Year 12 Physics Module 5 ⏱ ~45 min 5 MC · 3 Short Answer Lesson 10 of 18

Newton's Law of Universal Gravitation

Newton's law connects the barely measurable attraction between laboratory masses with the force that acts between planets. Learn when $F=GMm/r^2$ and $g=GM/r^2$ apply, how their directions are represented, and how mass and centre-to-centre distance shape a gravitational field.

Today's hook: A 70 kg student is pulled toward Earth with a force of about 686 N, yet two 70 kg students one metre apart attract each other with only $3.27\times10^{-7}$ N. The same law produces both forces. Which quantities create that enormous difference?

0/5TASKS

You’re here

Connect and orient

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Use the two checked lesson worksheets online, or open the complete worksheet and print it from your browser.

Build Foundations & guided practice Apply Application & data response Exam Complete printable worksheet

Before you read, estimate

Estimate the gravitational force between you (60 kg) and Earth. Then estimate the gravitational force between you and the person next to you, 1 m away.

Which force is larger, and by how much? Write your estimate before working through the lesson.

Warm-up, which statement about gravitational force is correct?

Learning Intentions

goals

Know, Newton's Law of Universal Gravitation

State that every mass attracts every other mass with force $F = \dfrac{GMm}{r^2}$
Identify each symbol and its units; recall $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

Understand, Inverse Square Law and Field Strength

Explain how force varies with distance ($F \propto 1/r^2$) and predict force changes when separation changes
Calculate gravitational field strength $g = GM/r^2$ at any distance from a planet

Can Do, Apply and Evaluate

Solve numerical problems using Newton's Law of Universal Gravitation
Use $r = R + h$ correctly (centre-to-centre distance) in field strength calculations

Scan these before reading

vocab

Newton's law of gravitationEvery particle of matter attracts every other particle with a force $F = GMm/r^2$, always attractive, acting along the line joining centres.

Gravitational constant $G$The universal constant $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$. Cavendish's 1798 torsion-balance result allowed its value to be inferred.

Inverse-square lawA relationship where a quantity decreases as the square of distance increases: doubling distance quarters the force ($F \propto 1/r^2$).

Spherical modelOutside a spherically symmetric body, its gravitational field is modelled as if all its mass were concentrated at its centre. This is why $r$ is measured from the centre.

Gravitational field strengthThe gravitational force per unit mass at a point: $g = F/m = GM/r^2$, measured in N/kg or m/s$^2$.

Universal gravitationThe principle that gravity is a universal property of all matter, not limited to Earth or large bodies, acting across any distance.

Cross-lesson links: Lessons 7–8 established radial acceleration and force for uniform circular motion. This lesson introduces the gravitational force and field models; Lesson 11 will combine them with circular motion to analyse satellites.

Misconceptions to fix

✗ Wrong: There is no gravity in space, astronauts on the ISS are in "zero gravity."

✓ Right: At 400 km altitude, gravitational field strength is still about 89% of its surface value. Apparent weightlessness in orbit is a Lesson 11 application; it does not mean Earth's gravitational field is zero.

✗ Wrong: Gravity is a property of Earth only, small objects do not attract each other.

✓ Right: Every mass attracts every other mass. Two 70 kg people 1 m apart attract each other with a (tiny) gravitational force. The force is simply too small to notice for everyday objects.

✗ Wrong: In the formula $F = GMm/r^2$, $r$ is the distance from the surface.

✓ Right: $r$ is always the centre-to-centre distance between the two masses. For a point at height $h$ above a planet: $r = R + h$, where $R$ is the planet's radius.

Astronauts on the ISS experience zero gravitational force.

In Newton's law, $r$ is the distance from the planet's surface to the object.

Every mass in the universe attracts every other mass.

Core Content

You’re here

State the law and its model conditions

Newton's Law of Universal Gravitation

+5 XP

Every mass attracts every other mass, always attractive, always mutual

Place two 1 kg masses 1 m apart: the gravitational attraction between them is only $6.67 \times 10^{-11}$ N and requires extremely sensitive apparatus to measure. Apply the same law to Earth's mass ($5.97 \times 10^{24}$ kg) and a student at its surface and the result is their weight. The difference comes from the product of the masses and the centre-to-centre separation.

Newton's Law of Universal Gravitation, force $F$ acts along the line joining the centres of masses $M$ and $m$.

