A satellite ils launched into a circular orbit `1600km` above the surface of the earth. Find the period of revolution if the radius of the earth is `R=6400km` and the acceleration due to gravity is `9.8ms^(-2)`. At what height from the ground should it be launched so that it may appear stationary over a point on the earth's equator?

Explanation

The orbiting period of a satellite at a height h from earth's surface is `T = (2pir^(3//2))/(gR^(2))` where `r =R +h`
then `T = (2pi (R+h))/(R) sqrt((R +h)/(g))`
Here `R = 6400km,h = 1600 km = R//4`
Then `T=(2pi(R+R/4))/Rsqrt(((R+R/4)/g))=2pi(5//4)^(3//2)sqrt(R/g)`
Putting the given values `T = 2 xx 3.14 xx sqrt((6.4 xx 10^(6)m)/(9.8m//s^(2))) (1.25)^(3//2) = 7092 sec = 1.97hours`
Now a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution round the earth is equal to the period of revolution of the earth round its own axis which is 24 hours Let us find the height h of such a satellite above the earth's surface in terms of the earth's radius Let it be `nR` then
`T = (2pi (R+nR))/(R) sqrt((R +nR)/(g)) = 2pi sqrt((R)/(g)) (1+n)^(3//2) = 2 xx 3.14 sqrt((6.4 xx 10^(6) "meter/sec")/(9.8 "meter/sec"^(2))) (1+n)^(3//2)`
`= (5075 sec) (1 +n)^(3//2) = (1.41 "hours") (1 +n)^(3//2)`
For `T =24` hours we have
(24hours) = (1.41) hours) `(1 +n)^(3//2)`
or `(1 +n)^(3//2) = (24)/(1.41) =17`
or `1 + n = (17)^(2//3) = 6.61` or `n = 5.61`
The height of the geo-stationary satellite above the earth surface is `nR = 5.61 xx 6400km = 3.59 xx 10^(4)km` .