The required centripetal force Fc is the gravitational force Fg
Important formulas Fc= (mv2)/r
Fg = G( M m) /r2
v = 2 π r /T
- m mass of the satellite in kg
- M mass of the earth ( Mearth = 5.976 x 1024 kg)
- G gravitational constant in N m2kg-2 ( G= 6.6726 x 10-11 Nm2 kg-2)
- T period of the satellite in s
- r distance between the center of mass of the earth and the satellite in m
The satellite is in a uniform circular motion.
Geostationary satellites always appear stationary at the same point above the equator
The period is equal to the rotational period of the earth (= 24 h)
Example
What is the height of a geostationary satellite ?
Fc= Fg
mv2/r = G (m M)/r2
v2 = G M/r
r = G M/v2 (1)
v = 2 π r /T (2)
We substitute v from equation (2) in equation (1) :
r3 = (GM T2)/4 π2
r3 = ( 6.6726 x 10-11)( 5.976 x 1024) (24 x 3600)2)/4 π2 = 7.5400 x 1022
r = 4.217 x 107 m = 42.17 x 103 km
Altitude satellite = r – Rearth = 42.17 x 103 – 6.378 x 103 = 35.8 x 103 km
Find g at this altitude
Fg = m g = (G M m) / r2
g = Fg/m = (GM)/r2 =( 6.6726 x 10-11)( 5.976 x 1024) / (4.217 x 107)2 = 0.22 m/s2