Predict quantitatively the orbital properties of planets and artificial satellites in a variety of situations, including near-Earth and geostationary orbits, using the relationship between orbital speed, radius, and period

What this dot point is asking

NESA wants you to apply gravitational and circular-motion principles to predict the speed, period, radius, and altitude of an artificial satellite, and to contrast different orbit types (near-Earth, geostationary, and others). This dot point combines Newton's Law of Universal Gravitation with $F = m v^2 / r$ and Kepler's Third Law, and is examined nearly every year.

The answer

A satellite in a stable circular orbit moves under the gravitational pull of the central body alone. Gravity provides the centripetal force.

The fundamental equation

For a satellite of mass $m$ orbiting a central body of mass $M$ at radius $r$:

Solving for the orbital speed:

The satellite's mass cancels. Orbital speed depends only on the central mass and the orbital radius.

Orbital period

Using $v = 2 \pi r / T$:

This is Kepler's Third Law in another form.

Common orbits used in HSC

Low Earth Orbit (LEO). Altitude 200 to 2000 km. Periods 90 to 130 minutes. Used by the ISS, Earth-observation satellites, and Starlink. High orbital speed (about 7 to 8 km/s).

Geostationary Earth Orbit (GEO). Altitude about 35 800 km (radius $4.22 \times 10^7$ m). Period exactly one sidereal day (about 23 h 56 min). The satellite sits over a fixed equatorial point. Used for television, weather imaging, and continuous communications.

Medium Earth Orbit (MEO). Altitude 2 000 to 35 800 km. Used by GPS satellites (about 20 200 km altitude, 12-hour period).

Why orbits stay stable

A satellite in orbit is constantly falling toward Earth, but its tangential velocity carries it sideways fast enough that it falls "around" the curvature of Earth rather than into it. The orbit is the geometric path where gravitational acceleration matches the centripetal requirement at every instant.

If the satellite were faster than orbital speed at a given radius, it would rise to a higher orbit (or escape if above $v_{\text{esc}}$). If slower, it would spiral in.

Atmospheric drag

In low orbits (below about $400$ km), residual atmosphere creates drag, slowly reducing orbital energy. Satellites must boost periodically (the ISS does this every few months) or eventually re-enter.

Worked example with numbers

A communications satellite is to be placed in a geostationary orbit. Calculate its orbital radius and speed. Use $M_E = 5.97 \times 10^{24}$ kg, $T = 86400$ s, $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Radius from Kepler's Third Law:

$r = \left(\frac{G M T^2}{4 \pi^2}\right)^{1/3} = \left(\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{39.48}\right)^{1/3}$

$r = (7.52 \times 10^{22})^{1/3} = 4.22 \times 10^7$ m.

Orbital speed:

$v = \sqrt{\frac{G M}{r}} = \sqrt{\frac{3.98 \times 10^{14}}{4.22 \times 10^7}} = \sqrt{9.43 \times 10^6} = 3070$ m/s.

Geostationary satellites travel at about $3.07$ km/s at a radius of $42 200$ km from Earth's centre.

Try it: Kepler's third law calculator for orbital period and radius, or the orbital energy calculator for $K$, $U$, $E$ at any altitude.

Common traps

Using altitude instead of orbital radius. $r$ is measured from the centre of Earth. For a $400$ km altitude orbit: $r = 6370 + 400 = 6770$ km.

Saying that a satellite is "outside gravity" or "weightless because there is no gravity." Gravity is what holds it in orbit. Astronauts feel weightless because they and the spacecraft are in free fall together, not because gravity is absent.

Forgetting that orbital speed decreases with altitude. Higher orbit means slower speed but longer period. Both period and orbital radius increase together, by $T^2 \propto r^3$.

Treating geostationary orbits as possible at any latitude. A geostationary orbit must be equatorial and prograde. A polar orbit cannot be geostationary.

Confusing geostationary and geosynchronous. Geosynchronous orbits have a 24-hour period but may be inclined to the equator. Geostationary is a special case in the equatorial plane.

In one sentence

A satellite in circular orbit has its gravitational attraction supplying the centripetal force, giving orbital speed $v = \sqrt{G M / r}$ and period $T = 2 \pi \sqrt{r^3 / (G M)}$, with low Earth orbits fast and short and geostationary orbits at $42 200$ km radius matching Earth's 24-hour rotation.