16.07 Dynamics Fall 2004 Version 1.1 Lecture d30- Orbit Transfers In this lecture, we will consider how to transfer from one orbit, or trajectory, to another. One of the ssumptions that we shall make is that the velocity changes of the spacecraft, due to the propulsive effects occur instantaneously. Although it obviously takes some time for the spacecraft to accelerate to the velocity of the new orbit, this assumption is reasonable when the burn time of the rocket is much smaller than the period of the orbit. In such cases, the Au required to do the maneuver is simply the difference between the relocity of the final orbit minus the velocity of the initial orbit When the initial and final orbits intersect, the transfer can be accomplished with a singe impulse. For more general cases, multiple impulses and intermediate transfer orbits may be required Given initial and final orbits, the objective is generally to perform the transfer with a minimum Au. In some situations, however, the time needed to complete the transfer may also be an important consideration. lost orbit transfers will require a change in the orbit's total specific energy, E. Let us consider the change in total energy obtained by an instantaneous impulse Av. If vi is the initial velocity, the final velocity, uf, will simply be Uf=v1+△U If we now look at the magnitude of these vectors we have v=2+△n2+2v;△cos, where B is the angle between vi and Au. The energy change will be △E==△n2+v;△ U cos B From this expression, we conclude that, for a given Au, the change in energy will be largest when U; and Au are co-linear(B=0), and For example, to transfer a satellite on an elliptical orbit to an escape trajectory, the most energy efficient impulse would be co-linear with the velocity and applied at the instant when the satellite is at the elliptical
J. Peraire 16.07 Dynamics Fall 2004 Version 1.1 Lecture D30 - Orbit Transfers In this lecture, we will consider how to transfer from one orbit, or trajectory, to another. One of the assumptions that we shall make is that the velocity changes of the spacecraft, due to the propulsive effects, occur instantaneously. Although it obviously takes some time for the spacecraft to accelerate to the velocity of the new orbit, this assumption is reasonable when the burn time of the rocket is much smaller than the period of the orbit. In such cases, the ∆v required to do the maneuver is simply the difference between the velocity of the final orbit minus the velocity of the initial orbit. When the initial and final orbits intersect, the transfer can be accomplished with a singe impulse. For more general cases, multiple impulses and intermediate transfer orbits may be required. Given initial and final orbits, the objective is generally to perform the transfer with a minimum ∆v. In some situations, however, the time needed to complete the transfer may also be an important consideration. Most orbit transfers will require a change in the orbit’s total specific energy, E. Let us consider the change in total energy obtained by an instantaneous impulse ∆v. If vi is the initial velocity, the final velocity, vf , will simply be, vf = vi + ∆v . If we now look at the magnitude of these vectors, we have, v 2 f = v 2 i + ∆v 2 + 2vi∆v cos β, where β is the angle between vi and ∆v. The energy change will be ∆E = 1 2 ∆v 2 + vi∆v cos β . From this expression, we conclude that, for a given ∆v, the change in energy will be largest when: - vi and ∆v are co-linear (β = 0), and, - vi is maximum. For example, to transfer a satellite on an elliptical orbit to an escape trajectory, the most energy efficient impulse would be co-linear with the velocity and applied at the instant when the satellite is at the elliptical 1
