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 red 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 retrore so that rendezvous will occur at B.

Explanation:

rA= 5000km+6378km=11378 km

rB= 500km+6378km=6878 km

Eccentricity, e=[11378-6878] / [11378 +6878] =0.24649

Evaluating the orbital equation at perigee yields the angular momentum

rB= h22/u * [1/[1+e]][1]

Pluging the values in equation[1]

6878 km= h22/398600* [1/[1+0.24649]]

we get h2= 58458 km2/s

Period of transfer ellipse, T_{2}=\frac{2\pi }{\mu ^{2}}( \frac{h_{2}}{\sqrt{1-e^{2}}})^{3}

Pugging the values we get , T_{2}=\frac{2\pi }{398600 ^{2}}( \frac{58458}{\sqrt{1-0.24649^{2}}})^{3}=\mathbf{8679.1s}

The period of circular orbit 3 is ,T_{3}=\frac{2\pi }{\mu }*r_{B}^{\frac{3}{2}}

T_{3}=\frac{2\pi }{398600 }*6878^{\frac{3}{2}}=\mathbf{5676.8s}

The time of flight from C to B on orbit 3 must be equal to  time of flight from A to B on orbit 2

\bigtriangleup t_{CB}=\frac{1}{2}T_{2}=\frac{1}{2}*8679.1=4339.5s

here orbit 3 is a circle, hence

i was thinking more like reflects green light, just not sure

reserve is the answer