F = m a
G m M/r^2 = m V^2/r
m, the mass of the satellite, is irrelevant, cancels.
G M = V^2 r
distance around circumference = 2 pi r
so t = 2 pi r / V
or V = 2 pi r/t
so
G M = (2 pi)^2 (r^3/t^2)
or
t^2 = (2 pi)^2 r^3/ (G M)
here
G = 6.67*10^=11
M = 5.98 *10^24 Kg
r = 6.38*10^6 + .2*10^6
= 6.58*10^6 meters
so
t^2 = (2 pi)2 *(6.58)^3*(10)^18/(6.57*5.98*10^13)
in seconds
A satellite of mass 225 kg is launched from a site on Earth's equator into an orbit at 200 km above the surface of Earth. (The mass of the Earth is 5.98 1024 kg, and the radius of the Earth is 6.38 103 km.) (a) Assuming a circular orbit, what is the orbital period of this satellite?
1 answer