Answer to Question #103456 in Astronomy | Astrophysics for mm

Question

Assume that the moon is at a distance of 300,000 kms from the earth and that it takes 28 days for
it to orbit the earth once. Geostationary satellites are those which are at a rest relative to earth.
Using these two statements derive the altitude of the geostationary satellite from the centre of the
earth.

Answer 1

According to the Kepler's Third Law, the ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun.

Thus, $\frac{p^2_{sat}}{p^2_{moon}}=\frac{d^3_{sat}}{d^3_{moon}}$

or $d_{sat}=d_{moon}(\frac{p^2_{sat}}{p^2_{moon}})^{\frac{1}{3}}$

Since, $p_{sat}=1\;day,\;\;d_{sat}=300000(\frac{1^2}{28^2})^{\frac{1}{3}}=32535\;km$

The altitude of the geostationary satellite from the centre of the earth is 32535 km.