which demonstrates the graphical solution method to this non-linear problem related to rocket design altitude. Fig. 5 A does clearly show the intersection points of h.B(jlp 7*2 U^) curves and hG1VEN= 400 [km], but Fig. 5B, having a logarithmic horizontal axis, gives roughly μf= 0.06, illustrating some of the difficulties in reaching higher altitudes.

[0462] The final altitude h can be solved by applying the energy conservation law

[0021] as,

[0463]

[0464] Where KE(hB} and PE(hB} are the kinetic and potential energies at the altitude hg, where the velocity of the rocket is the known value of VB. Similarly,andare the kinetic and potential energies at the final altitude / l, where the rocket velocity is zero, giving zero kinetic energy at apogee, which can be written explicitly as,

[0465]

[0466]

[0467]

[0468] Fig. 6A depicts the climb altitude h(jlp T2W} for the constant gravitational acceleration of g = 9.8 [m / s2] with thrust-to-weight ratio T2W = 1.1 and 8 where the vertical axis is in logarithmic scale. Since the Vfi( / Zp T2W} term in (5.64) becomes +oo for jlj = 0, h.(jlp T2W} also becomes +°° , as shown with an asymptote (dotted vertical line) in Fig. 6A at / 2y = 0.

[0469] The horizontal lines superimposed on hGiVEN= 40, 400, and 400,000 [km], and their intersection points with the hQlp T2W^ curves give the corresponding μfvalues that is the graphical solution to the non-linear problem h.(jZp T2W} — hGIVEN= 0. Fig. 6A shows that for / l(μf, T2W) = 40 [fcm], μfshould be roughly in the interval of 0.6 < (1^ < 0.75 for all the ranges of T2IV, which is an easily achievable task. This demonstrates the graphical solution method of solving this non-linear problem for rocket design altitude. Fig. 6A also clearly shows that the intersection points of the / l( / Zp T2W") curves and hGIVEN= 400 [km] are roughly in the interval of 0.275 < flj < 0.45. Fig. 6A shows the difficulty in reaching extreme altitudes of 400,000 [km] which is the approximate distance to the Moon, requiring very small μfvalues. Fig. 6A graphically demonstrates that reaching LEO orbital heights is only achievable with multi-stage rockets.

[0470] The first term in the brackets of (5.64) is the distance that the rocket “coasts” after depleting all its propellant under constant acceleration “g” and is represented as,

[0471]

[0472] Naming this distance h.c, as the “coast” distance, writing (5.64) in terms of the sum of h.Band hcbecomes handy in evaluating the derivatives and limits with respect to gy giving,

[0473]

[0474] Since (5.41) holds the limit of0 becomes,

[0475]

[0476]

[0477]

[0478] Due to (5.66), the limits of / l at at gy- = 0 and g^ = 1 give the same values as (5.67) and (5.68). Due to (5.67), g^ -> 0 becomes the asymptotes for the h.c(jlp T2VV) and h(jlp T2W^ curves, as shown in Figs. 6A and 6B.

[0479] Fig. 6B depicts bum height hB(jAp T2IV) and coast height hc(jJ.p T21V) for the constant gravitational acceleration of g = 9.8 [m / s2] with thrust-to-weight ratio T2W =1.1 and 8 where the vertical axis is in logarithmic scale. The asymptote for h.c(jlp T21V) is shown at g^ -» 0 with the dotted vertical line while hB(jJ.p T2W) has a finite value less than 600 [km],

[0480] 5.2 Solution of Tsiolkovsky’s Rocket Equation for Constant GravitationalAcceleration

[0481] When the Rocket Thrust is in the Same Direction of the Gravitational Force

[0482] The modification to accommodate the rocket thrust being in the same direction of the gravitational force is straightforward. The bum velocity Vy after consuming all the rocket propellant as given in initially at (5.28), (5.29) and finally in (5.40) can be written for gravitational force is in the opposite direction of the rocket thrust as,

[0483]

[0484] Superscript — g is not an exponent, it is a superscript indicating it is the velocity gained when the rocket thrust is in the opposite direction of the gravitational force symbolized as g . After some simple math, the velocity gained when the rocket thrust is in the same direction of gravitational force can be written as,

[0485]

[0486] Similar notation can be used for the hB, derived in (5.59) for the distance covered when the rocket thrust is in the opposite direction of the gravitational force symbolized as gas,

[0487]

[0488] Again, after some simple math, the velocity gained when the rocket thrust is in the same direction of gravitational force can be written as,

[0489]

[0490] Fig. 3H depicts the rocket thrust assisted velocity reached Vg, having rocket thrust being in the same direction of the constant gravitational acceleration shown as VB8(in rocket powered vertical descent) and rocket thrust being in the opposite direction of the constantgravitational acceleration shown as (in rocket powered vertical ascent) using the solution of modified Tsiolkovsky’s rocket equation for T2W = 1.1 and 4 versus In both casesvelocities are after all the propellant is consumed. Escape velocity and orbital velocities for 400[km] altitude is also shown for reference.

[0491] Fig. 31 depicts the rocket thrust assisted distance covered having rocket thrust being in the same direction of the constant gravitational acceleration shown as (in rocketpowered vertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as (in rocket powered vertical ascent) using the solutionof modified Tsiolkovsky’s rocket equation for T2W = 1.1 and 4 versus μfIn both cases distances are after all the propellant is consumed. The distances of 400 and 40 [km] are also shown for reference.

[0492] Fig. 3 J is same data depicted in Fig. 31 in logarithmic X axes to show the covered distances clearer for larger μfvalues.

[0493] Fig. 3K depicts the ratios between the velocities reachedand ratios of the distances coveredforT2W = 1.1.

[0494] Fig. 3L depicts the ratios between the velocities reachedand ratios of the distances coveredT2W = 4.

[0495] As can be seen the only differences are the sign change in the last terms in (5.69) and (5.71) but it has a very significant effect on the velocities gained and distances covered between the vertical rocket assisted descent described in this invention and prior art ascend.

[0496] With all the notations given above Fig. 3H - Fig. 3L displays the comparative analysis between (5-69) and (5.70) for velocities gained graphically for T2W = 1.1 and 4.As can be seen in in Fig. 3H for T2W = 1.1 the same payload can gain orbital velocity for orbital height of

[0497] h = 400 [km] (ISS orbital altitude) with jlj = 0.2 in rocket assisted decent, while the same velocity can be gained with μf= 0.05 giving a Aiij(T2W = 1.1) = 0.15 resulting in very significant propellant savings and reduction in rocket thrust for the same payload in descent versus ascent. Fig. 3H also shows the similar fact with less savings for T2W = 4. Precise numerical values of A[lf(h, 72147) are given in the third column of Table3B for all six orbital altitudes listed in Table 3 A for T2W = 1.1, 2, and 8.

[0498] Fig. 31 depicts the rocket thrust assisted distance covered having rocket thrust being in the same direction of the constant gravitational acceleration shown as hB9(in rocket powered vertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as hB9(in rocket powered vertical ascent) using the solution of modified Tsiolkovsky’s rocket equation for T2W = 1.1 and 4 versus p.y . In both cases distances are after all the propellant is consumed. The distances of 400 and 40 [km] are also shown for reference.

[0499] Fig. 3 J is same data depicted in Fig. 31 in logarithmic X axes to show the covered distances clearer for larger μfvalues.

[0500] Fig. 3K depicts the ratios between the velocities reached 7 / yB(μf) = VB9 / and ratios of the distances coveredfor72W = 1.1.

[0501] Fig. 3L depicts the ratios between the velocities reached ^VB( / Z / )= / and ratios of the distances covered T)2147) for72IV = 4.

[0502] As a conclusion, the main reason of achieving orbital velocities by going into smaller radiuses then initially calculated FQ given as,

[0503]

[0504] using energy conservation principle with the same mass as in (1.1). This results in defining a third radius with a scaling factor Sy as given in (1.2) as,

[0505]

[0506] It is important to point out that all these calculations are done for vertical descent and ascent using rocket propulsion based on mass ejection for constant gravitational force. Fig.3H and Fig. 3L clearly show that the thrust to weight ratio (T2W) plays an important part in the results. From the curves it is evident that as the T2W increases, the curves get piled on top of each other and are difficult to distinguish from each other. Table 3B gives a clearer view of the analysis done for satellites at different orbital altitudes showing the T2W effect on the results and the logic of coming up with a third radius rhand its relation to Tothrough the scaling factor Sy.

[0507] This is done in two steps: a. Find the μf8which will give the orbital velocity VORB / TC^) when initially stationary at the altitude h by solving the equation given in (5.70) as,

[0508]

[0509] ii) By substituting μf8for μfin (5.72) hg8(jlf, T2W) is obtained, which is the fall distance covered when the rocket thrust is in the same direction of gravitational acceleration while reaching VORBITQ1} from 811initial velocity of zero and is written on the 5thcolumn of Table 3B for all 6 orbital altitudes and for 4 different values of T2W. The arithmetic for calculating Sy is straightforward and is listed in column 7 of Table 3B. Column 6 gives the ratio. As can be seen all the way to orbital altitudes of 781 [km], the resultsare consistent, but the inaccuracy caused by assuming constant acceleration throughout the entire descent path becomes apparent.

[0510] Table 3B lists several calculations done on the 6 orbital altitudes given in Table3 A for the same rocket thrust in the same direction with respect to the gravitational force shown as superscripts and opposite with respect to direction to the gravitational force shown as superscripts.

[0511] The remaining columns of Table 3B are for comparative analysis between the rocket thrust against gravity and in the same direction as the gravity.

[0512] Solving (5.69) gives M-j8, which gives the orbital velocity VORB / TC^O when initially stationary at the altitude h. when the rocket thrust is against gravity as,

[0513]

[0514] The third column of Table 3B gives the differences between the solutions of(5.75) and (5.76) as,

[0515]

[0516] By substituting is obtained, which is theheight from which the payload is launched if rocket thrust is opposite to gravitational force given at fourth column of Table 3B.

