Newton's law of universal gravitation (SACE Stage 2 Physics)

Syllabus dot point

“Apply Newton's law of universal gravitation and the concept of gravitational field strength to interactions between masses.”

9 min readUpdated 2 Jun 2026

Quick answer

Every pair of masses attracts with a force $F = G\dfrac{m_1 m_2}{r^2}$ , where $G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ and $r$ is the centre-to-centre separation. The force is an inverse-square law. Gravitational field strength is the force per unit mass, $g = G\dfrac{M}{r^2}$ , which equals the free-fall acceleration.

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What this dot point is asking
The law
Gravitational field strength
Why it matters
How SACE assesses this
Exam-style questions

What this dot point is asking

You need to apply Newton's law of universal gravitation to calculate the attractive force between two masses, understand its inverse-square dependence, and relate it to gravitational field strength.

Force between Earth and a satellite

A 500 kg satellite orbits at a height of 2000 km above Earth's surface

Earth's mass $M = 5.97\times10^{24}\ \text{kg}$ and radius $= 6.37\times10^6\ \text{m}$ . Find the gravitational force on the satellite.

Orbital radius (centre to satellite):

r = 6.37\times10^6 + 2.00\times10^6 = 8.37\times10^6\ \text{m}.

Force:

F = G\frac{Mm}{r^2} = 6.67\times10^{-11}\times\frac{(5.97\times10^{24})(500)}{(8.37\times10^6)^2}

F = 6.67\times10^{-11}\times\frac{2.99\times10^{27}}{7.00\times10^{13}} \approx 2.8\times10^3\ \text{N}.

This force is the centripetal force that keeps the satellite in orbit.

The law

Newton's law of universal gravitation

F = G\frac{m_1 m_2}{r^2}

F

is the attractive force on each mass (N),

m_1

and

m_2

the masses (kg),

r

the distance between their centres (m), and

G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}

the universal gravitational constant.

Key features:

The force is always attractive and acts along the line joining the two centres.
It is mutual: by Newton's third law each mass pulls the other with the same magnitude, regardless of how different the masses are.
It follows an inverse-square law: doubling the separation reduces the force to a quarter; tripling it reduces it to a ninth.
It is directly proportional to each mass: doubling one mass doubles the force.

For spherical bodies such as planets, $r$ is measured from centre to centre, so for an object on Earth's surface $r$ is the radius of the Earth.

Gravitational field strength

A mass creates a gravitational field around it. The field strength $g$ at a point is the force per unit mass placed there.

Gravitational field strength

g = \frac{F}{m} = G\frac{M}{r^2}

g

has units

\text{N kg}^{-1}

, numerically equal to the free-fall acceleration in

\text{m s}^{-2}

M

is the mass producing the field. Field strength also obeys the inverse-square law, so it weakens with the square of the distance from the centre.

This is why $g \approx 9.8\ \text{N kg}^{-1}$ at Earth's surface, and falls off as you move to higher altitude or orbit. At twice Earth's radius from the centre, $g$ has fallen to a quarter of its surface value.

Why it matters

This single law explains the fall of an apple and the orbit of the Moon: the same gravitational interaction governs both. Combined with the centripetal force requirement, it underpins satellite and planetary orbits (covered in the next dot point), and the field-strength idea connects the abstract force law to the everyday acceleration of free fall.

How SACE assesses this

SACE Stage 2 gravitation questions are almost always direct substitutions into $F = \dfrac{Gm_1 m_2}{r^2}$ , usually with real astronomical bodies (a moon and its planet, or a probe and the Sun) and data given in scientific notation. The two recurring traps are squaring the separation correctly and keeping every quantity in SI units. A second common part asks you to explain the mutual, equal-and-opposite nature of the force using Newton's third law, where the symmetry of the formula in $m_1$ and $m_2$ is the key point. Quote the formula, substitute with units, square $r$ as a separate step, and state the answer to an appropriate number of significant figures.

Common errors

Using surface distance instead of centre-to-centre: For a body above a planet, $r$ is the planet's radius plus the height, measured from the planet's centre.
Forgetting the inverse square: It is $r^2$ in the denominator, not $r$ . A common error is to halve the force when the distance doubles, instead of quartering it.
Thinking the larger mass feels a larger force: The forces on the two masses are equal and opposite. The Earth pulls you with the same force you pull the Earth; you accelerate more only because your mass is far smaller.

Confusing $g$ as field strength with $g$ as acceleration. They are numerically equal and have equivalent units, but field strength ( $\text{N kg}^{-1}$ ) is the more general idea.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style4 marks

Titan has a mass of

1.35\times10^{23}\ \text{kg}

and Saturn has a mass of

5.68\times10^{26}\ \text{kg}

. (a) Calculate the magnitude of the gravitational force between Titan and Saturn when their centres are separated by

1.24\times10^9\ \text{m}

. (b) Explain how the gravitational forces are consistent with Newton's third law of motion.

Show worked answer →

(a) Apply Newton's law of universal gravitation, $F = \dfrac{Gm_1 m_2}{r^2}$ , with $G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ .

F = \dfrac{(6.67\times10^{-11})(1.35\times10^{23})(5.68\times10^{26})}{(1.24\times10^9)^2} = \dfrac{5.114\times10^{39}}{1.538\times10^{18}} = 3.33\times10^{21}\ \text{N}.

(2 marks)

(b) Newton's third law states forces occur in equal and opposite pairs. Titan pulls on Saturn and Saturn pulls on Titan with the same magnitude in the opposite direction. The expression $\dfrac{Gm_1 m_2}{r^2}$ is symmetric in the two masses, so each body feels the same magnitude force regardless of the mass difference. (2 marks)

SACE 2024-style2 marks

Tethys is a moon of Saturn with a mass of

6.17\times10^{20}\ \text{kg}

, orbiting Saturn (mass

5.65\times10^{26}\ \text{kg}

) at an orbital radius of

2.95\times10^8\ \text{m}

. Calculate the magnitude of the gravitational force between Tethys and Saturn.

Show worked answer →

Apply $F = \dfrac{Gm_1 m_2}{r^2}$ with $G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$ .

Numerator: $(6.67\times10^{-11})(6.17\times10^{20})(5.65\times10^{26}) = 2.325\times10^{37}$ .
Denominator: $(2.95\times10^8)^2 = 8.70\times10^{16}$ .

F = \dfrac{2.325\times10^{37}}{8.70\times10^{16}} = 2.67\times10^{20}\ \text{N}.

1 mark for correct substitution, 1 mark for the answer of about

2.67\times10^{20}\ \text{N}

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Stage 2 Physics subject outline (2025 onwards) — SACE Board
Stage 2 Physics: external assessment — SACE Board