Asked by Andrea
What is the period of revolution of a satellite with mass m that orbits the earth in a circular path of radius 7820 km (about 1450 km above the surface of the earth)?
Answers
Answered by
Damon
F = m a
Force of gravity = m v^2/r
G m Me /r^2 = m v^2/r
where G is Newton's gravitational constant and Me is mass of earth
r = 7,820,000 meters
G Me = v^2 r
v^2 = G Me/r
solve for v
(2 pi r)/v = period
Force of gravity = m v^2/r
G m Me /r^2 = m v^2/r
where G is Newton's gravitational constant and Me is mass of earth
r = 7,820,000 meters
G Me = v^2 r
v^2 = G Me/r
solve for v
(2 pi r)/v = period
Answered by
Gauri kumal
6979.13Second
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