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' surface is
`T=(2pir^(3/2))/(gR^(2))`
where `r=R+h`.
Then, `T=(2pi(R+h))/Rsqrt(((R+h)/g))`
Here, `R=6400km,h=1600km=R//4`
`T=(2pisqrt(R+R/4))/Rsqrt(((R+R/4)/g))=2pi(1.25)^(3/2)sqrt(R/g)`
Putting the given values,
`T=2xx3.14xxsqrt(((6.4xx10^(6)m)/(9.8ms^(-2))))(1.25)^(3/2)=7092s=1.97h`
Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is `24h`. Let us find the height `h` of such a satellite above the earth's suface in terms of the earth,'s radius.
Let it be `nR`.Then
`T=(2pi(R+nR))/Rsqrt(((R+nR)/g))`
`=2pisqrt((R/g))(1+n)^(3/2)`
`=2xx3.14sqrt(((6.4xx10^(6)m//s)/(9.8m//s^(2))))(1+n)^(3/2)`
`(5075s)(1+n)^(3/2)=(1.41h)(1+n)^(3/2)`
For `T=24h`, we have `(24h)=(1.41h)(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 geostationary satellite above the earth's surface is `nR=5.61xx6400km=6.59xx10^(4)km`.