Newton's Law of Gravitation

In the late 17th century, Isaac Newton proposed one of the most important laws in all of physics: a universal law that describes the gravitational attraction between any two masses in the universe. This law unified terrestrial gravity — the force that makes an apple fall — with celestial mechanics — the force that keeps the Moon in orbit around the Earth and the planets in orbit around the Sun.

Newton's Law of Universal Gravitation

Newton's law states that every point mass attracts every other point mass with a force that is:

Directly proportional to the product of their masses
Inversely proportional to the square of the distance between their centres

Mathematically:

$F = \frac{GMm}{r^2}$

where:

F is the gravitational force between the two masses (N)
G is the gravitational constant, 6.674 × 10⁻¹¹ N m² kg⁻² (sometimes called "big G")
M and m are the two masses (kg)
r is the distance between the centres of the two masses (m)

This is a universal law — it applies to every pair of masses in the universe, from atoms to galaxies. The same equation that describes the force between your body and the Earth also describes the force between binary star systems billions of light years away.

flowchart LR
    A["Mass M"] -- "F = GMm/r²" --> B["Mass m"]
    B -- "F = GMm/r² (equal & opposite)" --> A
    subgraph Key["Newton's Third Law Pair"]
        C["Same magnitude"]
        D["Opposite directions"]
        E["Act on different bodies"]
    end

The Gravitational Constant G

The gravitational constant G is one of the fundamental constants of nature. Its value is extremely small:

$G = 6.674 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

This tiny value explains why gravitational forces between everyday objects are imperceptible. Two 1 kg masses placed 1 metre apart attract each other with a force of just 6.674 × 10⁻¹¹ N — far too small to notice. It is only when at least one of the masses is astronomical (like a planet or star) that gravity becomes significant.

G was first measured by Henry Cavendish in 1798 using a torsion balance experiment. His measurement allowed the mass of the Earth to be calculated for the first time, which is why the experiment is sometimes called "weighing the Earth."

It is important to distinguish between G (the universal gravitational constant, always the same everywhere) and g (the gravitational field strength at a particular point, which varies with location).

Symbol	Name	Value	Units	Constant?
G	Universal gravitational constant	6.674 × 10⁻¹¹	N m² kg⁻²	Yes — same everywhere
g	Gravitational field strength	Varies (9.81 at Earth's surface)	N kg⁻¹	No — depends on location

Common exam mistake: Confusing G and g. The constant G is universal and never changes. The field strength g varies with position, mass, and radius of the body you are on. A question asking "what happens to G if you double the mass?" is testing whether you know G is a constant — the answer is nothing.

The Inverse Square Law

The inverse square relationship (F ∝ 1/r²) is one of the most important features of Newton's law. It means that:

If you double the distance, the force drops to 1/4 of its original value
If you triple the distance, the force drops to 1/9
If you halve the distance, the force increases by a factor of 4

The inverse square law arises because gravitational influence spreads out over the surface of an expanding sphere. At distance r, the "influence" is distributed over a sphere of area 4πr², so the force per unit area decreases as 1/r².

Distance multiplier	Force multiplier	Percentage of original
×1 (original)	×1	100%
×2	×1/4	25%
×3	×1/9	11.1%
×5	×1/25	4%
×10	×1/100	1%
×0.5 (halved)	×4	400%

This relationship is fundamental to understanding how gravity behaves over astronomical distances. The Sun's gravitational pull on Neptune (at 30 AU) is 900 times weaker than its pull on Earth (at 1 AU), because 30² = 900.

Calculating Gravitational Forces

Worked Example 1: Force Between the Earth and Moon

M (Earth) = 5.97 × 10²⁴ kg
m (Moon) = 7.35 × 10²² kg
r = 3.84 × 10⁸ m

$F = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^8)^2}$

$F = \frac{2.93 \times 10^{37}}{1.47 \times 10^{17}} = 1.99 \times 10^{20} \text{ N}$

This enormous force — about 2 × 10²⁰ N — is what keeps the Moon in orbit.

