Two satellites, A and B, are in different circular orbits about the earth. The orbital speed of satellite A is eighty-two times that of satellite B. Find the ratio (TA/TB) of the periods of the satellites.

Answer 1

F = G Mm/r^2
F = m a = m v^2/r
so
G M m/r^2 = m v^2/r

v^2 = G M/r
G M same for both

va^2 ra = vb^2 rb
but
va = 82 vb

6724 vb^2 ra = vb^2 rb
so
rb = 6724 ra

circumference = 2 pi r
period = circumference / v = 2 pi r/v

period a = 2 pi ra/82 vb
period b = 2 pi 6725 ra/vb

1/82 / 6725 = 1/551368 or 1.81*10^-6