A satellite orbits at radius 1.0 × 10^7\ m around a planet of mass 6.0 × 10^24\ kg. Find its orbital speed (G = 6.67 × 10^-11). [2 marks]

WJEC Eduqas Eduqas 2020-style practice: A satellite orbits the Earth in a circular orbit of radius 7.0 × 10^6\ m. Calculate its orbital speed and period. Take G = 6.67 × 10^-11\ N m^2\ kg^-2 and the mass of the Earth M = 6.0 × 10^24\ kg.

For a circular orbit, gravity provides the centripetal force: GMmr^2 = mv^2r, so v = GMr. v = (6.67 × 10^-11)(6.0 × 10^24)7.0 × 10^6 = 5.72 × 10^7 = 7.6 × 10^3\ m s^-1. Period from v = 2π rT: T = 2π rv = 2π(7.0 × 10^6)7.6 × 10^3 = 5.8 × 10^3\ s, about 96\ minutes. Markers reward equating gravity to the centripetal force giving v = sqrt((GM)/(r)), the speed about 7.6\ km s^-1, and the period about 5.8 × 10^3\ s.

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How does gravity govern the orbits of planets and satellites, and what is special about a geostationary orbit?

Orbits and the wider universe: circular orbits under gravity, Kepler's third law, the energy of an orbiting body, geostationary satellites, and escape velocity.

A focused answer to the Eduqas A-Level Physics Component 2 orbits content, covering circular orbits under gravity, the derivation of Kepler's third law, the energy of an orbiting body, geostationary satellites, and escape velocity.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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Quick answer

For a circular orbit, gravity provides the centripetal force, $\frac{GMm}{r^2} = \frac{mv^2}{r}$ , giving the orbital speed $v = \sqrt{\frac{GM}{r}}$ : bodies in lower orbits move faster. Eliminating $v$ gives Kepler's third law, $T^2 \propto r^3$ (specifically $T^2 = \frac{4\pi^2}{GM}r^3$ ). An orbiting body has kinetic energy and (negative) gravitational potential energy; its total energy is negative and it is bound. A geostationary satellite has a 24-hour period, orbits above the equator in the direction of the Earth's rotation, and so stays above a fixed point, ideal for communications. The escape velocity, the minimum speed to escape a body's gravity, is $v_\text{esc} = \sqrt{\frac{2GM}{r}}$ .

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What this dot point is asking

Eduqas wants you to analyse a circular orbit by equating gravity to the centripetal force, derive and use Kepler's third law, describe the energy of an orbiting body, explain geostationary satellites and their uses, and define escape velocity.

The answer

Circular orbits under gravity

Kepler's third law

Energy of an orbiting body

Geostationary satellites and escape velocity

Orbital radius of a geostationary satellite

Find the radius of a geostationary orbit around the Earth. Take $G = 6.67 \times 10^{-11}$ , $M = 6.0 \times 10^{24}\ \text{kg}$ and $T = 24\ \text{hours}$ .

step 1: Convert the period to seconds

$T = 24 \times 3600 = 8.64 \times 10^{4}\ \text{s}$ .

step 2: Rearrange Kepler's third law for the radius

From $T^2 = \dfrac{4\pi^2}{GM}r^3$ , $r^3 = \dfrac{GMT^2}{4\pi^2} = \dfrac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(8.64 \times 10^{4})^2}{4\pi^2}$ .

step 3: Evaluate

$r^3 = \dfrac{(4.0 \times 10^{14})(7.46 \times 10^{9})}{39.5} = 7.56 \times 10^{22}$ , so $r = (7.56 \times 10^{22})^{1/3} = 4.2 \times 10^{7}\ \text{m}$ , about $36\,000\ \text{km}$ above the surface.

Examples in context

Orbital mechanics underpins the entire space industry: GPS satellites in medium orbit, the International Space Station in low orbit (about 90 minutes per orbit), and the geostationary ring of communications and weather satellites. Kepler's third law lets astronomers weigh stars and planets from the orbits of their companions, and the same energy analysis governs the fuel needed to raise or de-orbit a spacecraft.

Try this

Q1. State Kepler's third law. [1 mark]

Cue. The square of the orbital period is proportional to the cube of the orbital radius, $T^2 \propto r^3$ .

Q2. A satellite orbits at radius $1.0 \times 10^{7}\ \text{m}$ around a planet of mass $6.0 \times 10^{24}\ \text{kg}$ . Find its orbital speed ( $G = 6.67 \times 10^{-11}$ ). [2 marks]

Cue. $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{1.0 \times 10^{7}}} = 6.3 \times 10^{3}\ \text{m s}^{-1}$ .

Q3. State two conditions for a satellite to be geostationary. [2 marks]

Cue. A 24-hour period and an equatorial orbit in the direction of the Earth's rotation.

Exam-style practice questions

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

Eduqas 20205 marksA satellite orbits the Earth in a circular orbit of radius

7.0 \times 10^{6}\ \text{m}

. Calculate its orbital speed and period. Take

G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}

and the mass of the Earth

M = 6.0 \times 10^{24}\ \text{kg}

Show worked answer →

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

$v = \sqrt{\dfrac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{7.0 \times 10^{6}}} = \sqrt{5.72 \times 10^{7}} = 7.6 \times 10^{3}\ \text{m s}^{-1}$ .

Period from $v = \dfrac{2\pi r}{T}$ : $T = \dfrac{2\pi r}{v} = \dfrac{2\pi(7.0 \times 10^{6})}{7.6 \times 10^{3}} = 5.8 \times 10^{3}\ \text{s}$ , about $96\ \text{minutes}$ .

Markers reward equating gravity to the centripetal force giving $v = \sqrt{\frac{GM}{r}}$ , the speed about $7.6\ \text{km s}^{-1}$ , and the period about $5.8 \times 10^{3}\ \text{s}$ .

Eduqas 20224 marksExplain what is meant by a geostationary orbit and state two conditions a satellite must satisfy to be geostationary.

Show worked answer →

A geostationary satellite orbits so that it remains directly above the same point on the Earth's surface at all times, appearing stationary from the ground.

The conditions are: it must have an orbital period of exactly 24 hours (matching the Earth's rotation, strictly one sidereal day); it must orbit directly above the equator in the same direction as the Earth's rotation (west to east); and consequently it sits at a fixed altitude of about $3.6 \times 10^{7}\ \text{m}$ (radius about $4.2 \times 10^{7}\ \text{m}$ from the centre).

Markers reward the definition (stays above the same point), and two valid conditions (24-hour period, equatorial orbit in the direction of rotation, fixed radius).

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)