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ratelena [41]
3 years ago
14

A spacecraft and a staellite are at diametrically opposite position in the same circular orbit of altitude 500 km above the eart

h. 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
Physics
1 answer:
Tanya [424]3 years ago
5 0

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  

<u>a) determine the increase in speed </u>

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

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