Newton's Law of Gravitation

Spec mapping: OCR H556 Module 5.4 — Gravitational fields (Newton's law of gravitation as an inverse-square attractive force between two point masses; the universal gravitational constant $G$G; the shell theorem for spherically symmetric bodies; gravitational field strength of a spherical mass $g = GM/r^{2}$g=GM/r2; the variation of $g$g with $r$r both inside and outside a uniform sphere). Refer to the official OCR H556 specification document for exact wording.

In 1687, in his Principia Mathematica, Isaac Newton wrote down what is still one of the most consequential equations in science: the inverse-square law of universal gravitation. Every pair of masses in the Universe attracts every other pair, with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres of mass. With this one law, plus his three laws of motion, Newton explained the fall of an apple in an orchard, the orbit of the Moon around the Earth, the orbit of the Earth around the Sun, the tidal bulges in the oceans, and the precession of the equinoxes — unifying terrestrial and celestial mechanics for the first time in human history.

This lesson is the technical heart of OCR Module 5.4. We state the law, motivate the inverse-square geometry, prove (in outline) the shell theorem that lets us treat extended spherical bodies as point masses, derive the spherical-mass field result $g = GM/r^{2}$g=GM/r2, sketch the $g$g-vs-$r$r graph through and outside a planet, and work several specimen calculations including the field at altitude, the mass of a planet from surface gravity, the (tiny) force between two students, and the location of the Earth-Moon null-field point. The A* discriminators land repeatedly: using $r$r from the centre of mass not the altitude, distinguishing $G$G (universal constant) from $g$g (local field), and recognising when the shell theorem licenses the point-mass shortcut.

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Statement of the Law

The gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is attractive and acts along the line joining the two masses.

In symbols, the magnitude of the force is

$F = \frac{GMm}{r^{2}}$F=r2GMm​

with the vector form that captures both magnitude and direction:

$\vec{F}_{\text{on } m \text{ by } M} = -\frac{GMm}{r^{2}} \, \hat{r}$Fon m by M​=−r2GMm​r^

where $\hat{r}$r^ is the unit vector pointing from $M$M to $m$m, and the minus sign records that the force on $m$m points back towards $M$M (i.e. opposite to $\hat{r}$r^). The corresponding Newton-3 pair acts on $M$M with the same magnitude $F$F in the opposite direction (towards $m$m).

Symbols:

• $F$F — magnitude of the gravitational force on each mass due to the other, in newtons.
• $M, m$M,m — the two masses, in kilograms.
• $r$r — the distance between their centres of mass, in metres. This is the load-bearing detail: $r$r is never the gap between the surfaces, never the altitude above a planet; it is always centre-to-centre.
• $G = 6.67 \times 10^{-11} \text{ N m}^{2} \text{ kg}^{-2}$G=6.67×10−11 N m2 kg−2 — the universal gravitational constant, provided in the OCR data booklet.

Gravity is always attractive. There is no "negative mass", no anti-gravity, no shielding — in classical physics, every pair of masses pulls on every other pair.

The Gravitational Constant $G$G

$G$G is a fundamental constant of nature, on the same footing as the speed of light $c$c and the Planck constant $h$h. It was first measured by Henry Cavendish in 1798 using a delicate torsion-balance experiment — a pair of lead spheres suspended on a fine wire was deflected by the gravitational pull of two large lead balls placed nearby; the deflection angle, combined with the known torsional stiffness of the wire, gave the force per unit deflection and hence $G$G. Cavendish's value of $6.74 \times 10^{-11}$6.74×10−11 was within 1% of today's value — a remarkable feat of nineteenth-century precision metrology.

The currently accepted value is

$G = 6.674 \times 10^{-11} \text{ N m}^{2} \text{ kg}^{-2}$G=6.674×10−11 N m2 kg−2

OCR uses $G = 6.67 \times 10^{-11}$G=6.67×10−11 N m² kg⁻² (3 sf) in calculations, which is what you should use throughout.