[0517] On the other hand, since the whole objective is to put a satellite into orbit, the final masses in both cases are equal and it is equal to the satellite mass or referred to as payloadTHp^yin earlier calculations. Therefore written as,

[0518]

[0519] Since most of the mass of a rocket is propellant (5.78) can be approximated as,

[0520]

[0521] Which gives the eighth (last) column in Table 3B. Having a large ratio shows how much propellant is saved for the same payload with rocket thrust in the same direction of gravitation versus against it.

[0522] Real calculations and trajectory optimizations are only possible with numerical solution of three-dimensional equation of motion, and the and related Sy calculation is payload and orbital altitude dependent, but it is clearly shown in Table 3B that Sy values can be even as low as 1.1.

[0523] It was earlier proven that the total energy of an object in a circular orbit at a radiusTORBIT is equal to the total energy of the same object, which is stationary at the radius of r0=Then when the same object is launched to a smaller radius, rhthen 2TORBIT, asgiven in (1.2) and (5.74), it will have less gravitational potential energy and since it is stationary its total energy, compared to the total energy when orbiting at a radius of TQRBIT.

[0524] Fig. 131 demonstrates the methodology employed for two different values of sf = 1.5 and 2 along with the related mass versus time plots of Fig. 13J and Fig. 13K.Similar graphical demonstration is given for 20,200 [km] (GPS) orbital launch in Fig. 13M with the related mass versus time plots Fig. 13N and Fig. 130.

[0525] Modification of Tsiolkovsky’s Rocket Equation for Inverse Square LawGravitational Field.

[0526] If the launch altitude is in the order of, or larger than the Earth’s radius,Tsiolkovsky’s Rocket Equation for constant gravitational acceleration must be modified for the inverse square law gravitational field. In this work this modification is done by applying the energy conservation law for inverse square law gravitational fields which gives much better correlation to reality and numerical solution of the equations of motion. In this approach, the non-linearities in the equation relating the final and initial rocket masses to final velocity and altitude get more complex and solving it with Newton’s method becomes more challenging, requiring more care and is explained in detail below.

[0527] The Inverse Square Law Gravitational Field Relation can be incorporated into the conservation of energy formulation for calculating a more accurate final altitude / l. If the final altitude / l, calculated by (5.61) is comparable to or larger than the radius of Earth rEARTH, the potential energy expression for constant gravitational acceleration Q becomes inaccurate. Fig 2B clearly shows the ratio of the inverse square law calculated acceleration versus constant acceleration value on the surface of tire Earth as a function of radius measured from the center of the Earth and at sea level. The marked altitude on the curve is hBMAX>which shows that highest altitude that any rocket can reach during its bum phase giving 85% of the gravitation acceleration of the surface value, giving a much smaller h. than the gravitational potential energyderived for the inverse square law gravitational field as shown in Fig. 2C and explained above.This problem can be fixed by replacing the potential energy with the formulation for the inverse square law gravitational field giving (6.1),

[0528]

[0529] Subscripts “INV2” represent the “inverse square law gravitational field formulation” quantities, where rBand T^INVI816radiuses corresponding to the altitudes hBand h]NV2given as,

[0530]

[0531] Since TEARTHis a constant, the following derivative relations are also valid,

[0532]

[0533] Solving 13D from (6.1), gives,

[0534]

[0535] Multiplying denominator and dominator of (6.4) with — 1 gives,

[0536]

[0537] Substituting in (6.5), which is the case of interest gives,

[0538]

[0539] Since a proper (Vg, Vg) combination can make the denominator of (6.6) zero, an infinite value for both ThINV2a°d h-iNV2 is possible with a finite Vg. The equation gives the “finite” Vg. The equation which gives the finite Vg, resulting with an infinite value for bothThlNV2anc^ ^INV2canbe solved by solving,

[0540]

[0541] Giving,

[0542]

[0543] This corresponds to the escape velocity from a spherical boundary with a radius of rBenclosing the mass of Tn EARTH *nit- The escape velocity from Earth vs. altitude is very clearly shown in Fig. ID along with markings of the orbital altitudes given in Table 2. This means that the rocket reaches infinite radius or altitude if it has a velocity of VgAsyat rBor atthe altitude h.B. As can be seen this makes perfect physical sense for the defined escape velocity, which the solution of the rocket equation under constant g formulation does not give.Writing (6.8) in terms of the previously calculated altitude Hg and velocity Vggives,

[0544]

[0545] The corresponding “coast height” hCINV2 becomes,

[0546]

[0547] Fig. 7 A depicts asymptote X intercept ^ / isyM(T2W) vs. T2W and / ZMjAf(T2W) by solving the non-linear equation (6.7). As can be seen, the asymptotes for any T2W are very close to zero, in the range of 0.016 < < 0.036, and getting closer to zero with decreasing thrust-to-weight ratio T2W. As given by (6.6) TMNV2 becomes +oo along with thealtitude A / JVV2 due to the linear relation given in relation (6.2), which causes plotting and interpolation to be very difficult, if not impossible for μf= PASYM(J2W\ TO avoid this problem, MMW(T2W) is defined by a small 8 value greater than the PASYM(T2W) with very large altitude values in the family of altitude vs / Incurves in Fig. 7B through 7E.

[0548] Fig. 7B depicts climb altitude h.iNV2^Uf> T’ZV / ) for the inverse square gravitational field with asymptotes where vertical axis is in logarithmic scale for thrust-to-weight ratio T2W = 1.1 and 8.

[0549] Fig. 7C depicts bum height hBINV2( / Zp T2W) and coast height h.ciNV2^M-f» T2W) at T2W for the inverse square gravitational field with asymptotes where the vertical axis is in logarithmic scale.

[0550] Fig. 7D depicts climb altitude hINV2^Pp 7*21V) for the inverse square gravitational field with asymptotes and hQlp T2W) for the constant gravitational acceleration of g = 9.8 [m / s2] where the vertical axis is in logarithmic scale for thrust-to-weight ratio T2W = 1.1 and 8. Although the differences are self-explanatory, they can be better seen in logarithmic horizontal axes.

[0551] Fig. 7E depicts climb altitude hINV2^Pp T2W^ for the inverse square gravitational field with asymptotes and hQlp T21F) for the constant gravitational acceleration of g = 9.8 [m / s2] where the vertical and horizontal axes are in logarithmic scale for thrust- to-weight ratio T2W = 1.1 and 8. Important to note is that the inverse square gravitational field assumption has to be used for solving μffor altitudes higher than 400 [km]. The constant gravitational acceleration assumption can be safely used for calculating μfas well as altitudes less than 400 [km],

[0552] Fig. 7F depicts climb altitude hINV2(p.p T2W^ for the inverse square gravitational field with asymptotes for p < PASYMO"2W) with thrust-to-weight ratio T2W = 1.1, 2, 4, and 8.

[0553] As can be seen very clearly in Fig.TF there are 2 asymptotes for each curve. The first one is at functionsapproach — , which is non-physical. The second asymptotes of everycurve has an asymptote where their X axes intercept, as shown in Fig.7A, at where the functions approach + co. This asymptotebrings also a “jump” type discontinuity where jumps from +oo to zerowhen μfcrosses the asymptote as can be seen clearly in Fig. 7E and Fig.TF. This can be worded as in the interval of function becomes nonphysical.

[0554] Fig. 8 depicts climb altitudef for the inverse square gravitational field and climb altitude with the constant gravitational accelerationthecurves become indistinguishable. Fig. 8 is another wayof showing that the constant gravitational acceleration assumption can be safely used for calculating μf, by simply looking at the calculated the constantgravitational acceleration assumption is accurate enough to use, with no need to employ the inverse square gravitational field formulation or to question the formulation used in the solution, both give very close results.

[0555] Introducing Propellant Tank to Propellant Mass Ratio k.

[0556] The propellant tank mass is related to the mass of the propellant that it stores, by introducing a parameter k as,

[0557]

[0558] The initial mass m0of the rocket can be written as,

[0559]

[0560] where mprop, mREand mPAYLOADare the mass of the propellant, rocket engine and payload, and where the parameter k is called the “propellant tank-to-propellant mass ratio”. Typical values of k should be a small number like 0.05 to 0.2, with a smaller ratio being more advantageous. The chemistry and the resulting k parameter for liquid and solid fuel rockets are different. Here the analysis for a very simplified formulation applicable of a liquid rocket having a propellant tank being made from only a single cylinder is presented. A similar approach for k parameter calculations for the solid rockets can also be derived.

[0561] In a liquid fuel rocket the mpropconsists of oxidizer plus the fuel masses given as,

[0562]

[0563] which are typically stored in two different cylindrical tanks having two semi- spherical caps at the top and bottom. Due to the chemistry of burning the fuel with the oxidizer, their masses, densities, and their resulting volumes are not necessarily equal and can be calculated with their reaction chemistry. The simplest example uses hydrogen; H2as fuel and oxygen; O2for the oxidizer giving the chemical reaction of,

[0564]

[0565] Applying stoichiometry analysis to the chemical reaction given at (7.4) shows that 2 kg of hydrogen reacting with 16 kg of oxygen (1 / 8 mass ratio) gives 18 kg of steam. This 18 kg of hot steam is ejected from the rocket nozzle with a velocity of VE- Assuming each are stored in liquid form, hydrogen has a density of 71 kg / m3at 20.28 K (-252.87°C) and liquid oxygen has a density of 1,141 kg / m3at 90.19 K (-297.33°C). As can be seen liquid oxygen is denser than water and approximately 16 times denser than liquid H2. The question becomes how to calculate the volumetric ratios of the liquid h2and O2satisfying the calculated 1 / 8 massratio. Using this example, 2 kg of hydrogen volume V = 2 / 71=0.02817 m3, reacts with an oxygen volume of V (02)=16 / 1,141=0.01402 m3. This shows that the H2tank must be volumetrically 2.009 times larger than the O2 tank to satisfy the calculated 1 / 8 mass ratio.