Newton's Law of Universal Gravitation

$F_g = \dfrac{GMm}{r^2}$ , magnitude of the gravitational force between two masses (N)

$G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ , universal gravitational constant

Model conditions and direction

Use this form directly for point masses. It also applies outside spherically symmetric bodies when $r$ is measured centre to centre. The equation gives a magnitude; each force vector points toward the other mass along the line joining their centres.

where:

$F$ = gravitational force (N), always attractive
$G$ = universal gravitational constant $= 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$
$M$, $m$ = masses of the two objects (kg)
$r$ = centre-to-centre distance between the masses (m)

The force acts along the line joining the centres of the two masses. By Newton's third law, if mass $M$ exerts a force $F$ on mass $m$, then mass $m$ exerts an equal and opposite force $F$ on mass $M$.

Worked example 1, Force between two people

Calculate the gravitational force between two 70 kg people standing 1.0 m apart (centre to centre).

Given. $M = m = 70 \text{ kg}$, $r = 1.0 \text{ m}$, $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.
Find. $F$ (gravitational force).
Method. Use $F = \dfrac{GMm}{r^2}$.
Solve. $F = \dfrac{(6.67 \times 10^{-11}) \times 70 \times 70}{(1.0)^2} = \dfrac{6.67 \times 10^{-11} \times 4900}{1.0} = 3.27 \times 10^{-7} \text{ N}$.

This force is far too small to feel, but it is not zero or fundamentally unmeasurable. A torsion balance can detect forces on this scale.

Newton's Law: $F = GMm/r^2$ (N), where $G = 6.67\times10^{-11}$ N m² kg⁻². Force is always attractive, acts along the line joining centres, and $r$ is the centre-to-centre distance, not surface-to-surface. Newton's 3rd Law: both masses feel equal and opposite forces.

Pause, copy the highlighted law, the value of $G$, and the centre-to-centre rule into your book before moving on.

Two masses of 500 kg and 200 kg have their centres 4.0 m apart. Using $F = GMm/r^2$, the gravitational force between them is closest to:

You’re here

Define gravitational field strength

Gravitational Field Strength

+5 XP

The gravitational force per unit mass at any point in space

We just saw that $F = GMm/r^2$ depends on both masses. That raises a question: how can we describe gravity at a location independently of what's placed there? This card answers it → gravitational field strength $g = GM/r^2$ (N/kg); use $r = R + h$ for altitude calculations.

Rather than describing force (which depends on the test mass), physicists define a field, a property of space itself. The gravitational field strength $g$ tells us the force that would act on each kilogram of mass placed at that point, regardless of what that mass is.

Gravitational field strength $g$ at Earth's surface and at altitude $h$. Note $r = R + h$ (centre-to-centre).

Gravitational field strength

$g = \dfrac{F_g}{m} = \dfrac{GM}{r^2}$ , magnitude of field strength (N/kg, numerically equal to m/s$^2$)

$g' = g \times \left(\dfrac{R}{R + h}\right)^2$ , field at altitude $h$ above surface

$r = R + h$ , always use centre-to-centre distance

Gravitational field strength is a vector. Around an isolated spherical mass, $\vec g$ points radially inward. The equation above gives its non-negative magnitude; a negative sign is used only after a coordinate direction has been defined.

At Earth's surface

At Earth's surface, $r = R_{\text{Earth}} = 6.37 \times 10^6 \text{ m}$:

$g = \dfrac{GM}{R^2} = \dfrac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{(6.37 \times 10^6)^2} \approx 9.83 \text{ m/s}^2$

This is close to the commonly quoted $9.8 \text{ N/kg}$. The measured effective free-fall acceleration varies slightly with latitude, altitude, Earth's rotation and local mass distribution.

Common HSC error

Always use $r = R + h$ (centre-to-centre distance) in gravitational calculations. A frequent mistake is to use the altitude $h$ alone instead of the distance from Earth's centre. The force depends on the distance from the centre of the attracting mass, not from its surface.

Worked example 2, Field strength at 300 km altitude

Calculate the gravitational field strength at an altitude of 300 km above Earth's surface. Use $R_{\text{Earth}} = 6371 \text{ km}$, $g_{\text{surface}} = 9.8 \text{ m/s}^2$.

