EnglandPhysicsSyllabus dot point

What keeps a satellite in orbit, and why do geostationary satellites sit at one fixed height?

Orbital motion under gravity, the link to centripetal force, Kepler's third law, the energy of an orbiting body, and synchronous and geostationary orbits.

A focused answer to AQA A-Level Physics 3.7.2.4, covering orbital motion under gravity, the link between gravitational and centripetal force, Kepler's third law, orbital energy, and synchronous and geostationary satellites.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
Gravity as the centripetal force
Kepler's third law
Energy of an orbiting body
Synchronous and geostationary orbits
Try this

What this dot point is asking

AQA specification point 3.7.2.4 wants you to treat an orbiting body using gravity as the centripetal force, derive and use Kepler's third law, discuss the energy of an orbiting body, and describe synchronous and geostationary orbits and their uses.

Gravity as the centripetal force

A satellite in orbit is in free fall: there is no outward force balancing gravity, and gravity continuously pulls it towards the planet, curving its straight-line path into a circle. For a circular orbit the gravitational force supplies exactly the centripetal force needed:

The satellite's mass cancels, so all satellites at a given radius orbit at the same speed regardless of their mass. Satellites further out move more slowly, because $v \propto \dfrac{1}{\sqrt{r}}$ .

Kepler's third law

Using $v = \dfrac{2\pi r}{T}$ (the orbital circumference divided by the period) in the orbital speed relation, and squaring, gives:

This is Kepler's third law: the square of the orbital period is proportional to the cube of the orbital radius. The constant of proportionality depends only on the mass of the central body, which is why measuring a moon's period and radius lets you find the planet's mass.

Energy of an orbiting body

Synchronous and geostationary orbits

Geostationary orbits are ideal for communications and broadcasting because ground aerials can point in a fixed direction. Low polar orbits, by contrast, have short periods and scan the whole surface as the Earth turns beneath them, which suits weather and mapping satellites.

Finding the radius of a geostationary orbit

Find the orbital radius for a period $T = 24 \text{ h} = 86400 \text{ s}$ , with $GM = 3.99 \times 10^{14} \text{ m}^3 \text{ s}^{-2}$ .

step 1: Rearrange Kepler's third law

From $T^2 = \dfrac{4\pi^2}{GM}r^3$ , the radius cubed is $r^3 = \dfrac{GMT^2}{4\pi^2}$ .

step 2: Substitute the values

$r^3 = \dfrac{(3.99 \times 10^{14})(86400)^2}{4\pi^2}$ .

step 3: Evaluate

$(86400)^2 = 7.46 \times 10^9$ , so the numerator is $2.98 \times 10^{24}$ and $r^3 = \dfrac{2.98 \times 10^{24}}{39.5} = 7.54 \times 10^{22} \text{ m}^3$ .

step 4: Take the cube root

$r = (7.54 \times 10^{22})^{1/3} = 4.23 \times 10^7 \text{ m}$ , about $36000 \text{ km}$ above the surface.

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. Give one reason a geostationary satellite is useful for communications. [1 mark]

Cue. It stays above a fixed point, so ground aerials can point in a constant direction.

Q3. State how the orbital speed of a satellite changes as the orbital radius increases. [1 mark]

Cue. It decreases, since $v \propto \dfrac{1}{\sqrt{r}}$ .

Exam-style practice questions

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

AQA 20195 marksA geostationary satellite orbits the Earth with a period of

24 \text{ hours}

. Taking

GM = 3.99 \times 10^{14} \text{ m}^3 \text{ s}^{-2}

, calculate the radius of its orbit and explain two features a geostationary orbit must have.

Show worked answer →

Use Kepler's third law $T^2 = \dfrac{4\pi^2}{GM}r^3$ , with $T = 24 \times 3600 = 86400 \text{ s}$ .

$r^3 = \dfrac{GMT^2}{4\pi^2} = \dfrac{(3.99 \times 10^{14})(86400)^2}{4\pi^2} = 7.54 \times 10^{22} \text{ m}^3$ .

$r = (7.54 \times 10^{22})^{1/3} = 4.2 \times 10^7 \text{ m}$ from the Earth's centre, about $36000 \text{ km}$ above the surface.

A geostationary orbit must be equatorial (directly above the equator) and the satellite must travel west to east in the same sense as the Earth's rotation, so it stays above one fixed point.

Markers reward the correct period in seconds, the rearrangement and cube root, and two valid features (equatorial plane, same rotational sense, 24 hour period).

AQA 20214 marksShow that for a satellite in a circular orbit the orbital speed is given by

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

, and state how the speed changes for a satellite in a higher orbit.

Show worked answer →

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

Cancel one factor of $m$ and one factor of $r$ : $\dfrac{GM}{r} = v^2$ , so $v = \sqrt{\dfrac{GM}{r}}$ .

Because $v \propto \dfrac{1}{\sqrt{r}}$ , a satellite in a higher orbit (larger $r$ ) moves more slowly.

Markers reward equating gravitational and centripetal force, the correct algebra to isolate $v$ , and stating that a higher orbit means a lower orbital speed.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)