[0566] In rocket design there are some other factors that are considered for maximizing thrust and cooling issues of the rocket engine. For H2 / O2rockets the highest impulse power, Igp is achieved when the H2 / O2mass ratio is 1 / 4 (leaving half of the H2unbumt), not when it is 1 / 8 corresponding to full bum of H2. In practice the mass ratio is kept as 1 / 6 for other reasons. As an example, the space shuttle liquid 02tank is 19,563 cubic feet (553.96 m3) and H2tank is 53,518 cubic feet (1,515.46 m3), having volumetric ratio of 2.7357. The full load oxygen and hydrogen mass that can be stored in these tanks are 632.068 and 107.597 tons, giving the 1 / 6 mass ratio, as given earlier. As can be seen, calculating optimal mass and volumetric propellant / oxidant ratios is not that simple.

[0567] In a liquid propellant rocket the fuel and oxidizer are stored in two separate tanks, with piping and some additional essential parts like turbo pumps, controls, etc. To simplify all the calculations, it is assumed that the oxidizer and propellant tanks are two cylinders with same radiuses and spherical caps both having a uniform skin thickness tskin. Once the oxidizer and fuel masses and volumes for the mission are calculated as shown in the ^2 / ^2 example above, calculating the parameter k simply becomes a trivial volume and mass calculation. The goal here is to quantify the significant dependency of the parameter k to the geometrical parameters of the rocket, like its height and radius under these assumptions. This can be achieved by doing the analysis for only one tank, named the propellant tank. The parameter k as given in (7.1 ) is defined as the mass ratio between the propellant tank and the propellant stored in it, k can be simply calculated by the propellant tank skin area Sskin, skin material density Pskin, propellant volume contained in the propellant tank, VpTOp, and its density, Pprop- A good estimate of k can be given as,

[0568]

[0569] Ignoring the masses of both end caps of the propellant tank gives k as,

[0570]

[0571] As can be seen in (7.6) k is a decreasing function of the rocket radius r, closer to a function inversely proportional to the radius T of the rocket. A better estimate of k can be obtained by adding the masses of the top and bottom caps of the propellant tank. Assuming the caps are semi-spherical and has the same skin thickness the skin volume thetwo semi-spherical shell regions on the top and the bottom of the propellant tank is,

[0572]

[0573] The volume of the cylindrical shell region with a height of and a skinthickness of

[0574]

[0575] The empty mass of the tank becomes,

[0576]

[0577] Assuming the propellant tank is filled completely prior to launch gives the internal volume of the propellant tank as,

[0578]

[0579] The propellant mass becomes,

[0580]

[0581] Fig. 11 depicts the skin of a propellant tank.

[0582] Fig. 12 depicts the mass of the propellant tank m = f(r, tskin) for the skin thickness of ts^n= 3, 6, and 10 cm carrying 20 tons of propellant mass. It is clear from Fig.12 that the mass of the propellant tank 771 is inversely proportional to the radius of the propellant tank for a desired volume or mass.

[0583] Fig. 13A depicts k = / (r, tsfcin) f°r Maluminum skin thickness of tskin=3, 6, and, 10 cm when carrying 20 tons of propellant mass. It is clear from Fig. 13A that k is inversely proportional to the radius of the propellant tank for a desired volume or mass. Section 8 below describes how k determines the maximum altitude that a rocket can reach.

[0584] Thrust, F-p and the Remaining Rocket Parameters Calculated from pf = lily J IHQ and k -™ ytltank / Rlprop*

[0585] The introduction of the parameter k leads into very elegant solutions for all the rocket parameters and very interesting design relations. Propellant with a mass of 77lpropin a rocket is stored in a cylindrical tank with a mass of TH-tank . Since most of the rocket mass is propellant mass, most of the volume of the rocket will also be the volume of the propellant tank.Since the density of the propellant is known, it is straightforward to calculate the cylindrical volume of the propellant tank for a given diameter. The mass of the tank can be calculated with a given skin thickness and density of the tank.

[0586]

[0587] Solving ITlprop from (8.1) gives,

[0588]

[0589] The initial mass m0is related to the thrust FTgiven as,

[0590] g T2W

[0591] Solving Hlprop from (8.2) gives,

[0592]

[0593] Resulting in,

[0594]

[0595] With the introduction of the parameter k, the propellant tank mass can be simply related to the propellant mass. Employing the parameter k the initial rocket mass mg becomes,

[0596]

[0597] As can be seen term in (8.6) gives the sum of the propellantmass and the mass of the tank with which it is stored. In general, propellant mass also is the main structure of the rocket, everything is basically attached to it, which means that the parameter k determines a significant portion of the rocket mass as a function of the propellant mass which is in the rocket equation.

[0598] Another parameter in equation (8.6) is which is the mass of the rocketengine that includes the mass of additional components such as pumps, electronics, guidance, etc. TTlpAY is the mass of the payload and associated additional mass related to its housing.Substituting (8.5) into (8.6) gives the thrust equation,

[0599]

[0600] Both sides of the equation are equal to the total initial mass m0and becomes the thrust equation. The thrust FT, required to put a payload TnPAy into an altitude of h, equation(8.7) can be solved giving,

[0601]

[0602] Relation (8.8) assumes that the rocket has a single rocket engine. Simplifying the denominator gives,

[0603]

[0604] Since there is only one rocket engine and its thrust and mass are already specified as F-p and TnRE, solving its thrust does not make much sense, as it is already known. Instead, one can calculate the payload TnPAYthat the rocket can put to an altitude h with that thrust,

[0605] g

[0606] Since FT, TH-PAY > 0and must be finite, the denominator of relation (8.9) puts some important restrictions between (if and k as,

[0607]

[0608]

[0609]

[0610] Fig. 13B depicts the plot of relation (8.12), which gives the minimum value of k = kMAx that needs to be satisfied to give a desired value of jij. Since a given value of gy also gives the height k / jvvzG1 / ' 7’2147) that the rocket reaches, a very important relationship exists between the height that the rocket can reach and the kMAXvalue, as shown in Fig. 13C.

[0611] Fig. 13C depicts the height hINV2Q<-MAX» 7’2147) function, an important relationship between the height that the rocket can reach and the kMAXvalue that the rocket design must satisfy, graphically permitting the determination of kMAXf°r^GIVEN=400km , 20,200km, and 400,000 km. As can be seen, reaching a given altitude requires a minimum value of kMAXto be satisfied. The higher the desired altitude, the smaller the valueof^MAX that needs to be satisfied. If k > kMAX, the rocket cannot reach the given altitude, h- GIVEN=^JVV2(^MAX» 7’2147).

[0612] Relations (8.11) and (8.12) give the necessary conditions to be satisfied between them. The solution of the rocket equation for the rocket to reach the desired altitude h for a given T2W gives T2 W) and (8.12) gives the minimum value of k needed. If this number is not a realizable quantity, then the launch will not be successfill, leaving a multi-stage rocket as the only option to employ for a smaller k, which is an original relationship derived herein. If k is already given, then (8.11) gives the maximum value of gy and moreover the maximum altitude the rocket can reach. Since Fig. 13A gives k = f(r, tgkin)» “dasshown in Fig. 13C there is a functional relation between kMjNand hjNV2(kM!N> T2W), the altitude that the rocket can reach, a functional relationship h{NV2QrMiN» 7'2147), the radius of the rocket can be derived which is of interest. In other words, to reach a desired altitude there is a minimum rocket radius T"MIN that enables it, which is a function of bum chemistry, density of the oxidizer and fuel, their mix ratios, thickness of skin, its density, and machining. Ideally a value can be given to the denominator of (8.9) which satisfies (8.11) and (8.12).

[0613] Fig. 13D depicts the SpaceX Starship Super Heavy (loaded with 3,400 tons of propellant) and the ULA Atlas first stage (loaded with 284 tons of propellant) booster heightswith a thickness of 3 [cm] with respective diameters of 9 [m] and 3.81 [m]. The relationship between the booster radius and the booster height is shown in Fig. 13D for theSpaceX Starship Super Heavy and the ULA (United Launch Alliance) first stage of Atlas V for known parameters such as their actual propellant mass and tank parameters. Both curves in Fig.13D show an excellent fit for rocket booster heights. For the SpaceX Super Heavy the curve gives 71 [m] of booster height with the actual diameter of 9 [m] loaded with 3,400,000 [kg] of propellant mass. For ULA’s first stage of Atlas V the curve gives 32.5 [m] of booster height at the actual diameter of 3.81 [m] loaded with 284,089 [kg] of propellant mass.

[0614] Fig. 13E depictswhere F is the rocket diameter of the SpaceXStarship Super Heavy loaded with 3,400 Tons of propellant having Fig. 13Eshows the relationship between the SpaceX Starship Super Heavy maximum altitude as a function of its radius loaded with 3,400,000 [kg] of propellant mass and same tank parameters for T2W=1.1, 2, 4, and 8. It is clear that having a larger diameter gives a lower k, resulting in a higher maximum altitude that can be reached. The lowest k can be obtained with a spherical tank geometry giving the smallest area enclosing a given propellant volume.