Worked Example 2: Force Between Two Students

Two students, each with mass 60 kg, stand 1.5 m apart:

$F = \frac{6.674 \times 10^{-11} \times 60 \times 60}{1.5^2} = \frac{2.40 \times 10^{-7}}{2.25} = 1.07 \times 10^{-7} \text{ N}$

This is about 0.1 micronewtons — completely undetectable without extremely sensitive equipment.

Worked Example 3: Force on the International Space Station

The ISS orbits at approximately 408 km above Earth's surface. Calculate the gravitational force on it (mass ≈ 420,000 kg).

The orbital radius is r = R_Earth + h = 6.37 × 10⁶ + 4.08 × 10⁵ = 6.778 × 10⁶ m

$F = \frac{GMm}{r^2} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 4.20 \times 10^5}{(6.778 \times 10^6)^2}$

$F = \frac{1.673 \times 10^{20}}{4.594 \times 10^{13}} = 3.64 \times 10^6 \text{ N} = 3.64 \text{ MN}$

The ISS still experiences about 89% of the gravitational force it would feel on the surface. Astronauts appear "weightless" not because gravity is absent, but because they are in free fall — the ISS and everything inside it accelerate together.

Common exam mistake: Students often say astronauts on the ISS are weightless because there is no gravity in space. This is wrong. At ISS altitude, g ≈ 8.7 N kg⁻¹ (about 89% of the surface value). The apparent weightlessness is because the ISS is in continuous free fall around Earth.

Newton's Third Law and Gravity

An essential point: the gravitational force between two objects is a Newton's third law pair. The Earth pulls on you with a force mg downward, and you pull on the Earth with an equal force mg upward. These forces are equal in magnitude, opposite in direction, and act on different bodies.

The reason the Earth does not noticeably accelerate towards you is that F = ma: the same force on the Earth's enormous mass produces an unimaginably tiny acceleration (about 10⁻²³ m s⁻²).

Superposition of Gravitational Fields

When more than two masses are present, the total gravitational force on any one mass is the vector sum of the individual forces from all other masses. This is the principle of superposition.

For example, a spacecraft travelling between the Earth and the Moon experiences gravitational pulls from both bodies simultaneously. At each point, the net force is the vector sum of the Earth's pull and the Moon's pull.

Worked Example 4: Three Masses in a Line

Three masses are arranged in a line: mass A (2.0 × 10³⁰ kg) at x = 0, mass B (5.0 × 10²⁰ kg) at x = 1.0 × 10¹¹ m, and mass C (3.0 × 10³⁰ kg) at x = 3.0 × 10¹¹ m. Find the net gravitational force on mass B.

Force on B due to A (towards A, in −x direction):

$F_{BA} = \frac{6.674 \times 10^{-11} \times 2.0 \times 10^{30} \times 5.0 \times 10^{20}}{(1.0 \times 10^{11})^2} = \frac{6.674 \times 10^{40}}{1.0 \times 10^{22}} = 6.674 \times 10^{18} \text{ N}$

Force on B due to C (towards C, in +x direction):

$F_{BC} = \frac{6.674 \times 10^{-11} \times 3.0 \times 10^{30} \times 5.0 \times 10^{20}}{(2.0 \times 10^{11})^2} = \frac{1.001 \times 10^{41}}{4.0 \times 10^{22}} = 2.503 \times 10^{18} \text{ N}$

Net force on B: F_BA − F_BC = 6.674 × 10¹⁸ − 2.503 × 10¹⁸ = 4.17 × 10¹⁸ N towards A.

Superposition is crucial in astrophysics. The orbit of any planet is influenced by the gravitational pull of the Sun and all other planets. These perturbations — small deviations from a perfect elliptical orbit — led to the discovery of Neptune in 1846 when astronomers noticed unexplained anomalies in the orbit of Uranus.

Linking F = GMm/r² to g = GM/r²

The connection between Newton's law and gravitational field strength is straightforward. If a test mass m is placed in the field of a large mass M at distance r:

$F = \frac{GMm}{r^2}$

Dividing both sides by m:

$\frac{F}{m} = \frac{GM}{r^2}$

But F/m is the definition of gravitational field strength g. Therefore:

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

This shows that the field strength at any point depends only on the source mass M and the distance r — it is independent of the test mass. This is a fundamental property of fields: the field exists regardless of whether a test mass is present to detect it.

