Radius of the Earth(R)=6.38×106m
Gravitation Constant(G)=6.67×10−11m3kg−1s−2
Mass of the Earth Me=5.97×1024 kg
Earth’s rotational period (one sidereal day)
T=24hrs=24×60×60s=86400s
Let the height of the orbit from Earth’s surface be h m and a satellite of mass m kg rotates with uniform angular velocity ω rad/s in the geosynchronous orbit.
Equating centripetal force with the gravitational force we can write
mω2(R+h)=GMem/(R+h)2
(2π/T)2(R+h)=GMe(R+h)2
(R+h)3=GMeT2/4π2
Height from the Earth’s surface that a satellite must attain in order to be in geosynchronous orbit
h=42241 − R = 42241 − 6380 = 35861km
And the satellite’s velocity
v = ω × (R+h) = 2π/T × 4224.1 × 104 ≈ 3070m/s