[0615] If the single rocket engine cannot give the desired thrust then several rocketengines must be deployed. For this case the thrust relation (8.9) must be modified. Assuming all the rocket engines have the same mass (mass Per Rocket Engine) and they all generatethe same thrust per enginethe number of rocket enginescan be calculated from the calculated . The equivalent rocket engine massand the total thrust in (8.6) and (8.7) become related to the number of rocket engines that gives the necessity of including this effect.As an example, for cases where the needed thrust F is larger than the number of rocketengines n#gthat is needed becomes,

[0616]

[0617] where, FTPRE the thrust per rocket engine. There can only be an integer number of rocket engines and this number also has to be greater than 1. The number NREas calculated by relation (8.13) does not necessarily give an integer number. It can even be a number smaller than 1 if the calculated thrust FTis less than the selected rocket engine thrust FTPRE. Therefore, nREis named the “engine thrust scaling factor” to represent its non-integer value. The engine thrust scaling factor nREis used to calculate the total equivalent rocket engine mass to match the needed thrust FT. Using FTPRE, the corresponding equivalent total rocket engine mass mRE, which appears in the FTthrust relation (8.9) becomes,

[0618]

[0619] A more widely used rocket engine parameter instead of mass per rocket engine mPREis the thrust-to-weight ratio of the rocket engine T2WPRE, a number like 100 to 200, a much larger number than any jet engine. As an example, the jet engine with the highest thrust- to-weight ratio is 8 for the GE J85 powering many airplanes like the F-5 and T-38, giving 13.1[kN] with a mass of 140 kg (afterburner versions give 22 [kN] with a mass of 230 [kg]). The F-l rocket engine that powered the Satum-5 had a thrust-to-weight ratio of 94. This is an important parameter in making the rocket engine selection. Using the T2WPREparameter the mass per rocket engine can be calculated as,

[0620]

[0621] Substituting (8.15) in (8.14) gives,

[0622]

[0623] Substituting (8.16) in (8.7) gives another term with FTdependency on the right- hand side as,

[0624]

[0625]

[0626]

[0627] As can be seen the thrust equation (8.9) incorporates another parameter involving the ratio between the rocket engine thrust-to-weight number and the rocket thrust-to-weight number. Since rocket mass is larger than the rocket engine mass, and the only thrust in the rocket is given by the rocket engine, this ratio is always given as less than 1 as,

[0628]

[0629] Solving thrust FTfrom the modified thrust equation (8.18) gives,

[0630]

[0631] To have Fj > 0 and a bounded value for it, the k relations derived at (8.11) and (8.12) must be modified as,

[0632]

[0633] and,

[0634]

[0635]

[0636]

[0637]

[0638]

[0639] A condition far “i stronger” than the relation given in (8.18), since jlf is generally much smaller than 1 for any orbital altitude, can be seen in Fig. 6A, Fig. 7E, Fig. 8, and Figs. 9A - 10B. T2W is a user defined parameter entered into the modified rocket equation, where the best value for the mission is searched. On the other hand, T2WpREis a given, and just depends on the rocket engine that is selected, not a user defined parameter, the larger the better along with other parameters for selecting the rocket engine. In general, the larger the T2WPREnumber, the smaller the ratio is in (8.23) for any given T2W.

[0640] As can be seen that the ratio (8.23) also determines the denominator in the thrust relation. Besides satisfying relation (8.22) for k, the numerical value of the denominator D is something which can be optimized for the minimum thrust needed for the mission. The denominator of equation (8.20) can be re-written with the denominator D represented explicitly as,

[0641]

[0642] Where D is,

[0643]

[0644]

[0645]

[0646]

[0647]

[0648] As can be seen the denominator D is always less than 1 and relation (8.28) gives an upper bound to T2W as,

[0649]

[0650] As can be seen the inequalities and relations given by (8.22) - (8.29) are very powerful relations. Once fl is solved, the equations give what k and T2W can be used with very simple relations. Setting all the parameters as a function of the calculated value of (1 and satisfying (8.27) - (8.29) becomes a simple task. A simple example can demonstrate it as,

[0651]

[0652]

[0653]

[0654] If this is not realizable, k then defines the need for multi-stage rockets and aids a very easy method of figuring each stage constraint. As can be seen in (8.30) for a realizable kdefining a set of targeted μffor each stage becomes a trivial task. Relation (8.29) is thus the key starting point for the rocket design.

[0655] To this point the theoretical and engineering aspects of rockets have been explained in a mathematical framework. At this point it is useful to examine actual technical data of some well-known rockets for comparison the calculations that have been made, derived, and formulated.

[0656] Historically, the most important rocket is probably the German V-2 of the World War n. The first successful V-2 launch was on 3rdof October, 1942 reaching an altitude of84.5km. It burned 55 [kg / s] alcohol with 25% water mixture and 68 [kg / s] of Liquid OXygen,LOX. Total bum time was 65 second and it carried a fuel mass was of 3,810 [kg] (%75 Ethanol,%25 Water) and 4,910 [kg] LOX as oxidizer. Its maximum range was 320 [km], reaching 88[km] altitude, 206 [km] if launched vertically. It carried a 910-1,000 [kg] Amatol high explosive warhead, and more than 3,000 of them were launched.

[0657] Space launches are expensive undertakings. An Ariane cost per launch ranges139,000,000 - 185,000,000 Euros, depending on the payload and launch orbit.

[0658] Table 4B gives some of the dimensional information for well-known rocket boosters.TABLE 4B specifications for well-known rockets

[0659] Quadratic Convergence Property of Newton’s Method for the Solution of theTsiolkovsky’s Rocket Equation in Gravitational Potential and All related Applications

[0660] Since Newton’s method is used for solving non-linear Tsiolkovsky’s RocketEquation derived for both gravitational field representations and followed by solving non-linear trajectory equations, always having a convergent solution for all these tasks is important. This is handled by applying Kantorovich’s theorem to find an initial approximation to the solution which guarantees the quadratic convergence of Newton’s method.

[0661] No rocket launches are ever completely vertical for reasons of safety reasons, they all have curved trajectories, and analytical solution of the rocket equation is only possible for a vertical flight trajectory in or opposing direction of gravitational force. In orbital launches,including the atmospheric air drag is not as important as the curved trajectory, but becomes very important in re-entry phase, therefore for both applications the numerical methods to solve the complete equation of motion are needed. This work uses specifically designed computer programs for rocket design and its trajectory calculations named “Rocket Designer” and “OrbitalLaunch” respectively, which are not included herein. In this work equations of motions are solved employing Runge-Kutta method with variable mass formulation to handle the rocket propulsion with an initial approximations obtained from the rocket equation and the use ofKantorovich’s theorem derived initial approximations for all the non-linear equation solutions using Newton’s method.

[0662] The computer program outputs of “Orbital Launch” and “Rocket Designer” are presented and compared to some existing rockets. The “Orbital Launch” program, as its name suggests, can simulate the any orbital launch solving the equation of motion in 3 dimensions using the Runge-Kutta [14-19, 27] method with any given launch parameters in an inverse square gravitational field. b. Newton’s Method of Solving Non-Linear Equations and Kantorovich’s Theorem forGuaranteeing Quadratic Convergence.

[0663] As can be seen in Fig. 4A - Fig. 9 A all the functions generated by the rocket equation for constant acceleration or for the inverse square gravitational field modification made in this work are non-linear functions in / If. The bulk of the engineering problem in rocketry is to design a rocket that can reach a given altitude or an orbit with desired orbital parameters and a given payload. This requires the solution of several non-linear equations. The task is to solve the nonlinear equations0 for the constant acceleration assumption, or hGIVEN— h.INV2(jlp T2W^ = 0 for the inverse square gravitational field assumption for a given T2W. Since the rocket design requires optimization, this task may be done many times.

[0664] Conventionally, the solutions are accomplished by graphical methods as explained above. The graphical solution method can be implemented into a computer program by generating n equally spaced sample points in the interval of PMIN PM AX calculating h^f, T2W) or hjNV2^Pf» 7*21V) functions at each sample point k, where 1 < k < n. Then the interval is found within hGIVENresides given as, / l[gy(k), T2IV] < hGjVEN< / i[μf(k + 1), T21V] for constant acceleration assumption or hINV2\.Pf(K)» 7'21V] < H-GIVEN — "*. 1)»T2W"} if the inverse square gravitational field assumption is employed. The only condition that needs to be satisfied is T2W) > hG1VEN> H(J1MAX» T2 IV) for constant acceleration assumption or h.1NV2(PMiN» 7*21V) > hGIVEN> hINV2(PMAX» T2W) if the inverse square gravitational field assumption is employed. For constant acceleration assumption PMIN PM AX816PMIN=6 and PM AX=1,and for the inverse square gravitational field assumption PMIN=PASYMCJ2W^ + 8 and PM AX=7 where 8 is a small enough value to satisfy the given simple conditions as explained above, as well as in Section 6.

[0665] Once the interval k, where 1 < k < n, and where hG]VENresides is found, linear interpolation, quadratic, or a cubic spline fit can be performed to get a better approximation of the solution. In this work Newton’s method is employed, which gives quadratic convergence if certain conditions are satisfied following the interpolation.

[0666] Newton’s method is the most widely used non-linear equation solution method and it requires an initial approximation. Also known as Newton-Raphson method, it is based onTaylor’s expansion of a non-linear function around an initial approximation. Newton used the method to solve a third order equation, a single non-linear equation, like in this work. Over the years Newton’s method was generalized for systems of equations as well. If the initial approximation meets certain criteria, it will converge quadratically, if not it can have a slow rate of convergence, or in some cases, it can even diverge. Like in any iterative method a good initial approximation can eliminate problems related to convergence.667] Since every non-linear problem is unique, before going into Newton’s method it is necessary to go over the initial approximation methodology employed for the solution of the non-linear equations faced in this work. The methodology is based on a computer program adaptation of the geometrical method of solving non-linear equations. The computer algorithm for solving the equation geometricallyis as follows: c. Find where h(jiMINlhavEN • THSisdone by solving the asymptote equation (6.7) with Newton’s method to get which is the X intercept of the asymptote along with an arbitrarily definedsmall enough 6.

