Orbits of planets and satellitesAQA A-Level Physics: Revision notes

A satellite in a circular orbit has its centripetal force provided by the gravitational attraction of the planet: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$. The satellite mass $m$ cancels, giving

$v = \sqrt{\dfrac{GM}{r}}$

so orbital speed depends only on the orbital radius and the mass of the planet. Larger orbits are slower. The period is $T = \dfrac{2\pi r}{v}$.

Worked example: at $r = 6.97\times10^{6}$ m, $v = 7.6\times10^{3}$ m s⁻¹.

Using $v = 2\pi r/T$ in $GM/r^2 = v^2/r$:

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

Since $4\pi^2/GM$ is constant for a given planet, $T^2 \propto r^3$. Plotting $T^2$ against $r^3$ gives a straight line through the origin with gradient $4\pi^2/GM$.

Kinetic energy: $E_k = \tfrac{1}{2}mv^2 = \dfrac{GMm}{2r}$. Potential energy: $E_p = -\dfrac{GMm}{r}$. Total energy: $E = E_k + E_p = -\dfrac{GMm}{2r}$.

The total is negative, showing the satellite is bound. Note $E_p = -2E_k$ and $E = -E_k$. Moving to a larger orbit increases the total energy (less negative) but decreases the speed and kinetic energy.

The escape velocity is the minimum speed at the surface needed to escape the gravitational field (reach infinity). The kinetic energy must at least cancel the potential energy so the total energy is zero: $\tfrac{1}{2}mv^2 = \dfrac{GMm}{R}$, giving

$v_{esc} = \sqrt{\dfrac{2GM}{R}}$

For the Earth $v_{esc} = 1.1\times10^{4}$ m s⁻¹. It is independent of the mass of the escaping body, and is $\sqrt{2}$ times the orbital speed just above the surface.

A synchronous orbit has a period equal to the Earth's rotation period. A geostationary orbit is a synchronous orbit that is in the equatorial plane, travels in the same direction as the Earth's rotation, and has radius about 4.2 × 10⁷ m (height about 3.6 × 10⁷ m). The satellite stays above a fixed point, which suits communications and broadcast television.

Low orbits (a few hundred km up, often polar) have short periods of about 90 minutes. They give higher-resolution images, and as the Earth rotates beneath them they cover the whole surface, which suits Earth observation and weather monitoring.