Given. $R = 6.371 \times 10^6 \text{ m}$, $h = 3.00 \times 10^5 \text{ m}$, $g = 9.8 \text{ m/s}^2$.
Find. $g'$ at altitude $h$.
Method. Ratio method: $g' = g \times \left(\dfrac{R}{R + h}\right)^2$; $r = R + h = 6671 \text{ km}$.
Solve. $g' = 9.8 \times \left(\dfrac{6371}{6671}\right)^2 = 9.8 \times (0.9550)^2 = 9.8 \times 0.9120 = 8.94 \text{ m/s}^2$.

At 300 km altitude, the field is still about 91% of its surface value. The field decreases with distance; it does not abruptly switch off.

Gravitational field magnitude: $g = GM/r^2$ (N/kg). Around a spherical mass, $\vec g$ points radially inward. At altitude $h$: $r = R + h$; $g' = g(R/(R+h))^2$. As $r\to\infty$, $g\to0$.

Add the highlighted field-strength formula and the altitude rule to your notes before the check below.

A satellite orbits at a distance of $2R_{\text{Earth}}$ from Earth's centre. Compared to surface gravity, the field strength there is:

You’re here

Investigate mass and distance relationships

The Inverse Square Law

+5 XP

How gravitational force decreases with distance, and why it is universal

We just saw that $g = GM/r^2$ decreases with distance. That raises a question: how can data reveal the mathematical pattern? This card answers it → compare ratios and linearise a force–distance graph using $1/r^2$.

Newton's law predicts $F_g\propto1/r^2$ when both masses are fixed. The geometry of a radial field is consistent with this relationship: spherical surfaces have area $4\pi r^2$, so a conserved radial field flux is distributed over an area proportional to $r^2$. This geometric picture supports the pattern; the measured force relationship is the evidence students test.

Visualising the inverse square law

Imagine concentric spherical surfaces around a spherically symmetric source. As $r$ increases, the same radial field flux crosses a larger area. Do not picture “force” as a substance being used up: the measurable statement is that the field magnitude and the force on a fixed test mass decrease as $1/r^2$.

Inverse square law, distance vs force

If distance becomes $2 \times$, force becomes $\dfrac{1}{4}$

If distance becomes $3 \times$, force becomes $\dfrac{1}{9}$

If distance becomes $\tfrac{1}{2} \times$, force becomes $4 \times$

Graphical representation

Graph of $F$ vs $r$: a hyperbolic curve that approaches zero asymptotically
Graph of $F$ vs $\dfrac{1}{r^2}$: a straight line through the origin with gradient $GMm$

The linear $F$ vs $1/r^2$ graph is powerful because it allows experimental verification, measured force plotted against $1/r^2$ should yield a straight line.

Other inverse square laws in physics

Gravitational force: $F = \dfrac{GMm}{r^2}$
Electrostatic force (Coulomb's law): $F = \dfrac{kq_1q_2}{r^2}$
Light intensity: $I = \dfrac{P}{4\pi r^2}$ where $P$ is the power of the source

Worked example 3, Weight at distance $2R$ from Earth's centre

A 60 kg astronaut weighs 600 N at Earth's surface. What would they weigh at a distance of $2R_{\text{Earth}}$ from Earth's centre?

Given. $F_{\text{surface}} = 600 \text{ N}$, $r_{\text{surface}} = R$, $r_{\text{new}} = 2R$.
Find. $F_{\text{new}}$ at $r = 2R$.
Method. Use the inverse square law: $\dfrac{F_{\text{new}}}{F_{\text{surface}}} = \left(\dfrac{R}{2R}\right)^2 = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$.
Solve. $F_{\text{new}} = \dfrac{600}{4} = 150 \text{ N}$.

At one Earth radius above the surface, the astronaut's weight is only one-quarter of their surface weight. Note: $g' = 9.8/4 = 2.45 \text{ m/s}^2$ at this altitude.

Inverse square law: $F \propto 1/r^2$. Distance $\times2 \Rightarrow$ force $\div4$; distance $\times3 \Rightarrow$ force $\div9$; distance $\div2 \Rightarrow$ force $\times4$. Graph of $F$ vs $1/r^2$ is a straight line through the origin. Same pattern applies to Coulomb's law and light intensity.

Add the highlighted inverse square rules to your notes before the check below.

Three of these statements about the inverse square law are correct. Pick the odd one out.

Investigate the force relationship

Working Scientifically

Use the explorer to vary one factor at a time. Hold $m_2$ and $r$ constant while changing $m_1$, then hold both masses constant while changing $r$.

