The space shuttle releases a satellite into a circular orbit 650 km above the earth. How fast must the shuttle be moving (relative to earth) when the release occurs.. The answer is 7530 m/s
mv2/(R+h)=GmM/(R+h)2v=MG/(R+h)−−−−−−−−−−−√
M - mass of Earth, value is 6.0 * 10^24 kg
G - gravitational constant, equals to 6.67*10^-11 N m2/kg2
R - radius of Earth, 6.4 *10^6 m
h - given height of satellite.
The shuttle must be moving
(relative to earth) when the at 7530
m/s when the release occurs. I am hoping that this answer has satisfied
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