Three observations about $G$G:

1. It is the smallest of the fundamental constants. Gravity is by far the weakest of the four fundamental forces. The ratio of the gravitational force between two protons to their electrostatic repulsion is about $10^{-36}$10−36 — gravity is 36 orders of magnitude weaker. Gravity wins on astronomical scales only because (i) all masses are positive (no cancellation) and (ii) the force is long-range ($1/r^{2}$1/r2 rather than the exponentially screened range of the weak and strong forces).
2. It is the only fundamental constant known to fewer than 5 significant figures. Measuring $G$G is fiendishly hard because gravity cannot be shielded and the laboratory apparatus itself contributes to the gravitational field. Modern measurements scatter by ~$10^{-4}$10−4, much worse than $c$c (defined exactly) or $h$h (known to ~$10^{-10}$10−10).
3. It is universal. The same $G$G applies to the apple in the orchard, the Moon's orbit, the Pioneer probe's escape, and the binary-neutron-star inspiral observed by LIGO — to within current experimental precision. Newton's claim that gravity is universal is one of the great empirical confirmations of physical law.

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Inverse-Square Geometry — Why $1/r^{2}$1/r2 and Not $1/r$1/r?

The $1/r^{2}$1/r2 fall-off is not arbitrary — it follows from a geometric fact about three-dimensional space. Imagine a closed sphere of radius $r$r enclosing a point mass $M$M. The "gravitational flux" $\Phi = g \times A$Φ=g×A through the sphere is constant (it depends only on the enclosed mass, not on $r$r — this is Gauss's law for gravity). The sphere's area is $A = 4\pi r^{2}$A=4πr2. For the flux to stay constant as $r$r increases, $g$g must fall as $1/r^{2}$1/r2.

graph TD
    A["Mass M at centre"] --> B["Sphere radius r<br/>area = 4π r²<br/>flux Φ = 4π G M"]
    A --> C["Sphere radius 2r<br/>area = 16π r²<br/>same Φ → g is 1/4"]
    A --> D["Sphere radius 3r<br/>area = 36π r²<br/>same Φ → g is 1/9"]

Quantitatively: doubling the distance from a point mass reduces the field (and force) to $1/4$1/4; tripling reduces it to $1/9$1/9; halving the distance quadruples it. This $1/r^{2}$1/r2 scaling is the geometric signature of any "radial spread" in three dimensions — Coulomb's law for electrostatics, the inverse-square law for light intensity from a point source, and even the inverse-square fall-off of sound intensity all share the same geometric origin.

A point worth pausing on: if space were two-dimensional, gravity would fall as $1/r$1/r (the perimeter of a circle is $2\pi r$2πr, so flux conservation would give $g \propto 1/r$g∝1/r). If space were four-dimensional, gravity would fall as $1/r^{3}$1/r3. The fact that planets have stable closed orbits at all is a consequence of $1/r^{2}$1/r2 being the unique exponent (along with $r$r for a Hooke-law spring) that supports them — Bertrand's theorem, an undergraduate result in classical mechanics.

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Force Between Extended Bodies — the Shell Theorem

Newton's law is stated for point masses, but real planets and stars are extended bodies thousands of kilometres across. How do we apply $F = GMm/r^{2}$F=GMm/r2 when the source has spatial extent?

Newton himself proved a theorem — using calculus methods he invented for the purpose — that handles the spherical case:

Shell theorem (Newton, 1687). A spherically symmetric body of total mass $M$M acts, on a point mass outside it, exactly as if all of its mass were concentrated at its centre.

A complementary result handles the inside case:

Shell theorem (inside). A spherically symmetric shell of mass exerts zero net gravitational force on a point mass anywhere inside the shell.

Two extraordinary simplifications follow. First: the gravitational pull of the Earth on an apple resting on the surface is exactly the same as the pull would be from a point mass of $5.97 \times 10^{24}$5.97×1024 kg located $6.37 \times 10^{6}$6.37×106 m below — at the Earth's centre. The whole multi-thousand-kilometre extent of the planet collapses, for gravitational purposes, to a single point. Second: a hollow spherical shell — say, a Dyson sphere enclosing a star — exerts no gravitational force on anything inside it, so a marble drifting inside such a shell would float freely with no gravitational tug whatsoever.

For non-spherical shapes (a cube of rock, an irregular asteroid, a dumb-bell), the shell theorem fails — you must in principle integrate Newton's law over the mass distribution. In OCR A-Level you only ever apply the shell theorem to spherical or near-spherical bodies.

