AQA AQA 2019-style practice: The Moon has mass 7.35 × 10^22 kg and radius 1.74 × 10^6 m. Calculate the gravitational field strength at its surface. Take G = 6.67 × 10^-11 N m^2 kg^-2.

Use the radial field formula g = GMr^2, with r the radius of the Moon. g = (6.67 × 10^-11)(7.35 × 10^22)(1.74 × 10^6)^2. The denominator is 3.03 × 10^12, so g = 4.90 × 10^123.03 × 10^12 = 1.6 N kg^-1, about one sixth of the value at the Earth's surface. Markers reward using the radius as r, correct substitution, and a sensible answer compared with Earth's gravity.

EnglandPhysicsSyllabus dot point

How can a mass exert a force on another mass across empty space?

The concept of a force field, Newton's law of gravitation, gravitational field strength as a vector, and the radial and uniform field models.

A focused answer to AQA A-Level Physics 3.7.2, covering the concept of a force field, Newton's law of gravitation, gravitational field strength g as force per unit mass, and the radial and near-uniform field models.

Generated by Claude Opus 4.89 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
Force fields
Newton's law of gravitation
Gravitational field strength
Radial and uniform fields
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What this dot point is asking

AQA specification point 3.7.2 wants you to understand the idea of a force field, state and use Newton's law of gravitation, define gravitational field strength as force per unit mass, and distinguish the radial field of a point mass from the near-uniform field close to a planet's surface.

Force fields

The field-line representation captures both direction (the lines point the way a test mass would be pushed) and strength (closer lines mean a stronger field). The field concept lets us explain action at a distance without invoking a physical contact.

Newton's law of gravitation

The force is always attractive and acts along the line joining the centres. It obeys an inverse square law, so doubling the separation quarters the force. For spherical masses, the law applies as if all the mass were concentrated at the centre, which is why $r$ is measured between centres. The constant $G$ is extremely small, which is why gravitational forces between everyday objects are imperceptible and only become significant for astronomical masses.

Gravitational field strength

The units $\text{N kg}^{-1}$ are equivalent to $\text{m s}^{-2}$ , which is why $g$ is numerically equal to the acceleration of free fall. For a point mass or uniform sphere of mass $M$ , combining $F = \dfrac{GMm}{r^2}$ with $g = \dfrac{F}{m}$ gives a radial field:

Radial and uniform fields

Around a planet the field lines are radial, pointing inwards towards the centre, and $g$ falls off as an inverse square with distance from the centre. Close to the surface, over distances small compared with the planet's radius, the field lines are effectively parallel and equally spaced, so $g$ is approximately constant and the field is treated as uniform. This near-surface approximation is what justifies using a constant $g = 9.81 \text{ N kg}^{-1}$ for projectile problems on Earth.

Field strength at the Earth's surface

Estimate $g$ at the Earth's surface using $M = 5.97 \times 10^{24} \text{ kg}$ and $r = 6.37 \times 10^6 \text{ m}$ .

step 1: Choose the radial formula

For a point or spherical mass, $g = \dfrac{GM}{r^2}$ .

step 2: Substitute the values

$g = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.37 \times 10^6)^2}$ .

step 3: Evaluate the denominator

$(6.37 \times 10^6)^2 = 4.06 \times 10^{13} \text{ m}^2$ .

step 4: Compute and interpret

$g = \dfrac{3.98 \times 10^{14}}{4.06 \times 10^{13}} = 9.81 \text{ N kg}^{-1}$ , which equals the familiar acceleration of free fall.

Try this

Q1. State how the gravitational force between two masses changes if their separation is tripled. [1 mark]

Cue. It becomes one ninth of the original (inverse square law).

Q2. Define gravitational field strength. [1 mark]

Cue. The gravitational force per unit mass acting on a small test mass at that point.

Q3. State the direction of the gravitational field lines around a planet. [1 mark]

Cue. Radially inwards, towards the centre.

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 20194 marksThe Moon has mass

7.35 \times 10^{22} \text{ kg}

and radius

1.74 \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 →

Use the radial field formula $g = \dfrac{GM}{r^2}$ , with $r$ the radius of the Moon.

$g = \dfrac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}$ .

The denominator is $3.03 \times 10^{12}$ , so $g = \dfrac{4.90 \times 10^{12}}{3.03 \times 10^{12}} = 1.6 \text{ N kg}^{-1}$ , about one sixth of the value at the Earth's surface.

Markers reward using the radius as $r$ , correct substitution, and a sensible answer compared with Earth's gravity.

AQA 20223 marksState Newton's law of gravitation in words, and explain what is meant by the statement that the field around a point mass obeys an inverse square law.

Show worked answer →

Newton's law of gravitation states that the gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres, and is always attractive.

An inverse square law means the field strength (or force) is proportional to $\dfrac{1}{r^2}$ , so if the distance from the mass is doubled, the field strength falls to a quarter; if tripled, to a ninth.

Markers reward the proportionality to the product of masses, the inverse square of separation, the attractive nature, and a correct description of the $\dfrac{1}{r^2}$ behaviour.

Related dot points

Sources & how we know this

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