Quick questions on Satellite and orbital motion: WACE Year 12 Physics Unit 3

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

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.

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.

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.