orbit's perigee, since at that point, the velocity is maximum. Of course, for many required maneuvers, the applied impulses are such that they cannot satisfy one or both of the above conditions. For instance, firing at the perigee in the previous example may cause the satellite to escape in a particular direction which may not be the requi Hohmann Transfer A Hohmann Transfer is a two-impulse elliptical transfer between two co-planar circular orbits. The transfer itself consists of an elliptical orbit with a perigee at the inner orbit and an apogee at the outer orbit It turns out that this transfer is usually optimal, as it requires the minimum AuT=Av+Ava to perform a transfer between two circular orbits. The exception for which Hohmann transfers are not optimal is for very large ratios of r2/r1, as discussed below The transfer orbit has a semi-major axis, a, which is T1+r2 Hence, the energy of the transfer orbit is greater than the energy of the inner orbit (a= r1), and smaller than the energy of the outer orbit (a= r2). The velocities of the transfer orbit at perigee and apogee are iven, from the conservation of energy equation, as n2=9(n-n1+n2 2 2 To +T2 Here, we have assumed that the orbits are around the earth and have replaced u by gR, where g is the acceleration due to gravity on the earth's surface and R is the radius of the earth
orbit’s perigee, since at that point, the velocity is maximum. Of course, for many required maneuvers, the applied impulses are such that they cannot satisfy one or both of the above conditions. For instance, firing at the perigee in the previous example may cause the satellite to escape in a particular direction which may not be the required one. Hohmann Transfer A Hohmann Transfer is a two-impulse elliptical transfer between two co-planar circular orbits. The transfer itself consists of an elliptical orbit with a perigee at the inner orbit and an apogee at the outer orbit. It turns out that this transfer is usually optimal, as it requires the minimum ∆vT = |∆vπ|+|∆vα| to perform a transfer between two circular orbits. The exception for which Hohmann transfers are not optimal is for very large ratios of r2/r1, as discussed below. The transfer orbit has a semi-major axis, a, which is a = r1 + r2 2 . Hence, the energy of the transfer orbit is greater than the energy of the inner orbit (a = r1), and smaller than the energy of the outer orbit (a = r2). The velocities of the transfer orbit at perigee and apogee are given, from the conservation of energy equation, as v 2 π = gR2 2 r1 − 2 r1 + r2 (1) v 2 α = gR2 2 r2 − 2 r1 + r2 . (2) Here, we have assumed that the orbits are around the earth and have replaced µ by gR2 , where g is the acceleration due to gravity on the earth’s surface and R is the radius of the earth. 2
The velocities of the circular orbits are vcl=VgR/Ti and vc2=vgR2/T2.Hence, the required perigee and apogee are, When the initial orbit has a radius larger than the final orbit, the same strategy can be followed but in this case, negative impulses will be required, first at apogee and then at perigee to decelerate the satellite. Example Hohman transfer [11 A communication satellite was carried by the Space Shuttle into low earth orbit(LEO)at an altitude of 322 km and is to be transferred to a geostationary orbit(GEO)at 35, 860 km using a Hohmann transfer. The characteristics of the transfer ellipse and the total Av required, AuT, can be determined as follows For the inner orbit. we have R+d1=6.378×10°+322×10°=6.70×106m E1 -2.975×107m2/s2(J/kg) Similarly, for the outer circular orbit R+d2=42.24×10°m =3072m/s -4718×10°m2/s2(J/kg) r1+r2 =-8.144×10°m2/32(J/kg) which shows that El< E E2. The velocity of the elliptical transfer orbit at the perigee and apogee can be determined from equations 1 and 2, as 1,067m/s Since the velocity at the perigee is orthogonal to the position vector, the specific angular momentum of the h=n1vx=6.787×1010m2/s
The velocities of the circular orbits are vc1 = p gR2/r1 and vc2 = p gR2/r2. Hence, the required impulses at perigee and apogee are, ∆vπ = vπ − vc1 = R r g r1 r 2r2 r1 + r2 − 1 ∆vα = vc2 − vα = R r g r2 1 − r 2r1 r1 + r2 . When the initial orbit has a radius larger than the final orbit, the same strategy can be followed but in this case, negative impulses will be required, first at apogee and then at perigee, to decelerate the satellite. Example Hohman transfer [1] A communication satellite was carried by the Space Shuttle into low earth orbit (LEO) at an altitude of 322 km and is to be transferred to a geostationary orbit (GEO) at 35, 860 km using a Hohmann transfer. The characteristics of the transfer ellipse and the total ∆v required, ∆vT , can be determined as follows: For the inner orbit, we have, r1 = R + d1 = 6.378 × 106 + 322 × 106 = 6.70 × 106 m vc1 = r µ r1 = s gR2 r1 = 7713 m/s E1 = − µ 2r1 = − gR2 2r1 = −2.975 × 107 m2 /s 2 (J/kg). Similarly, for the outer circular orbit, r2 = R + d2 = 42.24 × 106 m vc2 = 3072 m/s E2 = −4.718 × 106 m2 /s 2 (J/kg). For the transfer trajectory, 2a = r1 + r2 = 48.94 × 106 m E = − µ 2a = −8.144 × 106 m2 /s 2 (J/kg), which shows that E1 < E < E2. The velocity of the elliptical transfer orbit at the perigee and apogee can be determined from equations 1 and 2, as, vπ = 10, 130 m/s , vα = 1, 067 m/s . Since the velocity at the perigee is orthogonal to the position vector, the specific angular momentum of the transfer orbit is, h = r1vπ = 6.787 × 1010 m2 /s , 3
