Calculate the height from the Earth's surface that a satellite must attain in order to be in geosynchronous orbit and the satellite's velocity?

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