SACE Board SACE 2023-style practice: Skykraft satellites orbit Earth at an altitude of 5.3010^5\ m in a circular orbit. Earth's radius is 6.3710^6\ m and mass 5.9710^24\ kg. Calculate the orbital period of a Skykraft satellite.

First the orbital radius: r = 6.3710^6 + 5.3010^5 = 6.9010^6\ m. Equating gravity to the centripetal requirement gives Kepler's third law, T^2 = 4π^2 r^3GM. $T^2 = 4π^2(6.9010^6)^3(6.6710^-11)(5.9710^24) = (39.48)(3.28510^20)3.98210^14 = 3.25710^7.$ $T = 5.7110^3\ s\ (about 95\ minutes).$ 1 mark for the orbital radius, 1 mark for substitution into Kepler's third law, 1 mark for the answer.

SACE Board SACE 2025-style practice: BlackSky Global-4 passes above South Australia in a circular orbit of radius 6.8710^6\ m. Earth's mass is 5.9710^24\ kg. Calculate the orbital speed of the satellite.

For a circular orbit, gravity supplies the centripetal force, so GMmr^2 = mv^2r, giving v = GMr. $v = (6.6710^-11)(5.9710^24)6.8710^6 = 3.98210^146.8710^6 = sqrt(5.79610^7) = 7.6110^3\ m s^-1.$ The orbital speed is about 7.6\ km s^-1. 1 mark for the expression v = sqrt(GM/r), 1 mark for the answer.

§-Syllabus dot point

SAPhysicsSyllabus dot point

What determines the speed and period of a satellite, and why does the same gravitational force produce a stable orbit?

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

How gravity provides the centripetal force for a satellite, leading to expressions for orbital speed and period, Kepler's third law, and the idea of geostationary orbits.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
Gravity as the centripetal force
Orbital period and Kepler's third law
Geostationary orbits
"Weightlessness" in orbit
How SACE assesses this

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.

A low Earth orbit

This example finds the orbital speed and period of a satellite at $400\ \text{km}$ altitude, using $M = 5.97\times10^{24}\ \text{kg}$ , Earth radius $= 6.37\times10^6\ \text{m}$ , and $G = 6.67\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ .

Step 1: compute the orbital radius

Orbital radius is measured from the centre of Earth, not from the surface. Adding the altitude to Earth's radius gives the distance from the centre:

r = 6.37\times10^6 + 4.00\times10^5 = 6.77\times10^6\ \text{m}.

Step 2: compute the orbital speed

Equating the gravitational force to the centripetal requirement gives $v = \sqrt{GM/r}$ . The speed depends only on the central mass and the orbital radius, not on the satellite's own mass:

v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{6.77\times10^6}} = \sqrt{5.88\times10^7} \approx 7.7\times10^3\ \text{m s}^{-1}.

Step 3: compute the orbital period

The period is the circumference of the orbit divided by the orbital speed. Substituting the values from Steps 1 and 2:

T = \frac{2\pi r}{v} = \frac{2\pi(6.77\times10^6)}{7.67\times10^3} \approx 5.5\times10^3\ \text{s} \approx 92\ \text{minutes}.

Final answer: orbital speed $\approx 7.7\times10^3\ \text{m s}^{-1}$ and period $\approx 92\ \text{minutes}$ , matching the roughly 90-minute orbit of the International Space Station.

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:

v = \sqrt{\frac{GM}{r}}.

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 because the ground antenna can stay fixed.

"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. This apparent weightlessness is the same effect you feel for an instant at the top of a roller-coaster drop.

How SACE assesses this

SACE Stage 2 orbital questions typically give a satellite's altitude (which you must add to the planet's radius to get $r$ ) or its orbital radius directly, then ask for the orbital speed or period. The reliable method is to start from $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , cancel $m$ , and pick the right rearrangement: $v = \sqrt{\dfrac{GM}{r}}$ for speed, or $T^2 = \dfrac{4\pi^2 r^3}{GM}$ for period. Show the cube of the radius (or the substitution into $v$ ) as an explicit step, since the markers award method marks for the correct equation before the final value. A short closing sentence comparing the answer to a known orbit (such as the roughly 90-minute ISS orbit) demonstrates physical sense.

Exam-style practice questions

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

SACE 20233 marksSkykraft satellites orbit Earth at an altitude of

5.30\times10^5\ \text{m}

in a circular orbit. Earth's radius is

6.37\times10^6\ \text{m}

and mass

5.97\times10^{24}\ \text{kg}

. Calculate the orbital period of a Skykraft satellite.

Show worked answer →

First the orbital radius: $r = 6.37\times10^6 + 5.30\times10^5 = 6.90\times10^6\ \text{m}$ .

Equating gravity to the centripetal requirement gives Kepler's third law, $T^2 = \dfrac{4\pi^2 r^3}{GM}$ .

T^2 = \dfrac{4\pi^2(6.90\times10^6)^3}{(6.67\times10^{-11})(5.97\times10^{24})} = \dfrac{(39.48)(3.285\times10^{20})}{3.982\times10^{14}} = 3.257\times10^7.

T = 5.71\times10^3\ \text{s}\ (\text{about } 95\ \text{minutes}).

1 mark for the orbital radius, 1 mark for substitution into Kepler's third law, 1 mark for the answer.

SACE 20252 marksBlackSky Global-4 passes above South Australia in a circular orbit of radius

6.87\times10^6\ \text{m}

. Earth's mass is

5.97\times10^{24}\ \text{kg}

. Calculate the orbital speed of the satellite.

Show worked answer →

For a circular orbit, gravity supplies the centripetal force, so $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , giving $v = \sqrt{\dfrac{GM}{r}}$ .

v = \sqrt{\dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{6.87\times10^6}} = \sqrt{\dfrac{3.982\times10^{14}}{6.87\times10^6}} = \sqrt{5.796\times10^7} = 7.61\times10^3\ \text{m s}^{-1}.

The orbital speed is about

7.6\ \text{km s}^{-1}

. 1 mark for the expression

v = \sqrt{GM/r}

, 1 mark for the answer.