Question

Exercise 9.5 A geostationary satellite orbits the earth in 24 hours along an orbital path that is parallel to an imaginary plane drawn through the Earth's equator. Such a satellite appears permanently fixed above the same location on the Earth. How high above the earth's surface must it be located?

Answer 1

let h is the height above the surface of the earth the geostationary satellite revolves.

we know, Me = 5.97*10^24 kg

Re = 6.37*10^6 m

let h is the height of geostationary satellite.

The time period of geostationary satellite, T = 24 hours

= 24*60*60 s

= 86400 s

we know, T = 2*pi*(Re + h)^(3/2)/sqrt(G*Me)

T^2 = 4*pi^2*(Re + h)^3/(G*Me)

G*Me*T^2/(4*pi^2) = (Re + h)^3

==> Re + h = ( G*Me*T^2/(4*pi^2) )^(1/3)

==> h = ( G*Me*T^2/(4*pi^2) )^(1/3) - Re

= ( 6.67*10^-11*5.97*10^24*86400^2/(4*pi^2))^(1/3) - 6.37*10^6

= 3.59*10^7 m <<<<<<<--------------------Answer