What is 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).

What is low Earth Orbit?

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).

What is geostationary Earth Orbit?

Altitude about 35 800 km (radius 4.22 × 10^7 m). Period exactly one sidereal day (about 23 h 56 min). The satellite sits over a fixed equatorial point.

What is medium Earth Orbit?

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

A satellite is in a circular orbit of radius 9.0 × 10^6 m about Earth. Calculate its orbital speed and period. [4 marks]

Two satellites orbit Earth: A at r = 7000 km and B at r = 14000 km. (a) Identify which is faster and by what factor. (b) Find the ratio of orbital periods T_B / T_A.

NSWPhysicsSyllabus dot point

Inquiry Question 3: How does the force of gravity determine the motion of planets and satellites?

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

A focused answer to the HSC Physics Module 5 dot point on orbital motion of artificial satellites. The derivation of orbital speed from gravity-as-centripetal-force, low Earth and geostationary orbits, the worked LEO example, and the patterns markers look for.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-23

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

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 $r$ :

\frac{G M m}{r^2} = \frac{m v^2}{r}

Solving for the orbital speed:

v = \sqrt{\frac{G M}{r}}

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

Orbital period

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

T = \frac{2 \pi r}{v} = 2 \pi \sqrt{\frac{r^3}{G M}}

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.

Orbital radius and speed of a geostationary satellite

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 $^2$ .

Step 1: Rearrange Kepler's Third Law for the orbital radius

A geostationary satellite has the same orbital period as Earth's rotation: $T = 86400$ s (one day). Kepler's Third Law states $T^2 / r^3 = 4\pi^2 / (GM)$ . Rearranging for $r$ :

r = \left(\frac{G M T^2}{4 \pi^2}\right)^{1/3}

Step 2: Substitute and evaluate the cube root

Substitute all values into the bracket:

r = \left(\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{39.48}\right)^{1/3} = \left(7.52 \times 10^{22}\right)^{1/3}

Taking the cube root gives:

r = 4.22 \times 10^7 \text{ m}

This is the radius from Earth's centre, not the altitude above the surface. The altitude is about $35\,800$ km.

Step 3: Calculate the orbital speed

Once the radius is known, the orbital speed follows from equating gravity to the centripetal force, giving $v = \sqrt{GM/r}$ . Using $GM = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} = 3.98 \times 10^{14}$ N m $^2$ kg $^{-1}$ :

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 \text{ m/s}

Final answer: The geostationary orbital radius is $4.22 \times 10^7$ m ( $42\,200$ km from Earth's centre) and the orbital speed is $3070$ m/s, or about $3.07$ km/s.

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.

Examples in context

Example 1. Geostationary NBN Sky Muster satellite over central NSW. NBN's Sky Muster satellites orbit at geostationary altitude $h = 35{,}786 \text{ km}$ , so $r = 4.22 \times 10^7 \text{ m}$ . Required orbital speed is $v = \sqrt{G M_E / r} = \sqrt{6.674 \times 10^{-11} \times 5.97 \times 10^{24} / 4.22 \times 10^7} = 3.07 \times 10^3 \text{ m/s}$ . Period $T = 2\pi r / v = 2\pi \times 4.22 \times 10^7 / 3.07 \times 10^3 = 8.64 \times 10^4 \text{ s} = 23 \text{ h } 56 \text{ min}$ , exactly one sidereal day. This is why a single Sky Muster dish in Bourke or Broken Hill can track-free above the equator at longitude $145^{\circ}$ E and stay fixed.

Example 2. ISS low-Earth orbit visible over Sydney. The International Space Station orbits at $h = 408 \text{ km}$ , so $r = 6.78 \times 10^6 \text{ m}$ . Orbital speed: $v = \sqrt{G M_E / r} = 7.66 \times 10^3 \text{ m/s} = 7.66 \text{ km/s}$ . Period: $T = 2\pi r / v = 5562 \text{ s} = 92.7 \text{ minutes}$ . So it laps Earth roughly $15.5$ times a day. Sydney observers spot it crossing the sky in about $4$ minutes when the orbit favours an evening pass. The astronauts' orbital free-fall is the same as projectile motion, but with the curve of the Earth retreating beneath them at the same rate they fall.

