Analyse the orbital motion of satellites by equating gravitational force to the centripetal force requirement, deriving orbital speed and period.

What this dot point is asking

You need to analyse satellite orbits by equating the gravitational force to the centripetal force, and use the result to find orbital speed, period and radius.

Gravity as the centripetal force

A satellite in a circular orbit needs a net inward force of $mv^2/r$. The only force acting is gravity, so gravity must supply exactly this:

Cancelling and rearranging for orbital speed:

A lower orbit (smaller $r$) means a faster speed. A satellite cannot choose its orbital speed independently of its radius; each radius has exactly one circular-orbit speed.

Orbital period and Kepler's third law

The period is $T = \dfrac{2\pi r}{v}$. Substituting the orbital speed gives:

This means a satellite in a higher orbit has both a slower speed and a longer period.

Geostationary orbits

A geostationary satellite has a period of exactly one sidereal day so it stays above the same point on the equator. Solving $T^2 = \dfrac{4\pi^2 r^3}{GM}$ for $T = 24 \text{ h}$ gives an orbital radius of about $4.2\times10^7 \text{ m}$ (roughly 36 000 km altitude). These orbits are used for communications and weather satellites.

"Weightlessness" in orbit

Astronauts feel weightless not because gravity is absent - at the ISS gravity is still about 90% of its surface value - but because they and the spacecraft are in continuous free fall around the Earth, accelerating together at the same rate, so there is no normal force between them.