Derive an expression to show that for satellites in a circular orbit r^2 ∝ T^2 where T is the period of orbit and r is the radius of the orbit. - AQA - A-Level Physics - Question 7 - 2017 - Paper 2

The gravitational force acting on the satellite is given by Newton's law of gravitation:

$F_g = \frac{G M m}{r^2}$

where:

• $G$ is the gravitational constant
• $M$ is the mass of the planet around which the satellite orbits.

Since the centripetal force is provided by gravity, we set the two forces equal:

$\frac{mv^2}{r} = \frac{G M m}{r^2}$

On simplifying, we can cancel $m$ from both sides:

$\frac{v^2}{r} = \frac{G M}{r^2}$

The orbital speed $v$ can be related to the period $T$ by:

$v = \frac{2\pi r}{T}$

Substituting this into our previous equation gives:

$\frac{(\frac{2\pi r}{T})^2}{r} = \frac{G M}{r^2}$

After simplification, we find:

$\frac{4\pi^2 r}{T^2} = \frac{G M}{r^2}$

Rearranging this leads us to:

$T^2 = \frac{4\pi^2 r^3}{G M}$

Thus showing that:

$r^2 \propto T^2$