Model satellite motion as uniform circular motion and derive Kepler's third law

What this dot point is asking

WACE wants you to treat a satellite as undergoing uniform circular motion with gravity as the only force, then derive the relationships linking radius, speed and period. The orbiting body's own mass cancels, so a feather and a spacecraft at the same altitude orbit identically.

Gravity supplies the centripetal force

For a satellite of mass $m$ orbiting a body of mass $M$ at radius $r$, the gravitational force is the centripetal force:

Cancelling $m$ and rearranging gives the orbital speed,

Higher orbits are slower: a satellite further out moves more slowly because the field is weaker. This is counterintuitive but follows directly from the inverse-square law.

Orbital period and Kepler's third law

Since the satellite covers one circumference $2\pi r$ in one period $T$, $v=2\pi r/T$. Substituting into the speed expression and squaring gives

The right-hand side depends only on the central mass $M$, so for all satellites of the same body, $r^3/T^2$ is a constant. This is Kepler's third law, and it lets you compare two orbits with a clean ratio: $\dfrac{r_1^3}{T_1^2}=\dfrac{r_2^3}{T_2^2}$.

Geostationary satellites

A geostationary satellite stays above a fixed point on the equator, so its period equals one sidereal day (about $24$ hours) and it orbits west to east. Putting $T=86400\ \text{s}$ and $M=M_E$ into Kepler's law gives an orbital radius of about $4.2\times10^7\ \text{m}$, roughly $36000\ \text{km}$ above the surface. These orbits are used for communications and weather satellites because the antenna can point at one spot in the sky.

Low-Earth-orbit satellites, by contrast, sit only a few hundred kilometres up, complete an orbit in roughly ninety minutes, and sweep across the sky, which is why imaging and space-station orbits are low and fast. The choice of orbital radius is therefore a trade-off set entirely by the period required, and Kepler's third law is the tool that links the two for any central body.

Weighing a planet from its moon

A favourite exam application of Kepler's third law is finding the mass of a central body from the orbit of a satellite or moon. Rearranging $\dfrac{r^3}{T^2}=\dfrac{GM}{4\pi^2}$ gives $M=\dfrac{4\pi^2 r^3}{GT^2}$, so a single measured orbital radius and period determine the central mass without ever weighing it directly. This is how the masses of the Sun, Jupiter and other planets with moons are known. Note that the orbiting body's own mass never appears, which is why this method works equally well for a tiny natural moon or a large artificial satellite.

Energy changes between orbits

To move a satellite to a higher orbit, a rocket must do positive work against gravity, increasing the gravitational potential energy. Counter-intuitively the orbital kinetic energy decreases at the higher orbit (because $v=\sqrt{GM/r}$ falls), but the gain in potential energy outweighs the loss in kinetic energy, so the total mechanical energy rises. This is why reaching a higher orbit always costs fuel even though the satellite ends up moving more slowly, a subtlety examiners like to probe in extended-response questions.

Setting up the derivation

In an exam, always begin by equating the gravitational force to $mv^2/r$ and stating that gravity is the centripetal force. Showing this step earns the reasoning marks even if arithmetic slips later. Remember that $r$ is the orbital radius from the planet's centre, not the altitude.