Question

Consider a planet of mass M with an orbiting satellite at a distance of R from each of their centers. Show that Kepler's third law of gravitation is true.

Answer 1

Explanations & Calculations

• The satallite is orbiting the planet at a constant distance: at a circle of radius R.
• Apply Newton's second law towards the center of the orbit: towards center of the planet.

$\qquad\qquad \begin{aligned} \small F&= \small ma\\ \small G\frac{mM}{R^2}&= \small m\frac{v^2}{R}\to G\frac{\cancel{m}M}{R^{\cancel{2}}}=\cancel{m}\frac{v^2}{\cancel{R}}\\ \small G\frac{M}{R}&= \small v^2\cdots(1) \end{aligned}$

• Time spent on orbiting the planet can be written as $\large\frac{2\pi R}{v}$ as the orbiting speed stays constant due to the radius being constant.
• Ths gives,

$\qquad\qquad \begin{aligned} \small T&= \small \frac{2\pi R}{v}\\ \small v&= \small \frac{2\pi R}{T} \end{aligned}$

• Substituting this in (1) yeilds,

$\qquad\qquad \begin{aligned} \small G\frac{M}{R}&= \small \Big(\frac{2\pi R}{T}\Big)^2\\ \small (GM)\times T^2&= \small4\pi^2\times R^3\\ \small T^2&\propto \small R^3 (\text{as other terms are constants}) \end{aligned}$

• This proves the Kepler's third which denotes "The square of the orbital period directly proportional to the cube of the length of the semi-major axis of the orbit "
• It was originally denoted for the real elliptical orbital motions & all axes are identical in this problem as the orbit is a circle of radius R.