Gravitational Forces Across the Solar System

Pair of bodies	Force (N)	Comment
Sun – Earth	3.54 × 10²²	Keeps Earth in orbit
Sun – Jupiter	4.17 × 10²³	Largest planet-Sun force
Earth – Moon	1.99 × 10²⁰	Causes tides
Earth – ISS	3.64 × 10⁶	Station in free fall
You – Earth	~700	Your weight
You – the person next to you	~10⁻⁷	Undetectable

Summary

Newton's law of gravitation: F = GMm/r² applies universally to all masses
G = 6.674 × 10⁻¹¹ N m² kg⁻² is a universal constant; g varies with location
The inverse square law means doubling distance quarters the force
Gravitational forces obey Newton's third law: equal and opposite on different bodies
Superposition: the net force is the vector sum of all individual gravitational forces
The link g = GM/r² comes directly from dividing Newton's law by the test mass

A-Level Deep Dive: Newton's Law of Gravitation

Spec mapping

Edexcel 9PH0 specification Topic 12 — Gravitational Fields opens with Newton's law of universal gravitation $F = Gm_1 m_2 / r^2$ , the inverse-square dependence on separation, and the recognition that the law applies to every pair of point masses (and to spherically symmetric bodies via the shell theorem). It then develops field strength, potential and orbital motion (refer to the official specification document for exact wording). Newton's law is the foundation statement on which the rest of Topic 12 is built — $g = GM/r^2$ is obtained by dividing $F$ by the test mass, and $V = -GM/r$ is obtained by integrating $g$ from infinity. The law is examined in Paper 2 (9PH0/02), where it routinely appears in calculation questions about planetary, lunar and satellite forces, and in Paper 3 (9PH0/03) synoptic questions linking it to circular motion, energy conservation and Kepler's laws. The Edexcel formula booklet provides $F = Gm_1 m_2 / r^2$ explicitly, but candidates must remember the value of $G$ ( $6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ ) and that the law assumes point-mass (or shell-theorem) geometry.

Worked example with full mark scheme

Question (8 marks):

(a) Calculate the magnitude of the gravitational force between the Earth (mass $5.97 \times 10^{24}\ \text{kg}$ ) and the Moon (mass $7.35 \times 10^{22}\ \text{kg}$ ), given a centre-to-centre separation of $3.84 \times 10^{8}\ \text{m}$ . Use $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ . (3)

(b) The Moon orbits the Earth approximately in a circle of this radius with a period of $27.3$ days. Show that the centripetal force required is consistent with the gravitational force found in (a). (4)

Solution with mark scheme:

(a) Step 1 — apply Newton's law of gravitation.

$F = \dfrac{G m_1 m_2}{r^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^{8})^2}$

M1 — correct substitution into $F = Gm_1 m_2 / r^2$ with consistent SI units.

Step 2 — evaluate numerator and denominator separately to preserve precision.

Numerator: $(6.67 \times 10^{-11})(5.97 \times 10^{24})(7.35 \times 10^{22}) \approx 2.926 \times 10^{37}$ .

Denominator: $(3.84 \times 10^{8})^2 = 1.475 \times 10^{17}$ .

M1 — correct evaluation of $r^2$ . Common slip: forgetting the square, or inputting $r$ in km rather than m.

A1 — $F = 2.926 \times 10^{37} / 1.475 \times 10^{17} = 1.98 \times 10^{20}\ \text{N}$ (accept $1.97$ – $2.00 \times 10^{20}\ \text{N}$ , units required).

(b) Step 1 — express the centripetal force.

The centripetal force required for circular motion of the Moon (mass $m$ ) at radius $r$ with angular speed $\omega = 2\pi / T$ is:

$F_c = m \omega^2 r = m \left(\dfrac{2\pi}{T}\right)^2 r$

M1 — correct centripetal-force expression in terms of $T$ and $r$ .

Step 2 — convert the period to seconds.

$T = 27.3 \times 24 \times 3600 = 2.359 \times 10^{6}\ \text{s}$ .

Step 3 — evaluate.

