EAS 2011 Introduction to Aerospace Engineering
Prof. Amor Menezes
Homework 9 Solutions; December 10, 2018
1. An interplanetary spacecraft mission from Earth to Saturn is proposed. The spacecraft is initially in a 1.5 Earth-radii
circular parking orbit about the Earth and its initial orbital plane is the Earth’s heliocentric plane. It is assumed that
the heliocentric orbital planes of the Earth and Saturn are identical.
(a) Find the ∆V required to leave the Earth’s gravitation and to place the spacecraft on a Hohmann elliptical transfer
orbit that connects with the circular heliocentric orbit of Saturn. Assume a posigrade burn in the parking orbit.
(b) Find the ∆V required to leave the Earth’s gravitation and to place the spacecraft on a parabolic heliocentric
transfer orbit that intersects the circular heliocentric orbit of Saturn. Include in your analysis the effect of the
Earth’s gravitation and assume a posigrade burn in the parking orbit.
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Solution: We know that RE = 6, 378 km, µE = 3.986 × 105 km
, µS = 1.327 × 1011 km3 /s2 , rEarth/Sun = 1.4957 ×
s2
108 km, rSaturn/Sun = 1.43 × 109 km, and the initial orbital radius of the spacecraft with respect to the Earth is
p
r = 9, 567 km. Furthermore, the initial speed of the spacecraft with respect to the Earth is Vc = µE /r = 6.45 km/s.
(a) The Hohmann transfer elliptical orbit from Earth’s heliocentric orbit to Saturn’s heliocentric orbit has perihelion at rEarth/Sun and aphelion distance rSaturn/Sun . Hence, the transfer orbit is describe by e = (rSaturn/Sun −
rEarth/Sun )/(rSaturn/Sun + rEarth/Sun ) = 0.810, a = (rSaturn/Sun + rEarth/Sun )/2 = 7.90 × 108 km, p = a(1 − e2 ) =
√
2.72 × 108 km, and H = µS p = 6.00 × 109 km2 /s. Therefore, the spacecraft should have velocity Vp =
H/r p = 40.02 km/s with respect to q
the Sun at the perihelion of the transfer orbit. Since the Earth’s orbital speed
with respect to the Sun is VEarth =
µS /rEarth/Sun = 29.7 km/s and the burn is posigrade (Saturn is an outer
planet), the spacecraft should have an escape speed of V∞ = Vp −VEarth = 10.23 km/s with respect to the Earth.
Finally, to escape from Earth’s SOI with the given escape speed of 10.23 km/s, the spacecraft
should be placed
q
on a hyperbolic orbit with respect to the Earth with velocity at perigee of VSC/Earth =
2µE
2
r +V∞ = 13.74 km/s.
Hence, the required burn on the parking orbit is ∆V = VSC/Earth −Vc = 7.29 km/s . The location of the burn is
the perigee of the resulting hyperbolic orbit of the spacecraft with respect to the Earth.
q
(b) The parabolic heliocentric escape orbit to Saturn requires a velocity at perihelion of Vp = 2µS /rEarth/Sun =
42.1 km/s. Since the Earth’s circular orbital velocity with respect to the Sun is VEarth = 29.7 km/s, the spacecraft
should have an escape speed of V∞ = Vp −VEarth = 12.32 km/s with respect to the Earth (assuming a posigrade
exit from Earth’s SOI). To escape from the Earth with the given escape speedqof 12.32 km/s, the spacecraft
should be placed in a hyperbolic Earth orbit with velocity at perigee VSC/Earth = 2µr E +V∞2 = 15.33 km/s. The
perigee of this Earth hyperbolic orbit occurs at the location of the rocket engine burn in the parking orbit. Hence,
the burn on the parking orbit requires ∆V = VSC/Earth −Vc = 8.88 km/s .
2. A chemical rocket with a specific impulse of 350 s can be used to change the velocity vector of a spacecraft in a
circular LEO at an altitude of 300 km.
(a) What is the maximum ∆V that can be achieved using this rocket engine? Assume that the burn time of the rocket
engine is negligible and that the mass of the spacecraft in LEO is 50% propellant.
(b) Can the spacecraft escape from the sphere of influence of the Earth using this rocket engine? If so, what is the
spacecraft’s terminal velocity? If not, what is the maximum energy elliptical orbit that the spacecraft can reach,
and what is the apogee radius of this elliptical orbit?
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Solution: The specific impulse of the rocket engine is Isp = 350 s and the gravitational acceleration at the surface
of the Earth is g = 9.81 × 10−3 km/s2 , so the exhaust velocity of the rocket engine is Ve = Isp g = 3.43 km/s.
