Obtain an equation for the orbital time period of a satellite revolving around the Earth.

(N/A) The orbital period $(T)$ of a satellite is the time taken to complete one full revolution around the Earth.
The orbital velocity of a satellite at a height $h$ above the Earth's surface is given by:
$v_{0} = \sqrt{\frac{GM_{E}}{R_{E}+h}}$ ... $(1)$
The orbital velocity is also defined as the ratio of the circumference of the orbit to the time period:
$v_{0} = \frac{2\pi(R_{E}+h)}{T}$ ... $(2)$
Equating $(1)$ and $(2)$:
$\sqrt{\frac{GM_{E}}{R_{E}+h}} = \frac{2\pi(R_{E}+h)}{T}$
Rearranging for $T$:
$T = \frac{2\pi(R_{E}+h)}{\sqrt{\frac{GM_{E}}{R_{E}+h}}}$
$T = 2\pi \sqrt{\frac{(R_{E}+h)^{3}}{GM_{E}}}$ ... $(3)$
Squaring both sides:
$T^{2} = \frac{4\pi^{2}}{GM_{E}}(R_{E}+h)^{3}$ ... $(4)$
Using the relation $g = \frac{GM_{E}}{(R_{E}+h)^{2}}$,we can express the period in terms of acceleration due to gravity $g$ at height $h$:
$T = 2\pi \sqrt{\frac{R_{E}+h}{g}}$ ... $(5)$
From equation $(4)$,if we let $r = R_{E}+h$,then $T^{2} = Kr^{3}$,where $K = \frac{4\pi^{2}}{GM_{E}}$. This confirms Kepler's third law of planetary motion,$T^{2} \propto r^{3}$.