and the eccentricity can be determined as 2Eh e=1/1+-2=0.7265 Finally, the impulses required are Urx-tc1=10,130-7713=2414m/s U2-ta=3072-1607=1465m/s The sign of the Au's indicates the direction of thrusting(whether the energy is to be increased or decreased) and the total Au is the sum of the magnitudes. Thus △vr=|4r|+|△ta|=2417+1465=3882m Since the transfer trajectory is one half of an ellipse, the time of fight(TOF) is simply half of the period TOF=丌 7×109)=19.050s=529 V3986×104 In order to illustrate the optimal nature of the Hohmann transfer, we consider now an alternative transfer in which we arbitrarily double the value of the semi-major axis of the Hohmann transfer ellipse, and find he characteristics and Aur of the resulting fast transfer. The semi-major axis of the transfer ellipse will be2a=98×10°m,andE=-p/(2a)=-4.067×10°m2/s2(J/kg). The velocity of the transfer orbit at departure will be 3.986×101 2(-4.067×106+ 6.70×106 10.530m/s, △vr=10,530-7713=2817m/s The specific angular momentum is and the orbits eccentricity, e, is 0.863 We can now calculate, from the energy conservation equation, the velocity of the transfer orbit at the point of interception with the outer orbit, vint 3.986×10 int 2(-4.067×10 2.24×106
and the eccentricity can be determined as, e = s 1 + 2Eh2 µ2 = 0.7265 . Finally, the impulses required are, ∆vπ = vπ − vc1 = 10, 130 − 7713 = 2414 m/s ∆vα = vc2 − vα = 3072 − 1607 = 1465 m/s, The sign of the ∆v’s indicates the direction of thrusting (whether the energy is to be increased or decreased) and the total ∆v is the sum of the magnitudes. Thus, ∆vT = |∆vπ| + |∆vα| = 2417 + 1465 = 3882 m/s. Since the transfer trajectory is one half of an ellipse, the time of flight (TOF) is simply half of the period, T OF = π r (24.47 × 106) 3 3.986 × 1014 = 19, 050 s = 5.29 h . In order to illustrate the optimal nature of the Hohmann transfer, we consider now an alternative transfer in which we arbitrarily double the value of the semi-major axis of the Hohmann transfer ellipse, and find the characteristics and ∆vT of the resulting fast transfer. The semi-major axis of the transfer ellipse will be 2a = 98 × 106 m, and E = −µ/(2a) = −4.067 × 106 m2/s2 (J/kg). The velocity of the transfer orbit at departure will be vπ = s 2 −4.067 × 106 + 3.986 × 1014 6.70 × 106 = 10.530 m/s, and, ∆vπ = 10, 530 − 7713 = 2817 m/s . The specific angular momentum is, h = 10, 530 × (6.70 × 106 ) = 7.055 × 1010 m 2 /s, and the orbit’s eccentricity, e, is 0.863. We can now calculate, from the energy conservation equation, the velocity of the transfer orbit at the point of interception with the outer orbit, vint, vint = s 2 −4.067 × 106 + 3.986 × 1014 42.24 × 106 = 3277 m/s . 4
Since the angular momentum, h, is conserved, we can determine the component of vint in the circumferential direction (Uint)e 1670m/s and the elevation angle, is thus o=cos 1(vint)e Finally, from geometrical considerations Auint=022+ui2nt-2vc2Uint cos o which yields Avint=3142 m/s, so that r=2817+3142=5959m/s. Comparing to the value of the Hohmann transfer Aur of 3875 m/s, we see that the fast transfer requires a △ ur which is54% higher It can be shown that when the separation between the inner and outer orbits is very large(r2>11.9r1 (a situation which rarely occurs), a three impulse transfer comprising of two ellipses can be more energy efficient than a two-impulse Hohmann transfer. This transfer is illustrated in the picture below. Notice that the distance from the origin at which the two transfer ellipses intersect is a free parameter, which can be determined to minimize the total Au. Notice also that the final impulse is a Au which opposes direction of motion, in order to decelerate from the large energy ellipse to the final circular orbit. Although this transfer may be more energy efficient relative to the two-impulse Hohmann transfer, it often involves much larger travel times. Try it!
