Orbits and Satellites

For a satellite of mass $m$ in a circular orbit of radius $r$ around a mass $M$, the gravitational force provides the centripetal force.

The satellite’s mass cancels, so all satellites at the same radius move at the same speed. Larger orbits are slower.

Substituting $v=2\pi r/T$ into the orbital speed equation gives a relation between period and radius.

$\text{Kepler’s third law}$: the square of the orbital period is proportional to the cube of the orbit radius. The constant $4{\pi }^2/GM$ depends only on the central mass.

That makes orbits a way to $\text{weigh}$ astronomical bodies: measure the period and radius of any moon or planet, and you have the mass of what it orbits.

A $\text{geostationary}$ satellite stays above the same point on Earth’s surface. It must orbit above the equator, in the same direction as Earth rotates, with a period of 24 hours (strictly, one sidereal day of about 23 h 56 min).

Kepler’s law fixes its radius at about $4.2\times 10^7$ m, some 36 000 km above the surface. Ground dishes can point at it permanently, making it ideal for communications and weather monitoring.

$\text{Low Earth orbits}$, a few hundred kilometres up, have periods of about 90 minutes. Satellites there can pass over the whole planet as Earth turns beneath them, and they are close enough for high-resolution imaging.

Using $v^2=GM/r$, a satellite’s kinetic energy is $E_k=\frac{1}{2}m{v}^2=\frac{GMm}{2r}$.

Its potential energy is $E_p=-\frac{GMm}{r}$. So the total energy is $E=-\frac{GMm}{2r}$ — negative, because the satellite is bound to the planet.

Moving to a higher orbit makes the total energy less negative, so energy must be supplied. But kinetic energy falls, because the satellite moves more slowly in a larger orbit; potential energy rises by twice as much as kinetic energy falls.