Chapter 6
Q. 6.2
A spacecraft returning from a lunar mission approaches earth on a hyperbolic trajectory. At its closest approach A it is at an altitude of 5000 km, traveling at 10 km/s. At A retrorockets are fired to lower the spacecraft into a 500-km-altitude circular orbit, where it is to rendezvous with a space station. Find the location of the space station at retrofire so that rendezvous will occur at B (Fig. 6.6).
Step-by-Step
Verified Solution
The time of flight from A to B is one-half the period T_2 of elliptical transfer orbit 2. While the spacecraft coasts from A to B, the space station coasts through the angle \phi_{C B} from C to B. Hence, this mission has to be carefully planned and executed, going all the way back to lunar departure, so that the two vehicles meet at B.
According to Eq. (2.83), to find the period T_2 we need to only determine the semimajor axis of orbit 2. The apogee and perigee of orbit 2 are
r_A=5000+6378=11,378 km \quad r_B=500+6378=6878 km
Therefore, the semimajor axis is
a=\frac{1}{2}\left(r_A+r_B\right)=9128 km
From this we obtain
T_2=\frac{2 \pi}{\sqrt{\mu}} a^{3 / 2}=\frac{2 \pi}{\sqrt{398,600}} 9128^{3 / 2}=8679.1 s (a)
The period of circular orbit 3 is
T_3=\frac{2 \pi}{\sqrt{\mu}} r_B^{3 / 2}=\frac{2 \pi}{\sqrt{398,600}} 6878^{3 / 2}=5676.8 s (b)
The time of flight from C to B on orbit 3 must equal the time of flight from A to B on orbit 2.
\Delta t_{C B}=\frac{1}{2} T_2=\frac{1}{2} \cdot 8679.1=4339.5 s
Since orbit 3 is a circle, its angular velocity, unlike an ellipse, is constant. Therefore, we can write
\frac{\phi_{C B}}{\Delta t_{C B}}=\frac{360^{\circ}}{T_3} \Rightarrow \phi_{C B}=\frac{4339.5}{5676.8} \cdot 360=275.2^{\circ}
(The reader should verify that the total delta-v required to lower the spacecraft from the hyperbola into the parking orbit is 5.749 km/s. According to Eq. (6.1), that means over 85% of the returning spacecraft mass must consist of propellant!)
\frac{\Delta m}{m}=1-e^{\frac{\Delta v}{I_{ sp } g_0}} (6.1)