Try this

Q1. Define "geostationary orbit" and state its altitude above Earth's surface. [2 marks]

Cue. Circular orbit above the equator with a $24$ hr (sidereal) period; altitude $\approx 35{,}800 \text{ km}$ .

Q2. A satellite is in a circular orbit of radius $9.0 \times 10^6 \text{ m}$ about Earth. Calculate its orbital speed and period. [4 marks]

Cue. $v = \sqrt{G M_E / r} = 6.65 \times 10^3 \text{ m/s}$ ; $T = 2\pi r / v = 8500 \text{ s} = 142 \text{ min}$ .

Q3. Two satellites orbit Earth: A at $r = 7000 \text{ km}$ and B at $r = 14000 \text{ km}$ . (a) Identify which is faster and by what factor. (b) Find the ratio of orbital periods $T_B / T_A$ . (c) Explain qualitatively how Kepler's third law applies to satellites. [2+2+2 marks]

Cue. (a) A faster by $\sqrt{2}$ (since $v \propto 1/\sqrt{r}$ ). (b) $T \propto r^{3/2}$ , so $T_B / T_A = 2^{1.5} = 2.83$ . (c) Same central mass means $T^2 / r^3$ is constant for any Earth satellite.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC5 marksThe International Space Station orbits at an altitude of 400 km above Earth's surface. Calculate its orbital speed and orbital period. (Mass of Earth = 5.97 x 10^24 kg, radius of Earth = 6.37 x 10^6 m, G = 6.67 x 10^-11 N m^2/kg^2.)

Show worked answer →

Set gravity equal to the centripetal force for a circular orbit.

$\frac{G M m}{r^2} = \frac{m v^2}{r}$ , giving $v = \sqrt{\frac{G M}{r}}$ .

Orbital radius: $r = R_E + h = 6.37 \times 10^6 + 4.0 \times 10^5 = 6.77 \times 10^6$ m.

$v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^6}}$
$v = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^6}}$
$v = \sqrt{5.88 \times 10^7}$
$v = 7670$ m/s, or about $7.67$ km/s.

Orbital period: $T = \frac{2 \pi r}{v} = \frac{2 \pi \times 6.77 \times 10^6}{7670} = 5546$ s, about $92.4$ minutes.

Markers reward the derivation from setting gravity equal to centripetal force, the substitution with $r = R_E + h$ (not just $h$ ), and answers in correct SI units.

2018 HSC4 marksCompare a low Earth orbit (LEO, altitude ~400 km) and a geostationary orbit (altitude ~36000 km). Discuss the differences in orbital speed and period, and explain why geostationary orbits are useful for communications.

Show worked answer →

Orbital speed: $v = \sqrt{G M / r}$ , so speed decreases with increasing radius.

Orbit	Altitude	Radius (from Earth's centre)	Speed	Period
LEO	400 km	$6.77 \times 10^6$ m	7.67 km/s	92 min
Geostationary	36 000 km	$4.22 \times 10^7$ m	3.07 km/s	24 hours

A geostationary satellite orbits at the same angular rate as Earth's rotation (one revolution per sidereal day, in the equatorial plane). To a ground observer it remains fixed in the sky, allowing ground antennas to be aimed permanently without tracking. This is ideal for television broadcast, weather monitoring, and continuous communication links.

LEO satellites move rapidly across the sky and need tracking or large constellations (such as Starlink) to provide continuous coverage. Their lower altitude gives shorter signal delay (about 5 ms versus 240 ms for geostationary), which suits high-bandwidth and low-latency applications.

Markers reward the quantitative comparison, the geostationary geometry (24-hour equatorial), and a clear application-driven contrast.