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Also, we know that RE = 6, 378 km, µE = 3.986 × 105 km
, r = 6, 678 km, and the LEO circular orbit speed is
s2
p
Vc = µE /r = 7.73 km/s.
(a) If 50% of the mass of the spacecraft is propellant, we have M0 = 2M1 and the maximum delta-V is ∆V =
Ve ln(M0 /M1 ) = Ve ln(2) = 2.38 km/s .
p
(b) The escape speed from the circular LEO is given by Vesc = 2µE /r = 10.93 km/s. Since the spacecraft in LEO
has Vc = 7.73 km/s, the maximum ∆V will give V = Vc + ∆V = 10.11 km/s, which is less than the minimum
escape velocity. Hence, the spacecraft will be in an elliptical orbit and cannot escape the Earth’s influence .
The energy of the orbit is E = V 2 /2 − µE /r = −8.63 km2 /s2 , which gives a = −µE /(2E ) = 23, 106 km.
Therefore, the radius of apogee of the final orbit is ra = 2a − r p = 2a − r = 39, 534 km .
3. A spacecraft is to be launched from a low Earth circular orbit at an altitude of 600 km onto a hyperbolic orbit that
leaves the Earth’s gravitational sphere of influence. This is to be achieved using a single impulsive engine burn in the
low Earth orbit. Assume that a single stage rocket engine using a propellant with a specific impulse of 320 s provides
the thrust to achieve this orbital change. Assume that the engine burn time is negligibly small. The spacecraft
structure and payload have mass of 750 kg.
(a) What is the minimal ∆V magnitude and direction that is required to leave the sphere of influence of the Earth?
What is the propellant mass required to achieve this desired orbit change?
(b) What is the ∆V magnitude that is required to leave the sphere of influence of the Earth with a terminal velocity
magnitude of 1 km/s with respect to the Earth? What is the propellant mass required to achieve this desired orbit
change?
(c) What is the ∆V magnitude that is required to leave the sphere of influence of the Earth and the sphere of influence
of the Sun? What is the propellant mass required to achieve this desired orbit change?
Solution: The specific impulse of the rocket engine is Isp = 320 s, so the exhaust velocity of the rocket engine is
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Ve = Isp g = 3.14 km/s. Also, we know that RE = 6, 378 km, µE = 3.986 × 105 km
, r = 6, 978 km, the LEO circular
s2
p
orbit speed is Vc = µE /r = 7.56 km/s, and M1 = 750 kg.
(a) The minimum escape velocity from the Earth occurs at the perigee of a parabolic orbit around the Earth,
p
which is given by Vesc = 2µE /r = 10.69 km/s. Hence, ∆V = Vesc −Vc = 3.13 km/s in the êθ direction .
From the relation ∆V = Ve ln(M0 /M1 ), α = M0 /M1 = e∆V /Ve = 2.71. Hence, the required propellant mass is
M0 − M1 = (α − 1)M1 = 1, 283 kg .
(b) Similarly, to escape from the Earth with the given terminal speed ofpV∞ = 1 km/s, the spacecraft should
be in a hyperbolic orbit with the velocity at perigee given by Vp = 2µE /r +V∞2 = 10.74 km/s. Hence,
∆V = Vp −Vc = 3.18 km/s . From the relation ∆V = Ve ln(M0 /M1 ), α = M0 /M1 = e∆V /Ve = 2.75. Hence, the
required propellant mass is M0 − M1 = (α − 1)M1 = 1, 314 kg .
(c) The gravitational parameter of the Sun is µS = 1.327 × 1011 km3 /s2 and the distance betweenq
the Sun and
the Earth is rE/S = 1.5 × 108 km. Then, the speed of the Earth with respect to the Sun is VE/S =
µS /rE/S =
29.74 km/s. The minimum escape speed
q from the Sun occurs at the perihelion of a parabolic orbit around
the Sun, which is given by Vesc/S =
2µS /rE/S = 42.06 km/s. Assuming a posigrade exit from the Earth,
the terminal speed of the spacecraft with respect to the Earth should be V∞/E = Vesc/S − VE/S = 12.32 km/s.
The velocity of the spacecraft with respect to the Earth at the perigee of the hyperbolic orbit is Vp/E =
q
2
2µE /r +V∞/E
= 16.31 km/s. Hence, the rocket engine burn should provide ∆V = Vp/E −Vc = 8.75 km/s .
From the relation ∆V = Ve ln(M0 /M1 ), α = M0 /M1 = e∆V /Ve = 16.25. Hence, the required propellant mass is
M0 − M1 = (α − 1)M1 = 11, 438 kg .
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