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

To derive the expression, we start with the gravitational force acting on a satellite in circular orbit:

$F_g = \frac{GMm}{r^2}$

where:

• $F_g$ is the gravitational force,
• $G$ is the gravitational constant,
• $M$ is the mass of the central body (e.g., the planet),
• $m$ is the mass of the satellite,
• $r$ is the radius of the orbit.

For circular motion, the centripetal force is given by:

$F_c = \frac{mv^2}{r}$

where $v$ is the orbital speed of the satellite. Setting the two forces equal gives us:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

By canceling $m$ from both sides, we have:

$\frac{GM}{r^2} = \frac{v^2}{r}$

Multiplying both sides by $r$ yields:

$\frac{GM}{r} = v^2$

The orbital speed $v$ can also be expressed in terms of the period of orbit $T$ as:

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

Substituting this back into our previous equation results in:

$\frac{GM}{r} = \left( \frac{2\pi r}{T} \right)^2$

Rearranging gives:

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

This shows that $T^2 \propto r^3$, confirming the relationship derived.