Q.A geostationary satellite is to be placed in an equatorial orbit with a time period of exactly $24\ \text{h}$. Using $GM = 4.00 \times 10^{14}\ \text{m}^3/\text{s}^2$, derive and estimate the radius of its orbit, measured from the Earth's centre.

A satellite that appears stationary in the sky from the ground, requiring three conditions together: orbit in the equatorial plane, moving eastward (same sense as Earth's rotation), with orbital period exactly equal to the Earth's own rotation period ($\approx 24\ \text{h}$). Its orbital radius, from $r = (GMT^2/4\pi^2)^{1/3}$ using Kepler's third law, works out to about $4.2\times 10^4\ \text{km}$ from the Earth's centre -- a height of roughly $3.6\times 10^4\ \text{km}$ above …