Explanation:
The orbiting period of a satellite at a height h from earth' surface is
T=2πr32gR2
where r=R+h.
Then, T=2π(R+h)R(R+hg)−−−−−−−−√
Here, R=6400km,h=1600km=R/4
T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√
Putting the given values,
T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=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=2π(R+nR)R(R+nRg)−−−−−−−−−−√
=2π(Rg)−−−−−√(1+n)32
=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32
(5075s)(1+n)32=(1.41h)(1+n)32
For T=24h, we have (24h)=(1.41h)(1+n)32
or (1+n)32=241.41=17
or 1+n(17)23=6.61
or n=5.61
The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.