Since the angular momentum, h, is conserved, we can determine the component of vint in the circumferential direction (vint)θ = h r2 = 1670 m/s and the elevation angle, φ, is thus, φ = cos−1 (vint)θ vint = 59.36o Finally, from geometrical considerations, ∆v 2 int = v 2 c2 + v 2 int − 2vc2vint cos φ, which yields ∆vint = 3142 m/s, so that ∆vT = 2817 + 3142 = 5959 m/s. Comparing to the value of the Hohmann transfer ∆vT of 3875 m/s, we see that the fast transfer requires a ∆vT which is 54% higher. It can be shown that when the separation between the inner and outer orbits is very large (r2 > 11.9r1) (a situation which rarely occurs), a three impulse transfer comprising of two ellipses can be more energy efficient than a two-impulse Hohmann transfer. This transfer is illustrated in the picture below. Notice that the distance from the origin at which the two transfer ellipses intersect is a free parameter, which can be determined to minimize the total ∆v. Notice also that the final impulse is a ∆v which opposes direction of motion, in order to decelerate from the large energy ellipse to the final circular orbit. Although this transfer may be more energy efficient relative to the two-impulse Hohmann transfer, it often involves much larger travel times. Try it! 5
Interplanetary Transfers When more than one planet is involved, the problem is no longer a two-body problem. Nevertheless, it is common(at least to get a good approximation)to decompose the problem into a series of two body problems Consider, for example, an interplanetary sfer in the solar system. For each planet we define the sphere of influence(SOI). Essentially, this is the region where the gravitational attraction due to the planet is larger than that of the sun The mission is broken into phases that are connected by patches where each patch is the solution of a two body problem. This is called the patched conic approach. Consider, for instance, a mission to Mars. The first phase will consist of a geocentric hyperbola as the spacecraft escapes from earth sol. The second phase rould start at the edge of the earths SOl, and would be an elliptical trajectory around the sun while the spacecraft travels to Mars. The third phase would start at the edge of Mars'sol, and would be a hyperbolic approach trajectory with the gravitational field of mars as the attracting force For long missions, the situation can become very complex as one often tries to take advantage of the gravita tional fields of the planets encountered on the way, by entering into their SOIs with the objective of either changing direction or gaining additional impulse. This technique is often referred to as gravity assist. References [1 F.J. Hale, Introduction to Space Flight, Prentice-Hall, 1994
Interplanetary Transfers When more than one planet is involved, the problem is no longer a two-body problem. Nevertheless, it is common (at least to get a good approximation) to decompose the problem into a series of two body problems. Consider, for example, an interplanetary transfer in the solar system. For each planet we define the sphere of influence (SOI). Essentially, this is the region where the gravitational attraction due to the planet is larger than that of the sun. The mission is broken into phases that are connected by patches where each patch is the solution of a two body problem. This is called the patched conic approach. Consider, for instance, a mission to Mars. The first phase will consist of a geocentric hyperbola as the spacecraft escapes from earth SOI. The second phase would start at the edge of the earth’s SOI, and would be an elliptical trajectory around the sun while the spacecraft travels to Mars. The third phase would start at the edge of Mars’ SOI, and would be a hyperbolic approach trajectory with the gravitational field of Mars as the attracting force. For long missions, the situation can become very complex as one often tries to take advantage of the gravitational fields of the planets encountered on the way, by entering into their SOI’s with the objective of either changing direction or gaining additional impulse. This technique is often referred to as gravity assist. References [1] F.J. Hale, Introduction to Space Flight, Prentice-Hall, 1994. 6