SACE Board SACE 2024-style practice: Tethys is a moon of Saturn with a mass of 6.1710^20\ kg, orbiting Saturn (mass 5.6510^26\ kg) at an orbital radius of 2.9510^8\ m. Calculate the magnitude of the gravitational force between Tethys and Saturn.

Apply F = Gm_1 m_2r^2 with G = 6.6710^-11\ N m^2kg^-2. Numerator: (6.6710^-11)(6.1710^20)(5.6510^26) = 2.32510^37. Denominator: (2.9510^8)^2 = 8.7010^16. $F = 2.32510^378.7010^16 = 2.6710^20\ N.$ 1 mark for correct substitution, 1 mark for the answer of about 2.6710^20\ N.

§-Syllabus dot point

SAPhysicsSyllabus dot point

How does the gravitational attraction between two masses depend on their masses and separation?

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

Newton's inverse-square law of gravitation, how gravitational force depends on mass and separation, and the meaning of gravitational field strength, with worked examples.

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
The law
Gravitational field strength
Why it matters
How SACE assesses this

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

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.

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.

Exam-style practice questions

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

SACE 20234 marksTitan 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 20242 marksTethys 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}