Show that the orbital speed of a satellite is v = sqrt((GM)/(r)). [2 marks]

WJEC WJEC 2020-style practice: A geostationary satellite orbits the Earth with a period of 24\,hours. Given the Earth's mass is 6.0 × 10^24\,kg and G = 6.67 × 10^-11\,N m^2\,kg^-2, calculate the radius of its orbit.

Use Kepler's third law in the form T^2 = 4π^2GMr^3, rearranged for r. Convert the period: T = 24 × 3600 = 8.64 × 10^4\,s. Rearrange: r^3 = GM T^24π^2 = 6.67 × 10^-11 × 6.0 × 10^24 × (8.64 × 10^4)^24π^2. Numerator: 6.67 × 10^-11 × 6.0 × 10^24 = 4.0 × 10^14; times (8.64 × 10^4)^2 = 7.46 × 10^9 gives 2.99 × 10^24. Dividing by 4π^2 = 39.5 gives r^3 = 7.56 × 10^22. r = (7.56 × 10^22)^1/3 = 4.2 × 10^7\,m. Markers reward converting the period to seconds, the correct rearrangement, and the radius of about 4.2 × 10^7\,m (around 36\,000\,km above the surface).

WalesPhysicsSyllabus dot point

How does gravity govern orbits, and what does the wider universe reveal about mass we cannot see?

Circular orbits under gravity, Kepler's third law, satellites and escape, and evidence for dark matter from galactic rotation.

A focused answer to WJEC A-Level Physics Unit 4 orbits and the wider universe, covering circular orbits under gravity, Kepler's third law, satellites, and the evidence for dark matter from galactic rotation curves.

Generated by Claude Opus 4.811 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
The answer
Examples in context
Try this

What this dot point is asking

WJEC wants you to treat a circular orbit as gravity providing the centripetal force, derive and use Kepler's third law, describe satellites, and explain the evidence for dark matter from galactic rotation. This dot point takes the field laws of the previous topic and applies them to the cosmos, ending with one of the great open puzzles of modern physics, the dark-matter problem.

The answer

Circular orbits under gravity

Notice the orbiting body's own mass $m$ cancels: the orbital speed depends only on the central mass $M$ and the radius $r$ , not on how heavy the satellite is.

Kepler's third law

Combining the orbital condition with $v = \dfrac{2\pi r}{T}$ gives Kepler's third law: the square of the orbital period is proportional to the cube of the orbital radius.

Orbital speed of the Moon

The Moon orbits the Earth at a radius of $3.8 \times 10^{8}\,\text{m}$ . With the Earth's mass $6.0 \times 10^{24}\,\text{kg}$ and $G = 6.67 \times 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ , find its orbital speed.

step Use the orbital-speed formula

$v = \sqrt{\dfrac{GM}{r}}$ .

step Substitute

$v = \sqrt{\dfrac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{3.8 \times 10^{8}}}$ .

step Evaluate

The fraction is $\dfrac{4.0 \times 10^{14}}{3.8 \times 10^{8}} = 1.05 \times 10^{6}$ , so $v = \sqrt{1.05 \times 10^{6}} = 1.0 \times 10^{3}\,\text{m s}^{-1}$ , about $1\,\text{km s}^{-1}$ .

Satellites

A geostationary satellite orbits once every 24 hours above the equator, so it stays above a fixed point on Earth, useful for communications. Low polar orbits suit imaging and monitoring because they pass over different strips of the planet on each pass.

Dark matter

Examples in context

Example 1. GPS satellites: The satellites that power satellite navigation orbit at a radius of about $2.6 \times 10^{7}\,\text{m}$ , completing two orbits a day. Their position is found from Kepler's third law, and because $v = \sqrt{GM/r}$ they travel at nearly $4\,\text{km s}^{-1}$ . Knowing their speed and height precisely (and correcting for relativity) is what lets a receiver fix your position to a few metres.
Example 2. Weighing the galaxy: By measuring how fast stars orbit the centre of the Milky Way at a given radius and applying $v = \sqrt{GM(r)/r}$ , astronomers infer the mass enclosed within that radius. The surprisingly high orbital speeds far from the centre reveal far more mass than the visible stars provide, which is the local version of the dark-matter evidence described above.
Example 3. Comparing the planets: Kepler's third law, $T^2 \propto r^3$ , applies to every planet orbiting the Sun with the same constant of proportionality, because $M$ is the Sun's mass in each case. Earth orbits at $1\,\text{AU}$ in one year, so a planet four times further out would have a period of $4^{3/2} = 8$ years. This is how astronomers related the planets' distances to their easily observed orbital periods long before spacecraft confirmed the values.

Try this

Q1. Show that the orbital speed of a satellite is $v = \sqrt{\frac{GM}{r}}$ . [2 marks]

Cue. Equate gravity to centripetal force, $\frac{GMm}{r^2} = \frac{mv^2}{r}$ , then rearrange for $v$ .

Q2. State the evidence from galactic rotation curves for dark matter. [2 marks]

Cue. The curves stay flat at large radii rather than falling, implying extra unseen mass providing gravity.

Exam-style practice questions

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

WJEC 20206 marksA geostationary satellite orbits the Earth with a period of

24\,\text{hours}

. Given the Earth's mass is

6.0 \times 10^{24}\,\text{kg}

and

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

, calculate the radius of its orbit.

Show worked answer →

Use Kepler's third law in the form $T^2 = \dfrac{4\pi^2}{GM}r^3$ , rearranged for $r$ .

Convert the period: $T = 24 \times 3600 = 8.64 \times 10^{4}\,\text{s}$ .

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

Numerator: $6.67 \times 10^{-11} \times 6.0 \times 10^{24} = 4.0 \times 10^{14}$ ; times $(8.64 \times 10^{4})^2 = 7.46 \times 10^{9}$ gives $2.99 \times 10^{24}$ . Dividing by $4\pi^2 = 39.5$ gives $r^3 = 7.56 \times 10^{22}$ .

$r = (7.56 \times 10^{22})^{1/3} = 4.2 \times 10^{7}\,\text{m}$ .

Markers reward converting the period to seconds, the correct rearrangement, and the radius of about $4.2 \times 10^{7}\,\text{m}$ (around $36\,000\,\text{km}$ above the surface).

WJEC 20184 marksExplain why the observed rotation curves of spiral galaxies provide evidence for dark matter.

Show worked answer →

For a star orbiting at radius $r$ from the galactic centre, gravity provides the centripetal force, so the orbital speed is $v = \sqrt{GM(r)/r}$ , where $M(r)$ is the mass enclosed within that radius.

If the visible mass were all there is, then beyond the bright central region $M(r)$ would be roughly constant and the speed should fall as $v \propto 1/\sqrt{r}$ at large radii.

Observations instead show the rotation curve stays flat (roughly constant speed) far out into the dim outskirts. This requires the enclosed mass to keep increasing with radius, meaning there is a large amount of unseen matter, dark matter, providing the extra gravity. Markers reward the expected falling curve from visible mass and the flat observed curve implying extra unseen mass.

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)