[0668] It) Equally divide the region starting from PMIN to UMAX with 71 sample points, having the asymptote for thiscurve is at This makes it clear thatSimply a small enough 8 can be found that gives h(jlMiN> 7*2140 > hGIVEN,which is an important condition to satisfy. Since there is no computational value issue in calculation of , establishing a value for UMAX is morestraightforward, therefore settingdoes not create a computational issue.

[0669] til) Calculateat each sampling point, having 7l(l) =

[0670] iv) Find the interval k, where hG1VENresides with

[0671] Apply linear interpolation to calculate the initial approximation of μfat h-GIVEN noted as

[0672] The algorithm for solving the equation 0.

[0673] The only difference from the earlier algorithm given is in equally dividing the region starting from PMIN to PM AX with n sample points, having μf(l) = PMIN and The importance of finding a PMIN? Just asmall 8 larger thengivinghas been pointed out above in theexplanation of Fig. 7F, indicating that thefunction becomes non-physical for the interval

[0674] As can be seen having a larger number of sampling points gives a more accurate initial approximation at any point.

[0675] Fig. 9 A depicts climb altitude and inverse square gravitationalfield functions for T2W = 1.1 and 4 constructed over 400 uniformly spaced discretesampling points. The horizontal lines represent hGIVEN= 40, 400, and 20,200km altitudes, where the intersection points of these lines with the functions becomes the geometric solution to thecurves become indistinguishable. The IntersectionRegions between the functions and hGiVENlines are circled as IR 1, IR 2, and IR 3.

[0676] Fig. 9B depicts the functions plotted in Fig. 9 A constructed with 40 discrete sampling points marking the curves to show a numerical algorithm for implementing the geometrical solution method in a computer program based on interpolation between the discrete sampling points.

[0677] Fig. 9C depicts a detailed view of the IR 2 region showing the discrete sampling points in the neighborhood of intersection withGIVENm of the

[0678] functions. Fig. 10Adepicts a detailed view of the IR 1 and IR 2 regions showing discrete sampling points in the neighborhood of intersection between hGIVEN= 40 km and 400km for thefunctions.

[0679] Fig. 10B depicts a detailed view showing discrete sampling points in the neighborhood of the intersection region IR 1 between hGlVEN= 20,200km with the functions. In the left lower comer, asmall portion of the function is in the view. The question of finding simplesufficient conditions for the convergence of Newton’s method to a solution of a non-linear system was considered a difficult problem of numerical analysis until L. V. Kantorovich published a theorem in 1937 which guarantees the convergence of Newton’s method under very general circumstances, without even assuming the existence of a solution [26- 30], Before getting into Kantorovich theorem, it is useful to start with the Newton’s method, to prevent any confusion due to the notation [25, 27-30],

[0680] In Newton’s method for systems of non-linear equations, let the n equations,

[0681]

[0682] For the n unknowns y1(y2, • • ynare written in vector form as,

[0683] (9-2)

[0684] denote the matrix with elements,

[0685]

[0686] If the vector Y = F (0) is an initial approximation to a solution of the system (9.2) and if the matrix A [F^] is nonsingular, one may hope that the vector,

[0687]

[0688] Obtained by linearizing the system (9.2) at Y = is a better approximation to the solution. If the matrices involved continue to be nonsingular, one may hope toobtain a sequence of successively better approximations Y by thealgorithm:

[0689]

[0690] The vector and matrix norms used in Kantorovich’s theorem and throughout are defined as follows: Let V be a vector and

[0691]

[0692] and A be a matrix,

[0693]

[0694] In Kantorovich’s theorem, assume that the following conditions are satisfied: (i)the initial approximation, the matrix has an inverseand an estimate of its norm,

[0695]

[0696] (ii) The vector approximately satisfies the system of equations (9.2) in the sense that

[0697]

[0698] (ii) In the region defined by inequality (9.12) the vector is twicecontinuously differentiable with respect to components of Y and satisfies

[0699] y

[0700] (iv) The constants and K introduced above satisfy the inequality,

[0701]

[0702] Then the system of equations (9.2) has a solution which is in the cube,

[0703]

[0704] Moreover, successive approximations of Kvdefined by (9.5) exist and converge to y and the rate of convergence may be estimated by the inequality,

[0705]

[0706] From the application point of view, the most difficult problem is to evaluate the Boparameter in (9.8) for large systems. 7]0is the correction vector’s maximum magnitude norm for the first iteration and can be calculated very easily whenever Newton’s method is applied. K is not a difficult parameter to evaluate, because typically each equation is not a function of a large number of unknowns in any discrete variable method. Therefore, Bois the parameter that makes the Kantorovich’s theorem difficult to apply for large systems of nonlinear equations.

[0707] Kantorovich’s theorem can be applied to a single non-linear equation. Since we are only interested in finding sufficient condition for quadratic convergence for a single nonlinear equation, the Kantorovich’s theorem can be applied with no difficulty. Let the only non- linear equation that needs to be solved be represented as,

[0708]

[0709] where it’s first derivative with respect to X is,

[0710]

[0711] The second derivative of (9.14) with respect to X is,

[0712]

[0713] Let X(0) be the initial approximation to the solution and

[0714]

[0715] Similarly, (9.18) becomes the first correction in Newton’s method given as,

[0716]

[0717] Finally (9.10) takes the form of,

[0718]

[0719] Resulting in the Kantorovich’s quadratic convergence criterion for a single nonlinear equation becoming,

[0720] d. Application of the Kantorovich Quadratic Convergence Criteria to the Solution for the Rocket Equation Related Problems

[0721] As explained in detail above there are three non-linear equations that need to be solved in the rocket related problems. The first one is finding the asymptote X intercept for the inverse square gravitational field formulation given in this work for the rocket equation. This is represented as μASY(T2W), which is the solution of (6.7) explicitly given by (6.8), and which can be written in short with the introduction of a constant P,

[0722] where

[0723] The second non-linear equation is theequation for constant gravitational acceleration and the third one isthe inverse square gravitational field formulation given in this work for the rocket equation.

[0724] The goal is to find an initial approximation of the solution for every non-linear equation encountered in this work which is being solved to satisfy Kantorovich’s quadratic convergence criterion.

[0725] Finding an initial approximation which satisfies the Kantorovich Quadratic Convergence Criterion for solving μASY(T2W) from

[0726]

[0727] The solution of (9.21) corresponds to escape velocity from earth at an altitude of hB. Even at μf= 0 / 1B, as can be seen in Fig. 5A and Fig. 5B, which gives the maximum possible value of hBfor any T2W, it is small compared to the earth radius rEARTH• Therefore, the initial approximation of VBcan be closely approximated with escape velocity from the Earth as,

[0728]

[0729] Relation (5.40) which gives the V^can also be approximated as,

[0730]

[0731] for small p^ « 1 as it should be close to zero, where the asymptotes are located which gives an

[0732]

[0733] Using (9.24) as initial approximation for solving (9.21) satisfies Kantorovich quadratic convergence criterion for solving PASY(T2W) for any T2W.

[0734] For the other two equations the region starting from PMIN to PMAX with 71 sample points, having μf-(l) = μMINaud μf(n ) = μMAXis used to find the interval where the solution resides, as it is explained above. The solution is approximated by linear interpolation in the interval and the value is calculated. If h0> 0.5 theinterval is halved until h.Q < 0.5, before going into Newton’s method.

[0735] Fig. IOC shows the convergence properties of Newton’s method employed in solving rocket equations. As can be seen, the magnitude of the correction terms reduces quadratically with increasing iterations. These very small correction numbers, or in other words, the accuracy in the solution cannot be achieved with interpolation, cubic spline fits, or any other interpolation or search methods. This level of accuracy may potentially be seen as academic, as derived from mathematics. Basically, employing Kantorovich theorem, since the initial approximation to the solution satisfies the quadratic convergence criterion of Newton’s method, the solution is obtained in 3-4 iterations.

[0736] Since Newton’s method requires only the first derivatives of the related non-linear functions and Kantorovich criterion also needs their second derivatives, they both must be calculated at any given μfbetween PMIN to μM AXfor this method to be applied. Analytical derivation of the first and second derivatives can be very cumbersome as can be seen in Sections10 - 12. A discrete value representation of the first and second derivatives becomes a lot simpler way of evaluating this method as

[0737]

[0738]

[0739] where k represents the interval, and y and h. are the function values at the sampling points k, k - 1, and k + 1 and the uniform spacing between the adjacent sampling points respectively [22-30],e. Fast Algorithm for Approximating the Propellant Mass and Rocket DesignParameters for an Orbital Launch in “Rocket Designer”

[0740] Solving Tsiolkovsky’s Rocket Equation derived for both gravitational field representations gives the propellant mass and the initial thrust needed with a given thrust to weight ratio at the pad to launch a space craft to a given altitude for a vertical launch. The orbital launch trajectory is never vertical, it is curved and therefore the solution of theTsiolkovsky’s Rocket Equation is more an approximation.

[0741] Since relation (1.1) states that the energy of an orbiting spacecraft at a radiusTORBIT isequal to the spacecraft launched to r0= 2 ■ T0RB1T, solving Tsiolkovsky’s RocketEquation for r0will give the “minimum” propellant mass and thrust needed to put the spacecraft into orbit with better accuracy compared to prior approximations.Fig. 13F to Fig. 130 give 4thorder Runge-Kutta numerical solutions of the equation of motion results for various launch conditions to illustrate the disclosed method with real and popular examples. In Fig. 13F, 13H, 13L and 13M the rocket parameters like thrust, total and propellant mass, are calculated from very primitive physical and design parameters like given exhaust gas velocity Vg, T2W ratios and for k=0.01 launching a payload of 1,000 [kg] using Tsiolkovsky’sRocket Equation for inverse square law modification explained in this disclosure. As can be seen, even for launch altitudes many times larger than Earth’s radius, reaching 50,000 [km]Tsiolkovsky’s Rocket Equation with inverse square law modification gives excellent match withRunge-Kutta numerical simulations.

