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)2 v=MG/(R+h)−−−−−−−−−−−√ M - mass of Earth, value is 6.0 10^24 kg G - gravitational constant, equals to 6.6710^-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 your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.
