Asked by Anonymous
The space shuttle releases a satellite into a circular orbit 550 km above the Earth. How fast must the shuttle be moving (relative to Earth) when the release occurs?
Answers
Answered by
drwls
The earth's radius is Re = 6378 km, so the radius of the orbit is
R = Re + 550 = 6928 km
For a circular orbit, the centripetal force is the gravatational attraction force, so
V^2/R = G M/R^2
G is the universal constant of gravity and M is the mass of the earth.
Solve for V.
V = sqrt(GM/R) = sqrt[(GM/Re^2)*Re^2/R]
= sqrt(g Re)*sqrt(Re/R)
= 0.956*sqrt(g Re)
R = Re + 550 = 6928 km
For a circular orbit, the centripetal force is the gravatational attraction force, so
V^2/R = G M/R^2
G is the universal constant of gravity and M is the mass of the earth.
Solve for V.
V = sqrt(GM/R) = sqrt[(GM/Re^2)*Re^2/R]
= sqrt(g Re)*sqrt(Re/R)
= 0.956*sqrt(g Re)
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