Question

A spacecraft and a staellite are at diametrically opposite position in the same circular orbit of altitude 500 km above the earth. As it passes through point A, the spacecraft fires its engine for a short interval of time to increase its speed and enter an elliptical orbit. Knowing that the spacecraft returns to A at the same time the satellite reaches A after completing one and a half orbits, determine (a) the increase in speed required, (b) the periodic time for the elliptic orbit

Answer 1

Answer:

Hello the diagram related to your question is attached below

answer: a) 851 m/s

             b)  8506.1 secs

Explanation:

calculate the periodic time of the satellite using the equation below

t = $\frac{2\pi }{R} \sqrt{\frac{(R+h)^{3} }{g} }$  --  ( 1 )

where ; R = 6370 km

h = 500 km

g = 9.81 m/s^2

input given values into equation 1

t = 5670.75 secs

next calculate the periodic time taken by the space craft  

a) determine the increase in speed

V = v - $\sqrt{\frac{gR^2}{R + h} }$  

where ; v = 8463 m/s , R = 6370 km, h = 500 km

V = 851 m/s

b) Determine the periodic time for the elliptic orbit

τ = $\frac{3t}{2}$

 = $\frac{3*5670.76}{2}$  =  8506.1 secs

attached below is the remaining part of the detailed solution