Answer:
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