[0742] Fig. 13F and Fig. 13H depicts a strait launch of the payload to a given altitude, not an orbital launch!

[0743] Fig. 13G depicts propellant and total mass as a function of time for launching1,000 [kg] of payload to 40 [km] altitude for T2W=1.1 and 4, k = 0.01. As can be seen, since the bum rate is constant in all simulations showing the propellant and total mass versus time plots decreases linearly with time and is a function of T2W. In all other propellant and total mass as afunction of time plots, the x axes of the plots are in logarithmic scale to make comparison between cases easier.

[0744] The Analytical Calculation of the First Derivatives of the Functions Derived in the Rocket Equation under Constant Acceleration Inverse Square Law GravitationalField.

[0745] There are four functions of interest namely hB(μf), h.c(μf) and that all are functions of μfand that need to be differentiated with respect to in applying Newton’s method for the solution. The VB(μf) relation given in (5.40) has two terms in the square bracket with each a function of μf. The first term in the bracket of the VBexpression given at (5.40) is,

[0746]

[0747] Applying the basic differentiation rule of,

[0748]

[0749] to (10.1) with the following variable transformation,

[0750]

[0751] Along,

[0752]

[0753] Resulting in,

[0754]

[0755] The differentiation of the second term in the bracket given in (5.40) is straightforward and need not be shown, resulting in the final differentiation being,

[0756]

[0757] As can be seen there are four terms in the bracket of (5.59) that give All fourterms must be differentiated with respect to and summed up. The first and third terms in thebracket of (5.59) are 1 and and their derivatives with respect to are 0 and -1 respectively.The second term in the bracket of (5.59) can be differentiated by “differentiation of a product” rule giving,

[0758]

[0759]

[0760]

[0761] The last term in (5.57) can be differentiated with respect to as,

[0762]

[0763] Finally, the sum of the four differentials gives,

[0764]

[0765] Differentiation of hcas given in (5.65) is straightforward giving,

[0766]

[0767] Finally, differentiation of h as given in (5.66) can be written as the sums of(10.10) and (10.11) giving,

[0768]

[0769] Derivatives for the inverse square law height hINVand radius rhINVwith respect to μf- are more complex, but still analytically available.

[0770] The Analytical Calculation of the First Derivatives of the Functions derived in the Rocket Equation under the Inverse Square Law Gravitational Field.

[0771] Next, the first derivatives of the functions derived in the rocket equation under inverse square law gravitational field can be found. Re-writing (6.6) in short form and showing the explicit μfdependency is,

[0772]

[0773] Where β is,

[0774]

[0775] The expression rhlNV(μf) in (11.1) is a function of 2 variables, where both variables are functions of μf, explicitly given at (4.40) and (4.59), where the relation between rBand hBis given in (5.52).

[0776]

[0777] Using the Leibnitz’s chain rule for 2 variables VBand VBgreatly reduces the complexity of differentiation with respect to Pp compared to explicitly writing the rhlNV(μf) by substituting the open forms TB(μf) and VB(μf) in (11.3) and differentiating it. Leibnitz’s chain rule for 2 variables rBand VBcan be written as,

[0778]

[0779] Since derivatives of VB / lBwith respect to μfare already given at (10.6) and(10.10), and the derivative of hBis equal to the derivative of rBwith respect to μf- as given in (6.3), the partial derivatives at (11.4) can be calculated with standard differentiation rules for divisional functions.

[0780] The dominator of (11.1) and its derivative with respect to rBare,

[0781]

[0782]

[0783] the denominator of (11.1) P and its derivative with respect to rBare,

[0784]

[0785]

[0786] Using the standard differentiation rule for a division gives,

[0787]

[0788] Explicitly (11.9) is,

[0789]

[0790] Finally resulting in,

[0791]

[0792] Similarly, derivatives of the dominator and denominator with respect to Vg are,

[0793]

[0794]

[0795]

[0796] Explicitly (11.14) becomes,

[0797]

[0798] Applying the chain rule (11.4) explicitly gives,

[0799]

[0800] where the derivatives of VBand hBare given in (10.6) and (10.10) respectively and substituting the relation (6.3) in (11.16) gives,

[0801]

[0802] where every term in (11.18) is an explicitly derived function of μf. f. The Analytical Calculation of the First Derivative Needed for Solving the Asymptote Location μASY in the Rocket Equation under the Inverse Square Law Gravitational Field.

[0803] For solving the μASY, where the asymptote is located, on the horizontal axes of rB(μf) curve giving the denominator of (11.17) noted as P2must be equated tozero, giving the equation to be solved as,

[0804]

[0805] To apply Newton’s method to solve (11.18) needed is the first derivative of,

[0806]

[0807] and must be evaluated with respect to μf. Applying Leibnitz’ s chain rule for 2 variables, this time the function to be differentiated with respect to μfis represented as P(rB.VB),

[0808]

[0809]

[0810]

[0811]

[0812] Giving Leibnitz’s chain rule for 2 variables as,

[0813]

[0814] where the derivatives of rBand V#are given in (10.6) and (10.10) respectively.Substituting the calculated partial derivatives in (11.23) becomes explicitly,

[0815]

[0816] Again, substituting the relation (5.3) in (11.24) gives,

[0817] y f

[0818] where every function and derivative in (11.25) is an explicitly derived function of

[0819] The Analytical Calculation of the Second Derivatives of the FunctionsDerived in the Rocket Equation under Constant Acceleration and the Inverse Square LawGravitational Field.

[0820] The second derivatives are important for applying the Kantorovich’s theorem which guarantees the quadratic convergence of the Newton’s method if it is met [26-28],

[0821]

[0822]

[0823]

[0824]

[0825] The second derivative is needed for solving the asymptote location in therocket equation under the inverse square law gravitational field. Differentiating (11.25) with respect to μf,

[0826]

[0827]

[0828]

[0829]

[0830]

[0831]

[0832] where every function and derivative in (12.8) and (12.10) is an explicitly derived function of jl^ previous to calculating (12.6).

[0833]

[0834] The open form of the second derivative of (11.18) with respect to μfcan be written as,

[0835]

[0836]

[0837]

[0838]

[0839]

[0840]

[0841]

[0842]

[0843]

[0844]

[0845] The differentiation in the parenthesis in (12.17) has 4 terms as multipliers which are all functions of therefore splitting it as,

[0846]

[0847]

[0848]

[0849]

[0850] All the needed terms in (12.13) are given and the second derivative of r^ycan be calculated summing up (12.15) and (12.19).

[0851] Rocket Assisted Descent “Orbital Descent” Program

[0852] The rocket assisted descent is controlled by 4 parameters. Before going into the parameters, it is useful to re-write the descent algorithm given in (1.1) as,

[0853]

[0854] The spacecraft or satellite is launched to a radius ofdefined by the scaling factor Sy, giving (13.1) always greater than There are an infinite number of trajectoriesfor rocket assisted descent from a radius of having the satellitevelocity vector magnitude equal to the orbital velocity VQRBIT and at the same time being tangent to the desired orbital trajectory.

[0855] The rocket assisted descent is controlled by thrust of the descent rocket F(see Figs. 20A-20C), the descent direction of the thrust referenced to radial direction aFD, bum time of the descent thruster and radius at which descent rocket is fired. As an example, thespacecraft or satellite can first descend straight down from Fhto a certain radius (Fig. 20B or20C) satisfying,

[0856]

[0857] During this straight descent towards earth the spacecraft or satellite, referred to herein as a payload, picks up speed only in radial direction towards the earth. Then, the rocket fires at a given direction referenced to radial direction symbolized as dp with a thrust of Fp and descent bum time TBD, such that after a distance it falls on the orbital trajectory with orbital velocity V ORBIT is also tangent to the desired orbital trajectory. Finding the most energy efficient set of 4 variables for a set of scaling factor Sf, is the challenging mathematical problem which is solved by the program Orbital Descent.

[0858] Fig. 15 is a flowchart illustrating a method for efficient orbital launch trajectories.The method is supported by the explanations and drawings described above. In some aspects, some the methods steps may be performed of the depicted order, or performed simultaneous with other steps. The method begins at Step 1500.

[0859] Step 1502 launches a payload as high as to a first radius (FQ) altitude, with respect to the center of tire Earth, which is associated with a first altitude defined with respect to sea level. One advantage to this method is that the velocity needed to acquire the first radius altitude is not limited to a specific range of values. Unlike conventional launch methods that typically require hypersonic speeds to acquire a stable orbit (see Fig. 14), the rocket in the method described herein can be launched at the lowest velocity needed to acquire the initial (e.g., first) radius, which reduces wear-and-tear and minimizes failed launches. In Step 1504 the payload attains a gravitational first potential energy at the first radius (FQ). In Step 1506 the payload altitude decreases in response to a gravitational pull of the Earth. In Step 1508 the payload attains a stable orbit around the Earth to a second radius (r0RBIT\ with respect to the center of the Earth, which is associated with a second altitude defined with respect to sea level.As used herein, the term “payload” is defined as the object (e.g., spacecraft, satellite, or satellite with attached rocket) that ultimately attains the stable orbit at the second radius altitude.

[0860] Fig. 16 is a drawing depicting the relationship between the first radius (Fo) 1600, second radius (F0BB / T) 1602, the payload 1604, and the Earth 1606. In one aspect the payload islaunched in Step 1502 to a first radius twice as high as the second radius, where the payload acquires a gravitational first potential energy. Then, in Step 1508 the payload acquires a total energy that is equal to the sum of its second potential energy and its kinetic energy (orbital velocity), which is also equal to the gravitational first potential energy.

