Calculate the gravitational force between two 5.0 × 10^3\ kg masses 2.0\ m apart. Take G = 6.67 × 10^-11\ N m^2\ kg^-2. [2 marks]

OCR OCR 2018-style practice: The Earth has mass 6.0 × 10^24\ kg and radius 6.4 × 10^6\ m. Calculate the gravitational field strength at its surface. Take G = 6.67 × 10^-11\ N m^2\ kg^-2.

Radial field strength: g = (GM)/(r^2) = (6.67 × 10^-11)(6.0 × 10^24)(6.4 × 10^6)^2. Numerator: (6.67 × 10^-11)(6.0 × 10^24) = 4.00 × 10^14. Denominator: (6.4 × 10^6)^2 = 4.10 × 10^13. So g = 4.00 × 10^144.10 × 10^13 = 9.8\ N kg^-1. Markers reward g = (GM)/(r^2), the substitution, and the value about 9.8\ N kg^-1 (matching the familiar surface gravity).

OCR OCR 2021-style practice: A satellite is to be placed in a geostationary orbit around the Earth (mass 6.0 × 10^24\ kg). Calculate the radius of this orbit. Take G = 6.67 × 10^-11\ N m^2\ kg^-2 and a period of 24 hours.

Equate gravitational force to centripetal force: (GMm)/(r^2) = mω^2 r, giving r^3 = (GM)/(ω^2). Period: T = 24 × 3600 = 86\,400\ s, so ω = (2π)/(T) = 7.27 × 10^-5\ rad s^-1. r^3 = (6.67 × 10^-11)(6.0 × 10^24)(7.27 × 10^-5)^2 = 4.00 × 10^145.29 × 10^-9 = 7.57 × 10^22. r = (7.57 × 10^22)^1/3 = 4.2 × 10^7\ m. Markers reward equating the two forces, the period in seconds, and the radius about 4.2 × 10^7\ m.

EnglandPhysicsSyllabus dot point

How does gravity vary with distance, and what determines the orbits of planets and satellites?

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

A focused answer to the OCR H556 gravitational fields content, covering Newton's law of gravitation, gravitational field strength for radial and uniform fields, gravitational potential and potential energy, the motion of satellites with Kepler's third law, geostationary orbits, and escape velocity.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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Quick answer

Newton's law of gravitation gives the attractive force between two point (or spherical) masses, $F = \frac{GMm}{r^2}$ , with $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ . Gravitational field strength is the force per unit mass, $g = \frac{F}{m}$ , equal to $\frac{GM}{r^2}$ for a radial field and roughly uniform near a planet's surface. Gravitational potential is the work done per unit mass bringing a small mass from infinity to a point, $V = -\frac{GM}{r}$ (negative because the field is attractive), and potential energy is $E_p = -\frac{GMm}{r}$ . For an orbit, gravity provides the centripetal force, giving Kepler's third law $T^2 \propto r^3$ . A geostationary orbit has a period of 24 hours over the equator.

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What this dot point is asking

OCR wants you to state and use Newton's law of gravitation, define and calculate gravitational field strength for radial and uniform fields, define gravitational potential and potential energy, analyse the motion of satellites including Kepler's third law and geostationary orbits, and use the idea of escape velocity.

The answer

Newton's law of gravitation

Gravitational field strength

Field strength is a vector pointing towards the mass producing the field; field lines are radial for a point mass and parallel near a surface.

Gravitational potential and potential energy

The escape velocity is the minimum launch speed to reach infinity with zero kinetic energy. Equating kinetic energy to the change in potential energy gives $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$ .

Satellites, Kepler's third law and geostationary orbits

Orbital speed of a satellite

A satellite orbits the Earth (mass $6.0 \times 10^{24}\ \text{kg}$ ) at a radius of $7.0 \times 10^{6}\ \text{m}$ . Find its orbital speed. Take $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ .

step 1: Use the orbital speed expression

Gravity is the centripetal force, so $v = \sqrt{\frac{GM}{r}}$ .

step 2: Substitute

$v = \sqrt{\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{7.0 \times 10^{6}}} = \sqrt{\frac{4.00 \times 10^{14}}{7.0 \times 10^{6}}}$ .

step 3: Evaluate

$v = \sqrt{5.72 \times 10^{7}} = 7.6 \times 10^{3}\ \text{m s}^{-1}$ , about $7.6\ \text{km s}^{-1}$ .

Examples in context

Geostationary satellites carry television, weather imaging and communications, parked over the equator so dishes can point at a fixed spot in the sky. Low polar orbits give global coverage for Earth observation and the GPS constellation uses precisely known orbits to fix position. Kepler's third law lets astronomers weigh stars and planets from the orbital periods and radii of their companions. The concept of escape velocity sets the speed a rocket must reach to leave a planet without further propulsion.

Try this

Q1. State Newton's law of gravitation in words. [2 marks]

Cue. The attractive force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation.

Q2. Calculate the gravitational force between two $5.0 \times 10^{3}\ \text{kg}$ masses $2.0\ \text{m}$ apart. Take $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ . [2 marks]

Cue. $F = \frac{GMm}{r^2} = \frac{(6.67 \times 10^{-11})(5.0 \times 10^3)^2}{2.0^2} = 4.2 \times 10^{-4}\ \text{N}$ .

Q3. State two features that distinguish a geostationary orbit. [2 marks]

Cue. A period of 24 hours and an equatorial orbit in the same direction as the Earth's rotation (so the satellite stays above one point).

Exam-style practice questions

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

OCR 20184 marksThe Earth has mass

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

and radius

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

. Calculate the gravitational field strength at its surface. Take

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

Show worked answer →

Radial field strength: $g = \frac{GM}{r^2} = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{(6.4 \times 10^{6})^2}$ .

Numerator: $(6.67 \times 10^{-11})(6.0 \times 10^{24}) = 4.00 \times 10^{14}$ . Denominator: $(6.4 \times 10^{6})^2 = 4.10 \times 10^{13}$ .

So $g = \frac{4.00 \times 10^{14}}{4.10 \times 10^{13}} = 9.8\ \text{N kg}^{-1}$ .

Markers reward $g = \frac{GM}{r^2}$ , the substitution, and the value about $9.8\ \text{N kg}^{-1}$ (matching the familiar surface gravity).

OCR 20215 marksA satellite is to be placed in a geostationary orbit around the Earth (mass

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

). Calculate the radius of this orbit. Take

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

and a period of

24

hours.

Show worked answer →

Equate gravitational force to centripetal force: $\frac{GMm}{r^2} = m\omega^2 r$ , giving $r^3 = \frac{GM}{\omega^2}$ .

Period: $T = 24 \times 3600 = 86\,400\ \text{s}$ , so $\omega = \frac{2\pi}{T} = 7.27 \times 10^{-5}\ \text{rad s}^{-1}$ .

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

$r = (7.57 \times 10^{22})^{1/3} = 4.2 \times 10^{7}\ \text{m}$ .

Markers reward equating the two forces, the period in seconds, and the radius about $4.2 \times 10^{7}\ \text{m}$ .

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)