Newton's Law of Universal Gravitation
In 1798, Henry Cavendish used a torsion balance at the Royal Society in London to measure the gravitational attraction between two 1.61 kg and two 158 kg lead spheres. His result — $G = 6.74 \times 10^{-11}$ N m² kg⁻² — matched the modern value of $6.674 \times 10^{-11}$ to within 1%. He also calculated Earth's density as 5,448 kg/m³ (modern: 5,514 kg/m³), completing the first laboratory measurement of the gravitational constant.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
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
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
Every mass attracts every other mass — always attractive, always mutual
Place two 1 kg masses 1 m apart on a frictionless surface: the gravitational attraction between them is only $6.67 \times 10^{-11}$ N — so small it is unmeasurable without the exquisite precision of Cavendish's 1798 torsion balance. Yet the same equation, applied to Earth's mass ($5.97 \times 10^{24}$ kg) and your mass, gives your weight. The same law, applied to Earth and Moon, predicts the Moon's orbital speed within 0.3%. Gravity's strength depends entirely on mass and distance.
Newton's Law of Universal Gravitation — force $F$ acts along the line joining the centres of masses $M$ and $m$.
$F = \dfrac{GMm}{r^2}$ — gravitational force between two masses (N)
$G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ — universal gravitational constant
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 $M$.
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 negligible — about the weight of a single bacterium. It is far too small to feel, which is why we do not notice gravitational attraction between everyday objects.
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:
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).
$g = \dfrac{F}{m} = \dfrac{GM}{r^2}$ — field strength (N/kg or 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
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 $9.8 \text{ m/s}^2$. Small differences arise from Earth's rotation and its non-spherical shape.
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.
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. $g' = 9.8 \times \left(\dfrac{6371}{6671}\right)^2 = 9.8 \times (0.9550)^2 = 9.8 \times 0.9120 = 8.94 \text{ m/s}^2$.
Even at ISS altitude (~400 km), gravitational field strength is still ~88% of its surface value. Astronauts are in free fall, not "zero gravity."
Gravitational field strength: $g = GM/r^2$ (N/kg or m/s²). At altitude $h$: $r = R + h$ (centre-to-centre); $g' = g(R/(R+h))^2$. $g$ decreases with altitude but never reaches zero. Astronauts experience free fall — not zero gravity.
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:
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: what is the mathematical pattern — and why does gravity share it with light and electric force? This card answers it → $F \propto 1/r^2$ because the effect spreads over a spherical surface of area $4\pi r^2$.
The inverse square law is one of the most important patterns in physics. Whenever a quantity radiates uniformly from a point source in three-dimensional space, its intensity decreases as the inverse square of the distance. Gravity, electric force, and light intensity all follow this pattern for exactly the same geometric reason: the same total effect spreads over a surface area that grows as $r^2$.
Visualising the inverse square law
Imagine gravitational force spreading out uniformly over the surface of an expanding sphere centred on the source mass. The surface area of a sphere is $4\pi r^2$. As the sphere expands, the same total force is spread over an area that grows as $r^2$ — so the force per unit area (and hence the force on any one test mass) decreases as $1/r^2$.
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
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.
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.
British scientist Henry Cavendish performed one of the most elegant experiments in physics history. Using a torsion balance — a thin wire with small lead spheres that twisted in response to tiny forces — he measured the gravitational attraction between laboratory-scale masses.
The forces involved were incredibly small: around $10^{-7}$ N — comparable to the gravitational force between two people 1 m apart. Yet Cavendish measured them with remarkable precision, allowing him to calculate $G$ and consequently "weigh the Earth": $$g = \frac{GM}{R^2} \implies M = \frac{gR^2}{G} = \frac{9.8 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = 5.97 \times 10^{24} \text{ kg}$$
Modern CODATA value: $G = 6.6743(15) \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ — consistent with Cavendish's result to within about 1%.
The Cavendish experiment was significant because it was the first to:
Misconceptions — final check
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 extends to infinity — no cutoff
- Orbit = free fall (not "zero gravity")
- Cavendish (1798) first measured $G$
Which set correctly summarises what happens when the separation between two masses is halved?
Activities
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).
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.
- Explain why astronauts in low Earth orbit feel weightless even though $g$ at that altitude is approximately $8.7 \text{ m/s}^2$.
Why do astronauts in low Earth orbit feel weightless?
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.
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): $F = \dfrac{GMm}{r^2} = \dfrac{(6.67 \times 10^{-11}) \times 1 \times 2}{(0.5)^2} = \dfrac{1.334 \times 10^{-10}}{0.25} = 5.3 \times 10^{-10} \text{ N}$ (1 mark). This force is extraordinarily small — roughly equivalent to the weight of a single bacterium; with ordinary equipment the statement would be true (1 mark). Cavendish used a torsion balance (1798): a thin wire with small lead spheres twisted by the gravitational attraction of large spheres, allowing measurement of forces of order $10^{-7}$ N (1 mark). The statement is therefore conditional on available equipment — the force is not unmeasurable in principle, as Cavendish proved; modern experiments measure $G$ to parts per million (1 mark).
Five timed questions on Newton's Law of Universal Gravitation. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaGravitational field puzzles — apply the inverse square law under pressure. Lighter than the boss — pure concept-drill practice that hammers home $r^2$ dependence.
At the start you were asked about Henry Cavendish's 1798 torsion balance experiment at the Royal Society London — measuring $G = 6.74 \times 10^{-11}$ N m² kg⁻² from the attraction between 1.61 kg and 158 kg lead spheres, and deriving Earth's density as 5,448 kg/m³.
Verify Earth's density: if $G = 6.674 \times 10^{-11}$, $M_E = 5.97 \times 10^{24}$ kg, and $R_E = 6.371 \times 10^6$ m, then $\rho = M/V = M/(\tfrac{4}{3}\pi R^3) = 5.97 \times 10^{24} / (1.083 \times 10^{21}) = 5,514 \text{ kg/m}^3$. Cavendish's 1798 value of 5,448 kg/m³ was only 1.2% low — remarkable precision for the first measurement of $G$. He never called it "weighing the Earth" — that description came later. His goal was measuring $G$; the Earth's mass followed as a consequence. Has your understanding of Newton's law of gravitation changed?