[0861] Fig. 17 is a diagram depicting the relationships shown in Fig. 16, as referenced toEarth’s sea level, with the relationship expressed as:

[0862]

[0863] where ho is the first altitude associated with the first radius, defined with respect to sea level of Earth;

[0864] where TE is the Earth’s radius; and,

[0865] where h is the second altitude associated with the second radius, defined with respect to sea level.

[0866] Although the launches shown in the figures are depicted as vertical, practically that cannot be done for safety reasons. So even in the best case, Step 1502 may include launching the payload at a non-zero angle of attack with a kinetic energy that is greater than the gravitational first potential energy attained when the payload attains the first radius.

[0867] Fig. 18 is a plan view of the Earth centered on the polar axis. Attaining the stable orbit in Step 1508 includes potentially attaining a stable orbit at any orbital inclination angle in the range between 0 and 360 degrees. Exemplary inclination angles of 6\ and 62 are shown in the figure. Further, launching the payload in Step 1502 includes potentially launching the payload from any latitude on a surface of the Earth in the range between 90 and -90 degrees.Launch point 1 at Latitude 1 1800 and launch point 2 at latitude 2 1802 are shown.

[0868] In more detail, the payload attaining the gravitational first potential energy at the first radius in Step 1502 includes the gravitational first potential energy being expressed as:

[0869]

[0870] where G is Newton’s gravitational constant;

[0871] where 1TI is the mass of the payload; and.

[0872] where M is the mass of the Earth.

[0873] Similarly, attaining the stable orbit for the payload in Step 1508 includes the second potential energy being expressed as:

[0874]

[0875] As a result, if the first radius is twice the height of the second radius, the payload’s total energy being equal to the gravitational first potential energy in Step 1508 includes the relationship being expressed as:

[0876]

[0877] where VORBIT is the velocity of the payload in the stable orbit.

[0878] In another variation, Step 1502 launches the payload with a booster rocket using a first kinetic energy. Step 1503 separates the booster rocket from the payload at the first radius altitude. Then, the payload attaining the gravitational first potential energy of the payload at the first radius altitude additionally in Step 1504 includes the gravitation potential energy of the booster rocket being distinguishable from the payload’s gravitational first potential energy.

[0879] In one aspect subsequent to the payload attaining the first radius, Step 1505a initiates a rocket (payload) assisted descent maneuver. In a conventional gravity turn the rocket applies force, in the form of thrust, at least partially against the force of gravity. In contrast, in a rocket assisted descent maneuver as defined herein, the payload applies no force to counter the force of gravity, but rather, relies upon the force of gravity to send the payload into a stablelower altitude orbit. Thus, decreasing the payload altitude in response to the gravitation pull of the Earth in Step 1506 may include decreasing the payload altitude at a first angle of descent tangential to the stable orbit. The rocket assisted descent maneuver can be assisted by gimbaling the rocket engines, using auxiliary directional thrusters, or the use of adjustable aerodynamic surfaces (in an atmosphere). In one aspect, subsequent to the payload attaining the first radius, the performance of the rocket assisted descent maneuver may be accompanied with a payload roll maneuver as part of Step 1505b. Then, decreasing the payload altitude at the first angle of descent in Step 1506 includes the first angle of descent being a first angle of orbital inclination attained in response to the roll maneuver.

[0880] Fig. 19 is a diagram the relationship between (To= 2r0RB1T^ 1600, second radius (ToRBIT) 1602, the payload 1604, the Earth 1606, and (F^) 1900.

[0881] Figs. 20A through 20C are drawings depicting a velocity adjustment applied to the payload during or after the rocket assisted descent maneuver. In some aspects, launching the payload as high as the first radius altitude includes launching the payload to a third, typically lesser, radius (r^) altitude, where

[0882]

[0883] Subsequent to the rocket assisted descent maneuver in Step 1505a, Step 1505b applies a velocity adjustment force to the payload. So that decreasing the payload altitude at a first angle of descent in Step 1506 includes modifying a descending velocity of the payload in response to the velocity adjustment force. Note that when Sj = 2, then Fh=r0. At the third radius the payload has a total energy equal to its gravitational potential energy. Then, applying the velocity adjustment force to the payload in Step 1505b includes the applied velocity adjustment force also being responsive to the gravitation potential energy at the third radius.That is, the velocity force needed to attain a stable orbit is a calculation dependent upon the altitude at which the rocket assisted descent maneuver is performed.

[0884] In Fig. 20A the rocket assisted descent maneuver occurs at the point S at the third radius altitude. The velocity adjustment force is also applied at point S. Therefore, at point S the payload is subject to the force of gravity, as well as a force Fn orthogonal to the force of gravity in this example.

[0885] Figs. 20B and 20C depict the application of the velocity adjustment force after the payload is permitted to free-fall (only under the force of gravity). In this aspect, subsequent to the rocket assisted descent maneuver, Step 1505c permits the payload to fall, under only the force of gravity, to a fourth radius altitude ZD higher than the second radius altitude. Then, applying the velocity adjustment force to the payload in Step 1505b includes applying the velocity adjustment force when the payload reaches the fourth radius. Fig. 20C is essentially the same as Fig. 20B, showing a descent along a different inclination angle. The method is not limited to a particular range of inclination angles, as suffered by conventional launch methods. It should be understood that when third radius is a relatively great distance from the second radius, the velocity adjustment may be in the retro direction (against the direction of payload travel) on account of the payload’s gravitational potential. On the other hand, if the third radius is relatively close to the second radius, the velocity adjustment force is more likely to be in the direction of payload travel, as shown in the figures.

[0886] Fig. 21 is a flowchart illustrating a method for minimizing the energy required for an orbital launch. The method begins at Step 2100. Step 2102 launches a payload to at least a first radius, with respect to the center of the Earth, higher than a desired stable orbit at a second radius, defined with respect to the center of the Earth. In some aspects the payload can be launched to a radius that is even greater than the first radius. Advantageously, Step 2102 may launch the payload from any latitude on a surface of the Earth in the range between 90 and -90 degrees. Subsequent to the payload attaining the first altitude, in Step 2104 the altitude of payload decreases in response to the gravitational pull of the Earth. Then, in Step 2106 the payload attains a stable orbit around the Earth at the second radius. In one aspect, decreasing thealtitude of the payload in Step 2104 includes decreasing the altitude of the payload at a first inclination angle. Then, in Step 2106 the stable orbit is attained at the first inclination angle.The orbital inclination angle may be any inclination angle between zero and 360 degrees.

[0887] In one aspect, launching the payload in Step 2102 includes substeps. Step 2102a launches to a first radius 2 times the height of the second radius, and in Step 2102b the payload attains a gravitational first potential energy at the first radius. Then, attaining the stable orbit inStep 2106 includes the payload having a total energy that is the sum of the second potential energy at the second radius and the kinetic energy of the payload in the stable orbit. The payload total energy is also equal to the gravitational first potential energy.

[0888] In Step 2102b, where the payload attains the gravitational first potential energy at the first radius, the first gravitational potential energy may be expressed as:

[0889]

[0890] where G is Newton’s gravitational constant;

[0891] where TH is the mass of the payload;

[0892] where M is the mass of the Earth; and,

[0893] 7*0 is a first radius.

[0894] Similarly, attaining the stable orbit in Step 2106 includes the second potential energy being expressed as:

[0895]

[0896] where T ORBIT is a second radius.

[0897] As a result, if the first radius is twice the height of the second radius, the payload total energy at the second radius altitude in Step 2106 includes the relationship being expressed as:

[0898]

[0899] where V ORBIT is the velocity of the payload in the stable orbit.

[0900] In one aspect, decreasing the altitude of payload in response to the gravitational pull of the Earth in Step 2104 includes the following substeps. Step 2104a performs a rocket (or payload) assisted descent maneuver. Step 2104b causes the payload to descend along a first angle of descent tangential to the stable orbit.

[0901] In another aspect, launching the payload in Step 2102 includes launching the payload to a third radius (F^), where

[0902]

[0903] Subsequent to the rocket assisted descent maneuver in Step 2104a, Step 2104c applies a velocity adjustment force to the payload. Then, decreasing the payload altitude at a first angle of descent in Step 2104b includes the first angle of descent being responsive to the velocity adjustment force and the gravitational potential energy of the payload at the third radiusF^. Step 2104c may also include a roll maneuver.

[0904] Fig. 22 is a diagram depicting a variety of exemplary stable orbits that can be obtained using the disclosed method. Shown are both circular and elliptical orbits. The orbits are created by controlling the angle of descent created during rocket assisted descent maneuvers, as well as making adjustments to the velocity during descent.

[0905] Methods have been provided for the efficient launch of orbital satellites and spacecraft. Examples of particular method steps and hardware units have been presented to illustrate the invention. However, the invention is not limited to merely these examples. Other variations and embodiments of the invention will occur to those skilled in the art.

[0906] References on Rocket Equation and Space

[0907] “Lecture L14 - Variable Mass Systems: The Rocket Engine”, J. Peraire, S.Widnal, 16.07 Dynamics, Fall 2008, Version 2.0.

[0908] “Lecture 14.2. The Rocket Equation”, https: / / web.mit. edu / 16.unified / www / SPRING / DroDulsion / notes / nodel03.html

[0909] “Tsiolkovsky Rocket Equation”,

[0910] https: / / en.wikipedia.org / wiki / Tsiolkowskv rocket equation.

[0911] 4. “Free Fall”, https: / / en.wikiDedia.ore / wiki / Free fall

[0912] 5. “From Moon-Fall to Motions Under Inverse Square Laws”, S. K. Foong, EuropeanJournal of Physics, 2008, Vol 29, Number 5, pp. 987-1003.

