The Earth has mass 6.0 × 10^24\ kg and radius 6.4 × 10^6\ m. Find the gravitational field strength at its surface. Take G = 6.7 × 10^-11\ N m^2\,kg^-2.

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How does gravity act over distance, and what governs the orbits of planets?

Newton's law of gravitation, gravitational field strength and potential, the motion of satellites including geostationary orbits, and Kepler's third law.

A CCEA A-Level Physics answer on Newton's law of gravitation, gravitational field strength and gravitational potential, the motion of satellites including geostationary orbits, and how Kepler's third law follows from gravity.

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

CCEA wants you to state and apply Newton's law of gravitation, define gravitational field strength and gravitational potential, analyse the motion of satellites including geostationary orbits, and derive and use Kepler's third law. Geostationary "show that" questions and potential calculations recur frequently.

The answer

Newton's law of gravitation

Field strength and potential

Field strength follows an inverse-square law and is a vector (a field line direction); potential follows an inverse law and is a scalar with a sign. The two are linked: $g = -\frac{\mathrm{d}V}{\mathrm{d}r}$ , so the field strength is the negative gradient of the potential, and surfaces of constant potential (equipotentials) are everywhere perpendicular to the field lines.

Satellites and Kepler's third law

For a satellite in a circular orbit, gravity supplies the centripetal force:

\frac{GMm}{r^2} = \frac{mv^2}{r}.

A geostationary satellite orbits in the plane of the equator with a period of exactly $24$ hours, in the same direction as the Earth's rotation, so it stays above the same point on the ground. This is ideal for communications and broadcasting because a ground dish can point at a fixed spot in the sky.

Worked example: orbital speed of a low satellite

Speed and period of a satellite just above the surface

A satellite orbits at a radius of $7.0 \times 10^{6}\ \text{m}$ from the Earth's centre. Using $M = 6.0 \times 10^{24}\ \text{kg}$ and $G = 6.67 \times 10^{-11}$ , find its orbital speed and period.

step Equate gravity to the centripetal force

\frac{GMm}{r^2} = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{GM}{r}}.

step Substitute the numbers

v = \sqrt{\frac{(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}.

step Find the period

T = \frac{2\pi r}{v} = \frac{2\pi \times 7.0 \times 10^{6}}{7.6 \times 10^{3}} = 5.8 \times 10^{3}\ \text{s} \approx 96\ \text{minutes}.

This is the familiar low-Earth-orbit period, close to that of the International Space Station.

Examples in context

Example 1. GPS satellites. The Global Positioning System uses satellites at a radius of about $2.66 \times 10^{7}\ \text{m}$ (a $12$ hour period), higher than the ISS but lower than geostationary. Kepler's third law fixes the period once the radius is chosen, and the system relies on knowing each satellite's position to within metres, which is why relativistic corrections to the on-board clocks matter.

Example 2. Escape and the negative potential. To escape Earth a probe must gain enough kinetic energy to reach zero total energy, where the negative gravitational potential energy $-\frac{GMm}{r}$ is exactly cancelled. Setting $\tfrac{1}{2}mv_{\text{esc}}^2 = \frac{GMm}{r}$ gives $v_{\text{esc}} = \sqrt{\frac{2GM}{r}} \approx 11.2\ \text{km s}^{-1}$ at the surface, independent of the probe's mass.

Try this

Q1. The Earth has mass $6.0 \times 10^{24}\ \text{kg}$ and radius $6.4 \times 10^{6}\ \text{m}$ . Find the gravitational field strength at its surface. Take $G = 6.7 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ . [2 marks]

Cue. $g = \frac{GM}{r^2} = \frac{(6.7 \times 10^{-11})(6.0 \times 10^{24})}{(6.4 \times 10^{6})^2} \approx 9.8\ \text{N kg}^{-1}$ .

Q2. State two features of a geostationary orbit. [2 marks]

Cue. A period of $24$ hours and an orbit above the equator, in the same direction as the Earth's rotation.

Q3. Two planets orbit a star with radii in the ratio $1 : 4$ . Find the ratio of their orbital periods. [2 marks]

Cue. $T^2 \propto r^3$ , so $T_2 / T_1 = 4^{3/2} = 8$ ; the periods are in the ratio $1 : 8$ .

Exam-style practice questions

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

CCEA 20196 marksA satellite is to be placed in a geostationary orbit around the Earth. Take the mass of the Earth as 6.0 times 10 to the 24 kg and G as 6.67 times 10 to the minus 11. Show that the orbital radius is about 4.2 times 10 to the 7 m, and hence determine the satellite's orbital speed.

Show worked answer →

For a circular orbit, gravity supplies the centripetal force:

$\frac{GMm}{r^2} = m\omega^2 r$ , so $r^3 = \frac{GM}{\omega^2}$ .

A geostationary satellite has a period equal to one day, $T = 24 \times 3600 = 8.64 \times 10^{4}$ s, so

$\omega = \frac{2\pi}{T} = \frac{2\pi}{8.64 \times 10^{4}} = 7.27 \times 10^{-5}$ rad per second.

Therefore

$r^3 = \frac{GM}{\omega^2} = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{(7.27 \times 10^{-5})^2} = \frac{4.00 \times 10^{14}}{5.29 \times 10^{-9}} = 7.57 \times 10^{22}$ ,

so $r = (7.57 \times 10^{22})^{1/3} = 4.23 \times 10^{7}$ m, as required.

The orbital speed is $v = \omega r = 7.27 \times 10^{-5} \times 4.23 \times 10^{7} = 3.1 \times 10^{3}$ m per second.

Markers reward the centripetal condition, correct use of the 24 hour period, and a clear cube root giving the stated radius.

CCEA 20224 marksDefine gravitational potential at a point in a field. Explain why gravitational potential is always negative, and calculate the work done in moving a 500 kg probe from the Earth's surface (r equals 6.4 times 10 to the 6 m) to a point where r equals 1.3 times 10 to the 7 m.

Show worked answer →

Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.

It is always negative because gravity is attractive: the field does positive work on a mass moving inward from infinity, so the potential energy decreases below the zero defined at infinity. Hence $V = -\frac{GM}{r}$ is negative everywhere except at infinity, where it is zero.

The work done equals the change in potential energy, $W = m\,\Delta V = m\left(-\frac{GM}{r_2} + \frac{GM}{r_1}\right)$ :

$V_1 = -\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{6.4 \times 10^{6}} = -6.25 \times 10^{7}$ J per kg.

$V_2 = -\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{1.3 \times 10^{7}} = -3.08 \times 10^{7}$ J per kg.

$W = m(V_2 - V_1) = 500 \times (-3.08 \times 10^{7} + 6.25 \times 10^{7}) = 500 \times 3.17 \times 10^{7} = 1.6 \times 10^{10}$ J.

Markers reward the definition referencing infinity, the attractive-force reasoning, and a positive work value from the potential difference.

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Sources & how we know this

CCEA GCE Physics specification — CCEA (2016)