Canonical Lesson 10 explorer. Open fullscreen ↗

Record at least four values for the mass trial and four values for the distance trial.
Use the graph shape to identify $F_g\propto m_1$ and $F_g\propto1/r^2$.
Explain why plotting $F_g$ against $1/r^2$ should produce a straight line with gradient $Gm_1m_2$.

Evidence-based conclusion

Essential formulae, universal gravitation

Newton's Law: $F = \dfrac{GMm}{r^2}$

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

Field strength: $g = \dfrac{GM}{r^2} = \dfrac{F}{m}$ (N/kg or m/s$^2$)

Field at altitude: $g' = g \times \left(\dfrac{R}{R + h}\right)^2$

Centre-to-centre distance: $r = R + h$, always!

Inverse square law: $F \propto \dfrac{1}{r^2}$

Fill the gap. If the distance between two masses is tripled, the gravitational force becomes $\dfrac{1}{\_\_}$ of the original value.

You’re here

Evaluate evidence and limitations

Real world, the Cavendish experiment (1797–1798)

In 1797–1798, Henry Cavendish used a torsion balance designed from earlier work by John Michell. Small lead spheres on a suspended rod twisted toward larger lead spheres, allowing the tiny gravitational attraction between laboratory masses to be measured.

The forces were of order $10^{-7}$ N. Cavendish reported Earth's mean density rather than a numerical value of $G$ in modern notation; his measurement later allowed $G$ and Earth's mass to be inferred: $$g = \frac{GM}{R^2} \implies M = \frac{gR^2}{G} = \frac{9.8 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = 5.97 \times 10^{24} \text{ kg}$$

Modern CODATA value: $G = 6.67430(15) \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$. The uncertainty notation shows why $G$ is less precisely known than many other fundamental constants.

Which statement most accurately describes the significance of Cavendish's torsion-balance experiment?

You’re here

Consolidate the gravitational model

Wrap-up, Misconceptions and Summary

Misconceptions, final check

✗ "$g$ is constant everywhere at $9.8 \text{ m/s}^2$."

✓ $g$ varies with altitude (decreases with height), latitude (Earth's rotation means $g$ is slightly less at the equator), and local geology (density variations). The standard $9.8 \text{ m/s}^2$ is an average at sea level at mid-latitudes.

✗ "If $r$ doubles, the force halves."

✓ This is the most common inverse-square-law error. Doubling $r$ means $r^2$ quadruples, so the force becomes $\dfrac{1}{4}$, not $\dfrac{1}{2}$. The force is inversely proportional to the square of the distance.

Copy into your books

Key Definitions

Newton's Law: every mass attracts every other mass
$G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$
$g$ = gravitational field strength (force per unit mass)
Inverse square law: $F \propto 1/r^2$

Key Equations

$F = GMm/r^2$
$g = GM/r^2 = F/m$
$g' = g \times (R/(R+h))^2$
$r = R + h$ (centre-to-centre)

Inverse Square Summary

$r \times 2 \Rightarrow F \div 4$
$r \times 3 \Rightarrow F \div 9$
$r \times 10 \Rightarrow F \div 100$
$r \div 2 \Rightarrow F \times 4$

Key Principles

Gravity is always attractive
Gravity has no finite cutoff; its magnitude tends toward zero as distance increases
$F_g$ and $g$ equations give magnitudes; vectors point inward
Cavendish measured laboratory gravitational attraction, allowing $G$ to be inferred

Which set correctly summarises what happens when the separation between two masses is halved?

You’re here

Respond and check

Activities

Activity 1, Gravitational Force Between Spheres

ApplyBand 4

Apply Newton's Law of Universal Gravitation to solve force problems

Calculate the gravitational force between two spheres of mass 1000 kg and 500 kg whose centres are 2 m apart.
The distance between the spheres is now doubled to 4 m. Without recalculating fully, state the new force using the inverse square law.
Calculate the gravitational force between Earth ($M = 5.97 \times 10^{24} \text{ kg}$) and the Moon ($m = 7.35 \times 10^{22} \text{ kg}$, $r = 3.84 \times 10^8 \text{ m}$).

Activity check, for the 1000 kg and 500 kg spheres 2 m apart (Activity 1, Q1), the gravitational force is $F = \_\_\_ \times 10^{-6} \text{ N}$ (to 2 significant figures).

