Question

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

Answer 1

The orbiting period of a satellite at a height h from earth's surface is $T=\frac{2\pi r^{\frac{3}{2}}}{g{R}^2}$
where $r=R+h$.
Then, $T=\frac{2\pi (R+h)}{R}\sqrt{\left(\frac{R+h}{8}\right)}$
Here, $R=6400$km, $h=1600$km $=R/4$
$T=\frac{2\pi (R+\frac{R}{4})}{R}\sqrt{\left(\frac{R+\frac{R}{r}}{g}\right)}=2\pi (1.25{)}^{\frac{3}{2}}\sqrt{\frac{R}{g}}$
Putting the given values,
$T=2\times 3.14\times \sqrt{\left(\frac{6.4\times {10}^{6}m}{9.8m{s}^{-2}}\right)}(1.25{)}^{\frac{3}{2}}=7092s=1.97h$
Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earth is equal to the period of revolution of the earth up around its own axis which $24$h. 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
$T=\frac{2\pi (R+nR)}{R}\sqrt{\left(\frac{R+nR)}{g}\right)}$
$=2\pi \sqrt{\left(\frac{R}{g}\right)}(1+n{)}^{\frac{3}{2}}$
$=2\times 3.14\sqrt{\left(\frac{6.4\times {10}^{6}m/s}{9.8m/s^2}\right)}(1+n{)}^{\frac{3}{2}}$
$=\left(5075s\right)(1+n{)}^{\frac{3}{2}}=(1.41h\left)\right(1+n{)}^{\frac{3}{2}}$
For $T=24h$, we have $\left(24h\right)=\left(1.41h\right)(1+n{)}^{\frac{3}{2}}$
or $(1+n{)}^{\frac{3}{2}}=\frac{24}{1.41}=17$
or $1+n=(17{)}^{\frac{2}{3}}=6.61$
or $n=5.61$
The height of the geostationary satellite above the earth's surface is $nR=5.61\times 6400km=3.59\times {10}^4$km.