Question

A satellite is in circular orbit about Earth at an altitude of 300km. What are its: a) orbital period. b) orbital frequency and c) linear speed?

Answer 1

c)
mass of earth , Me = 5.972 * 10^24 kg
radius of earth, RE = 6.371*10^6 m
m ---> mass of satellite
h-->300 km = 300000 m
Use centripetal force = Gravitational force
m*v^2/(RE+h) = G*Me*m / (RE+h)^2
v^2= G*Me/ (RE+h)
v^2 = (6.673*10^-11)*( 5.972 * 10^24 ) / (6.371*10^6 + 300000)
v=7729 m/s
Answer: Linear speed : 7729 m/s

a)
T = 2*Pi*(RE+h) / v
= 2*pi*(6.371*10^6 + 300000) / 7729
=5423 s
=1.51 hr
Answer: Period : 1.51 hr

b)
1 revolution in 1.5 hr
frequency = 1/1.5
= 0.67 revolution per hour