$F_c = (7.35 \times 10^{22}) \left(\dfrac{2\pi}{2.359 \times 10^{6}}\right)^2 (3.84 \times 10^{8})$

$\omega = 2\pi / 2.359 \times 10^{6} = 2.664 \times 10^{-6}\ \text{rad s}^{-1}$ , so $\omega^2 = 7.10 \times 10^{-12}\ \text{rad}^2\ \text{s}^{-2}$ .

$F_c = (7.35 \times 10^{22})(7.10 \times 10^{-12})(3.84 \times 10^{8}) \approx 2.00 \times 10^{20}\ \text{N}$ .

M1 — correct numerical substitution.

A1 — $F_c \approx 2.0 \times 10^{20}\ \text{N}$ , in agreement with (a) to within $\sim 1\%$ .

A1 (consistency) — explicit comparison statement: "the gravitational force from (a) and the required centripetal force agree to within rounding error, confirming that gravity supplies the centripetal force for the Moon's orbit".

(c) B1 — by Newton's third law, the Moon exerts an equal and opposite force on the Earth: $1.98 \times 10^{20}\ \text{N}$ directed from the centre of the Earth toward the centre of the Moon.

Total: 8 marks (M4 A3 B1, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A spacecraft of mass $m = 1.20 \times 10^{4}\ \text{kg}$ lies on the line joining the centres of the Earth and the Moon. The Earth–Moon centre-to-centre distance is $d = 3.84 \times 10^{8}\ \text{m}$ . Take $M_E = 5.97 \times 10^{24}\ \text{kg}$ and $M_M = 7.35 \times 10^{22}\ \text{kg}$ .

(a) Determine the distance $x$ from the centre of the Earth at which the net gravitational force on the spacecraft is zero. (4)

(b) State and justify whether the spacecraft, placed at rest at this point, would remain there. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO2.1) — setting the magnitudes of the two gravitational forces equal: $GM_E m / x^2 = GM_M m / (d - x)^2$ , with the test mass $m$ and the constant $G$ cancelling cleanly.
M1 (AO2.1) — rearranging to $(d - x)^2 / x^2 = M_M / M_E$ , hence $(d - x)/x = \sqrt{M_M / M_E}$ .
M1 (AO1.1b) — evaluating the mass ratio: $M_M / M_E = 7.35 \times 10^{22} / 5.97 \times 10^{24} = 1.231 \times 10^{-2}$ , so $\sqrt{M_M / M_E} = 0.1109$ .
A1 (AO2.1) — solving $(d - x)/x = 0.1109$ to give $x = d / 1.1109 = 3.46 \times 10^{8}\ \text{m}$ from the Earth's centre (accept $3.4\text{–}3.5 \times 10^{8}\ \text{m}$ ).

(b)

B1 (AO3.1) — the point is one of unstable equilibrium along the Earth–Moon line: a small displacement toward the Moon increases $F_M$ and decreases $F_E$ , producing a net force away from the equilibrium.
B1 (AO3.2) — concluding that the spacecraft would not remain there in practice; any perturbation grows. (Credit also given for noting that this analysis ignores the Sun's field and the orbital motion of the Moon — a real Lagrange-point treatment requires the rotating frame.)

Total: 6 marks split AO1 = 1, AO2 = 3, AO3 = 2. Edexcel uses null-force questions of this kind to combine routine algebra (AO1/AO2) with a higher-tariff stability argument (AO3) — the AO3 mark separates A from A*.

Synoptic links

Connects to:

Topic 7 — Electric Fields (Coulomb's law analogy): Coulomb's law $F = kQ_1 Q_2 / r^2$ is structurally identical to Newton's gravitational law, with charge $Q$ in place of mass $m$ and $k = 1/(4\pi\varepsilon_0)$ in place of $G$ . The crucial physical difference is that gravity is always attractive (no "negative mass") whereas Coulomb forces can repel. Edexcel routinely asks candidates to compare and contrast — name the analogy and the breaking-point.
Topic 12 — Gravitational Field Strength: $g = GM/r^2$ is Newton's law divided by the test mass. The radial-field formula and the inverse-square decay are inherited directly from $F \propto 1/r^2$ . A confident grasp of Newton's law makes field-strength questions almost mechanical.