[0913] 6. Everydayastraunat.com, Tim Dodd

[0914] 7. “Rocket Factory”, Smarter everyday you tube program by, Destin Sandlin with ToryBruno, CEO of ULA ( United Launch Alliance),

[0915] https: / / www.voutube.com / watch?v=DOaPOIOLEUo

[0916] 8. “Launch Azimuth, Orbital Inclination, Launch Site Latitude Rocket Trajectories 6”,Alfanso Gonzales, https: / / www.voutube.com / watch?v=UIXiK41vl w

[0917] 9. Lecture 14: “Orbits and Escape Velocity” 8.01 Classical Mechanics, Fall 1999,Walter Lewin https: / / www.voutube.com / watch7wCW353dacbtA

[0918] 10. “Gravity Turn”, https: / / en.wikiDedia.org / wiki / Gravitv turn

[0919] 11. ULA’s Transformation and Vision for the Future of Space Launch”, Tory Bruno,CEO of ULA,

[0920] https : / / www. youtube. com / watch?v=6fpwd6K6L Y

[0921] References on Guidance and Equation of Motion

[0922] 12., “Solar Timer Using GPS Technology”, O. E. Akcasu and Ibrahim Akcay, USPatent 9,551,980 B2, January 24, 2017.

[0923] “Method and Apparatus for Controlling Electrical Power Usage Based on exact SunElevation Angle and Measured Geographical Location”, Osman E. Akcasu, USPTO 9,949,339B2, April 17, 2018.

[0924] “Vehicle for Aerial Delivery of Fire Retardant”, O. E. Akcasu, US Patent Number7,975,774 B2, July 12, 2011.

[0925] “System for Aerial Delivery of Fire Retardant”, O. E. Akcasu, US Patent Number8,165,731 B2, April 24, 2012.

[0926] “System for Aerial Delivery of Fire Retardant", O. E. Akcasu, European Union PatentNumber EP 2,163,279, February 25, 2013.

[0927] “Vehicle for Aerial Delivery of Fire Retardant”, O. E. Akcasu, European Union PatentNumber EP 2,163,844 Bl, August 12, 2015.

[0928] “Rocket or Artillery Launched Smart Reconnaissance Pod”, O. E. Akcasu, IbrahimOnur Uslu and Yusuf Leblebici, US Patent 9,234,728 B2, January 12, 2016.

[0929] “Rocket or Artillery Launched Smart Reconnaissance Pod”, O. E. Akcasu, IbrahimOnur Uslu and Yusuf Leblebici, European Patent EP 2871 438 B 1, 11 / 4 / 2018.

[0930] “JPL and the Space Age: Hunt for Space Rocks". https: / / www.voutube.com / watch7vMwNzTvu36WA

[0931] References on Math and Physics in General

[0932] “Gravitational Potential Energy”, https: / / en.wikipedia.org / wiki / Gravitational energy

[0933] “Capacitor with High Aspect Ratio Silicon Cores”, O. E. Akcasu, US 10,510,828 B2,December 17, 2019.

[0934] “The Feynman Lectures on Physics”, Richard P. Feynman, Robert B. Leighton,Matthew L. and Sands, Copyright 1963, 1989 California Institute of Technology, ISBN 0-201-51003-0, Vol. n, pp.7.9.

[0935] “Mathematical Handbook of Formulas and Tables,” Murray R. Spiegel, Schaum’sOutline Series, Copyright 1952 by McGraw-Hill, Inc.

[0936] “Handbook of Mathematical Functions,” Edited by Milton Abramowitz and Irene A.Stegun, Dover Publications, Inc., New York, 1972, Library of Congress Catalog Card Number:65-12253.

[0937] “Handbook of Chemistry and Physics,” 66* Edition, 1985-1986, Copyright CRCChemical Rubber Publishing Company, ISBN-0-8493-0466-0.

[0938] “Applied Numerical Analysis,” C.F. Gerald, Addison-Wesley Publishing Co., 1980,ISBN 0-201-02696-1.

[0939] “Functional Analysis and Applied Mathematics”, L. V. Kantorovich, Usepkhi Mat.Nauk., Vol. 3, pp. 89-185, 1948.

[0940] “Discrete Variable Methods in Ordinary Differential Equations”, P. Henrici, Wiley,New York, 1962.

[0941] "Convergence Properties of Newton's Method for the Solution of SemiconductorCarrier Transport Equations and Hybrid Solution Techniques for Multidimensional Simulation ofVLSI Devices", O. E. Akcasu, Solid-State Electron. Vol. 27, pp. 319-328, April 1984.

[0942] “Brachistochrone Curve”, https: / / en. wikiDedia.org / wiki / Brachistochrone curve

[0943] “Brachistochrone Curve” httos: / / www.voutube.com / watch?v=3HXCv4dmR7A

[0944] “Brachistochrone Curve” httDs: / / www.voutube.com / watch?v=zYOAUG8PxvM&t=2s

[0945] I claim:

Claims

CLAIMS1. A method for efficient orbital launch trajectories, the method comprising: launching a payload, tangential to, and as high as a first radius heightdefined with respect to the center of the Earth, associated with a first altitude, defined with respect to sea level; the payload attaining a gravitational first potential energy at the first radius height; decreasing the payload height in response to a gravitational pull of the Earth; and, the payload attaining a stable orbit around the Earth at a second radius height(TORBIT\ defined with respect to the center of the Earth, associated with a second altitude, defined with respect to sea level.

2. The method of claim 1 wherein launching the payload includes launching the payload to a first radius height equal to 2 times the height of the second radius height.

3. The method of claim 2 wherein the payload attaining the stable orbit includes the payload having a total energy that is the sum of a gravitational second potential energy and a second kinetic energy, with the payload total energy being equal to the gravitational first potential energy.

4. The method of claim 1 wherein the payload attaining the stable orbit includes attaining a stable orbit at any orbital inclination angle in the range between 0 and 360 degrees.

5. The method of claim 1 wherein launching the payload includes launching the payload from any latitude on a surface of the Earth in the range between 90 and -90 degrees.

6. The method of claim 3 wherein the payload attaining the gravitational first potential energy at the first radius height includes the gravitational first potential energy being expressed as:where G is Newton’s gravitational constant; where m is the mass of the payload; and. where M is the mass of the Earth.

7. The method of claim 6 wherein attaining the stable orbit includes the gravitational second potential energy being expressed as:

8. The method of claim 7 wherein the payload attaining the stable orbit and having the total energy equal to the gravitational first potential energy includes the relationship being expressed as:where VORBIT is the velocity of the payload in the stable orbit.

9. The method of claim 8 further comprising:subsequent to the payload attaining the first radius height, initiating a rocket(payload) assisted descent maneuver; wherein decreasing the payload height in response to the gravitational pull of theEarth includes decreasing the payload height at a first angle of descent tangential to the stable orbit.

10. The method of claim 9 further comprising: subsequent to the payload attaining the first radius height, performing a payload roll maneuver; and, wherein decreasing the payload height at the first angle of descent includes the first angle of descent being in a first angle of orbital inclination responsive to the roll maneuver.

11. The method of claim 9 wherein launching the payload as high as the first radius height includes launching the payload tangential to a third radius height (7^) defined with respect to the center of the Earth, wherethe method further comprising: subsequent to the rocket assisted descent maneuver, applying a velocity adjustment force to the pay load; and, wherein decreasing the payload height at the first angle of descent includes modifying a descending velocity of the payload in response to the velocity adjustment force.

12. The method of claim 11 wherein the payload attaining the third radius height includes the payload having a total energy at the third radius height equal to its gravitational potential energy; and,wherein applying the velocity adjustment force to the payload includes the applied velocity adjustment force being responsive to the gravitation potential energy at the third radius height.

13. The method of claim 12 further comprising: subsequent to the rocket assisted descent maneuver, permitting the payload to fall, under only the force of gravity, to a fourth radius height, defined with respect to the center of theEarth, higher than the second radius height; and, wherein applying a velocity adjustment force to the payload includes applying the velocity adjustment force as the payload approaches the fourth radius height.

14. A method for efficient orbital launch trajectories, the method comprising: launching a payload from a surface of the Earth; the payload attaining a non-orbiting first radius heightdefined with respect to the center of the Earth; the payload attaining a gravitational first potential energy and a first kinetic energy at the first radius height, where the first gravitational energy is greater than the first kinetic energy; decreasing the payload height in response to the gravitational first potential energy; performing a payload roll maneuver; creating a first angle of descent responsive to the roll maneuver; and, the payload attaining a stable orbit around the Earth at a second radius heightOORBZTX defined with respect to the center of the Earth.

15. The method of claim 14 wherein the payload attaining the first radius height includes creating a relationship between the first radius height and second radius height expressed as:

16. The method of claim 15 wherein the payload gravitational first potential energy at the first radius height is expressed as:wherewhere VORBIT is the velocity of the payload in the stable orbit; where G is Newton’s gravitational constant; where m is the mass of the payload; and. where M is the mass of the Earth.

17. A method for obtaining efficient orbital launch trajectories, the method comprising: for a payload, identifying a first radius heightwith respect to the center of the Earth; identifying a second radius height (r0BBIT) with respect to the center of the Earth, less than the first radius height; identifying a first height difference between the first radius height and the second radius height;launching the payload tangential to the first radius height from a surface of the Earth; using an Orbital Descent software program, identifying a direction aF, thrust , and descent burn time TBDfrom the first radius height; decreasing the payload height in response to the first height difference, direction thrust FD, and descent bum time TBD; and, the payload obtaining a stable orbital velocity at the second radius height.

18. The method of claim 17 wherein the first height difference is expressed with a scaling factor Sf, where:

19. The method of claim 18 further comprising: the payload attaining a gravitational first potential energy at the first radius height expressed as:where Sf = 2; where VORBIT is the velocity of the payload in the stable orbit; where G is Newton’s gravitational constant; where in is the mass of the payload; and. where M is the mass of the Earth.

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