Activity 2, Field Strength at Altitude

ApplyBand 5

Calculate gravitational field strength above Earth's surface

Find the gravitational field strength at 1000 km above Earth's surface. Use $R_{\text{Earth}} = 6371 \text{ km}$ and $g_{\text{surface}} = 9.8 \text{ m/s}^2$.
Calculate the gravitational field strength on the surface of Mars. Use $M_{\text{Mars}} = 6.42 \times 10^{23} \text{ kg}$, $R_{\text{Mars}} = 3.40 \times 10^6 \text{ m}$, and express your answer as a fraction of Earth's surface gravity.
State the direction of the gravitational field vector at both locations and identify the spherical-body assumption used.

When $g=GM/r^2$ is used outside a planet, which interpretation is correct?

You’re here

Apply universal gravitation independently

Quick recall, Newton's law of universal gravitation

+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 9 marks

+5 XP

ApplyBand 4(2 marks) 1. Calculate the gravitational force between Earth ($M = 5.97 \times 10^{24} \text{ kg}$) and the Moon ($m = 7.35 \times 10^{22} \text{ kg}$). The average distance between their centres is $3.84 \times 10^8 \text{ m}$.

1 mark: correct substitution into $F = GMm/r^2$ · 1 mark: correct answer with units

ApplyBand 5(3 marks) 2. Calculate the gravitational field strength on the surface of Mars ($M = 6.42 \times 10^{23} \text{ kg}$, $R = 3.40 \times 10^6 \text{ m}$). Compare your answer with Earth's surface gravity ($g = 9.8 \text{ m/s}^2$).

1 mark: correct formula $g = GM/R^2$ with working · 1 mark: correct numerical answer · 1 mark: correct comparison (approx 38% of Earth's $g$)

EvaluateBand 6(4 marks) 3. Evaluate the statement: "The gravitational force between two everyday objects is too small to measure." Use a calculation for a 1 kg and 2 kg mass separated by 0.5 m, and discuss the significance of the Cavendish experiment.

1 mark: correct calculation of $F$ between 1 kg and 2 kg · 1 mark: analysis that the force is extremely small · 1 mark: description of Cavendish torsion-balance method · 1 mark: evaluation, the statement is conditional on available equipment; Cavendish proved it is measurable

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (2 marks): $F = \dfrac{GMm}{r^2} = \dfrac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (7.35 \times 10^{22})}{(3.84 \times 10^8)^2} = \dfrac{2.93 \times 10^{37}}{1.47 \times 10^{17}} = 1.99 \times 10^{20} \text{ N}$ (1 mark method, 1 mark answer).

Q2 (3 marks): $g_{\text{Mars}} = \dfrac{GM}{R^2} = \dfrac{(6.67 \times 10^{-11}) \times (6.42 \times 10^{23})}{(3.40 \times 10^6)^2} = \dfrac{4.28 \times 10^{13}}{1.16 \times 10^{13}} = 3.71 \text{ m/s}^2$ (2 marks). Comparison: $g_{\text{Mars}}/g_{\text{Earth}} = 3.71/9.8 = 0.38$, so Mars gravity is approximately 38% of Earth's (1 mark).

Q3 (4 marks): $F = \dfrac{GMm}{r^2} = \dfrac{(6.67 \times 10^{-11}) \times 1 \times 2}{(0.5)^2} = 5.3 \times 10^{-10} \text{ N}$ (1 mark). This force is extraordinarily small and ordinary classroom force sensors would not resolve it (1 mark). Cavendish used a torsion balance: the attraction between laboratory masses produced a measurable twist in a suspended wire (1 mark). The statement is therefore too absolute—the force is difficult to measure, not unmeasurable in principle; specialised torsion balances can detect gravitational attraction between laboratory masses (1 mark).

You’re here

Retrieve and reflect

Check what actually stuck

Take the full module quiz

quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →

Arcade practice · Asteroid Blaster

Gravitational field puzzles, apply the inverse square law under pressure. Lighter than the boss, pure concept-drill practice that hammers home $r^2$ dependence.

How did your thinking change?

At the start you compared the strong Earth–student attraction with the tiny attraction between two students. Cavendish's torsion balance showed that laboratory-scale gravitational forces are difficult—but not impossible—to measure.

Cavendish reported Earth's mean density. In modern notation, a measured value of $G$ combines with $g=GM/R^2$ to give Earth's mass, and $M/(\tfrac{4}{3}\pi R^3)$ gives its mean density. Explain how the same inverse-square model accounts for both everyday weight and tiny laboratory forces.

I have completed this lesson

← Previous: Vertical Circular Motion: Extension Next: Gravitational Orbits →

Module overview · Physics · Checkpoint 2 · Module quiz