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How does the inverse-square law govern the gravitational force between two masses?

Apply Newton's law of universal gravitation to calculate the force between masses

A focused answer to the WACE Year 12 Physics Unit 3 content point on Newton's law of universal gravitation. The inverse-square force law, the gravitational constant, treating bodies as point masses, and how surface gravity relates to mass and radius.

Generated by Claude Opus 4.87 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

WACE wants you to calculate gravitational forces and to reason about how they scale with mass and distance. This is the foundational law for the whole gravity strand; orbital motion and field strength all follow from it.

The law itself

F=\frac{Gm_1 m_2}{r^2}.

The force is attractive, acts along the line joining the centres, and forms a Newton's third law pair: the Earth pulls you down with exactly the force you pull the Earth up with. The distance $r$ is measured centre to centre, not surface to surface, because a uniform sphere acts gravitationally as if all its mass sits at its centre.

The inverse-square behaviour

Because $F\propto 1/r^2$ , the way the force fades with distance is steep. Tripling the distance cuts the force to one ninth; halving it multiplies the force by four. Many exam questions are ratio questions: change one quantity and ask for the factor by which $F$ changes. Set up the ratio $F_2/F_1$ and let the constants cancel.

Why the constant is tiny

The gravitational constant $G$ is extremely small, which is why gravity is negligible between everyday objects and only becomes significant when at least one mass is astronomical. Two $1\ \text{kg}$ masses one metre apart attract with about $7\times10^{-11}\ \text{N}$ , far too weak to notice.

Surface gravity

Putting one mass as a planet $M$ and the other as a test mass $m$ at the surface (radius $r$ ), the weight is $mg=GMm/r^2$ , so

g=\frac{GM}{r^2}.

This explains why a more massive planet has stronger surface gravity and why a larger planet of the same mass has weaker surface gravity. It also lets you compare $g$ on different planets without knowing any test mass.

Treating bodies as point masses

The law as written applies strictly to point masses, yet WACE happily uses it for planets, moons and satellites. The justification is Newton's shell theorem: a body with a spherically symmetric mass distribution produces, at any external point, exactly the field it would if all its mass were concentrated at its centre. This is why $r$ is always the centre-to-centre distance and why we can treat Earth as a point of mass $M$ at its core when computing surface gravity or orbital forces. The approximation fails only when the separation is comparable to the bodies' own sizes or when the mass distribution is markedly non-spherical, neither of which arises in standard WACE problems.

Reading ratio questions

Many gravitation marks come from ratio reasoning rather than full substitution. The trick is always to write the target quantity as a formula, form the ratio of the two cases, and let every unchanged quantity cancel. For force, $\dfrac{F_2}{F_1}=\dfrac{m_{1,2}m_{2,2}}{m_{1,1}m_{2,1}}\left(\dfrac{r_1}{r_2}\right)^2$ ; for surface gravity, $\dfrac{g_2}{g_1}=\dfrac{M_2}{M_1}\left(\dfrac{r_1}{r_2}\right)^2$ . Doing this avoids ever needing $G$ and slashes arithmetic errors. A typical question changes mass by one factor and radius by another and asks for the net change; handle each factor separately, remembering radius is squared.

Distances and centres

When a question gives an altitude above a surface, add the planet's radius to get $r$ . Using altitude alone is a common slip. Likewise, the masses in the formula are the two interacting bodies, not the planet plus its atmosphere or anything else.

Exam-style practice questions

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

WACE 20216 marksThe planet Mars has a mass of

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

and a radius of

3.39\times10^{6}\ \text{m}

. (a) Calculate the gravitational field strength at the surface of Mars. (b) An astronaut and equipment have a combined mass of

95\ \text{kg}

. Calculate their weight on Mars and state how this compares with their weight on Earth.

Show worked answer →

A 6 mark calculation rewards correct substitution, working and units in both parts.

(a) Surface field strength. Use $g=\dfrac{GM}{r^2}$ with $G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ :

g=\frac{(6.67\times10^{-11})(6.42\times10^{23})}{(3.39\times10^{6})^2}=\frac{4.28\times10^{13}}{1.149\times10^{13}}=3.7\ \text{N kg}^{-1}.

(b) Weight on Mars. $W=mg=95\times3.7=3.5\times10^{2}\ \text{N}$ (about $354\ \text{N}$ ). On Earth the same mass weighs $95\times9.8=931\ \text{N}$ , so the Martian weight is roughly $0.38$ of the Earth weight, consistent with Mars having about $38\%$ of Earth's surface gravity.

Markers reward the correct formula, the squared radius, the value near $3.7\ \text{N kg}^{-1}$ , and a sensible comparison using $W=mg$ on both planets.

WACE 20234 marksTwo satellites of equal mass orbit a planet. Satellite A is at distance

r

from the planet's centre and satellite B is at distance

3r

. Explain, with reference to Newton's law of universal gravitation, how the gravitational force on satellite B compares with that on satellite A, and justify your answer quantitatively.

Show worked answer →

A 4 mark explain answer needs the inverse-square reasoning plus the numerical factor.

Set up the ratio: From $F=\dfrac{GMm}{r^2}$ with the same $M$ and $m$ , form $\dfrac{F_B}{F_A}=\left(\dfrac{r_A}{r_B}\right)^2$ .
Substitute: $\dfrac{F_B}{F_A}=\left(\dfrac{r}{3r}\right)^2=\dfrac{1}{9}$ .
Conclusion: The force on satellite B is one ninth of the force on satellite A, because the gravitational force follows an inverse-square law: tripling the separation reduces the force by a factor of $3^2=9$ .

Markers reward an explicit statement of the inverse-square law, the cancelling of constants, and the correct factor of $\tfrac{1}{9}$ .

WACE 20225 marksThe Moon (

m=7.35\times10^{22}\ \text{kg}

) orbits Earth (

M=5.97\times10^{24}\ \text{kg}

) at an average centre-to-centre distance of

3.84\times10^{8}\ \text{m}

. (a) Calculate the gravitational force between Earth and the Moon. (b) State the size and direction of the force the Moon exerts on Earth.

Show worked answer →

A 5 mark answer rewards a correct substitution and a Newton's third law statement.

(a) Force. Use $F=\dfrac{GMm}{r^2}$ :

F=\frac{(6.67\times10^{-11})(5.97\times10^{24})(7.35\times10^{22})}{(3.84\times10^{8})^2}=\frac{2.93\times10^{37}}{1.47\times10^{17}}=2.0\times10^{20}\ \text{N}.

(b) Force on Earth. By Newton's third law the Moon pulls Earth with an equal and opposite force, $2.0\times10^{20}\ \text{N}$ , directed from Earth toward the Moon.

Markers reward the squared distance, the value near $2.0\times10^{20}\ \text{N}$ , and the equal-and-opposite third-law statement.