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Gravitational Field Strength of a Spherical Mass

Combining Newton's law $F = GMm/r^{2}$F=GMm/r2 with the field definition $g = F/m$g=F/m (from Lesson 7):

$g = \frac{GM}{r^{2}}$g=r2GM​

This is the field strength at distance $r$r from the centre of a spherically symmetric body of mass $M$M, evaluated outside the body ($r \ge R$r≥R where $R$R is the body's radius). It is one of the most heavily used equations in OCR Module 5.4 and the rest of this lesson is built around it.

Two immediate consequences:

• $g \propto 1/r^{2}$g∝1/r2, so doubling your distance from the Earth's centre drops the field to a quarter.
• $g$g is independent of the test mass $m$m. Field is a property of the source, not the probe.

Calculating $g$g at the Earth's Surface

Let us verify the standard value $g_{s} = 9.81 \text{ N kg}^{-1}$gs​=9.81 N kg−1 from first principles. The Earth has mass $M_{E} = 5.97 \times 10^{24}$ME​=5.97×1024 kg and radius $R_{E} = 6.37 \times 10^{6}$RE​=6.37×106 m.

$g = \frac{GM_{E}}{R_{E}^{2}} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.37 \times 10^{6})^{2}}$g=RE2​GME​​=(6.37×106)26.67×10−11×5.97×1024​

Numerator: $6.67 \times 5.97 = 39.82$6.67×5.97=39.82, so $39.82 \times 10^{13} = 3.98 \times 10^{14}$39.82×1013=3.98×1014 N m² kg⁻¹.

Denominator: $(6.37)^{2} = 40.6$(6.37)2=40.6, so $(6.37 \times 10^{6})^{2} = 4.06 \times 10^{13}$(6.37×106)2=4.06×1013 m².

$g = \frac{3.98 \times 10^{14}}{4.06 \times 10^{13}} = 9.81 \text{ N kg}^{-1}$g=4.06×10133.98×1014​=9.81 N kg−1

The agreement is exact to 3 sf. This is not a coincidence — it is empirical confirmation that Newton's law plus the measured mass and radius of the Earth are mutually consistent. Historically, the chain was: measure $G$G (Cavendish, 1798), measure surface $g$g (Galileo-era pendulum experiments), then infer $M_{E}$ME​ from $M_{E} = gR_{E}^{2}/G$ME​=gRE2​/G. Cavendish's experiment was famously described as the first that "weighed the Earth".

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Worked Example 1 — Field Strength at the ISS Orbit

Calculate $g$g at the altitude of the International Space Station, 400 km above the Earth's surface. Use $R_{E} = 6.37 \times 10^{6}$RE​=6.37×106 m and $M_{E} = 5.97 \times 10^{24}$ME​=5.97×1024 kg.

Distance from the Earth's centre: $r = R_{E} + h = 6.37 \times 10^{6} + 0.40 \times 10^{6} = 6.77 \times 10^{6}$r=RE​+h=6.37×106+0.40×106=6.77×106 m.

$g = \frac{GM_{E}}{r^{2}} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.77 \times 10^{6})^{2}} = \frac{3.98 \times 10^{14}}{4.58 \times 10^{13}} = 8.69 \text{ N kg}^{-1}$g=r2GME​​=(6.77×106)26.67×10−11×5.97×1024​=4.58×10133.98×1014​=8.69 N kg−1

Only 11% less than the surface value. The ISS astronauts are emphatically not in "zero gravity" — they are in free fall alongside the station, with nearly the same gravitational field they would feel on the ground. They feel weightless because both the station and the astronauts are accelerating identically; there is no contact force between them, and "weight" as a felt sensation is the contact force, not the gravitational force. Lesson 11 develops this carefully.

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Worked Example 2 — Mass of the Moon from Surface Gravity

The gravitational field strength at the lunar surface is $g_{\text{Moon}} = 1.62$gMoon​=1.62 N kg⁻¹ and the Moon's radius is $R_{\text{Moon}} = 1.74 \times 10^{6}$RMoon​=1.74×106 m. Calculate the Moon's mass.

Rearranging $g = GM/R^{2}$g=GM/R2:

$M = \frac{gR^{2}}{G} = \frac{1.62 \times (1.74 \times 10^{6})^{2}}{6.67 \times 10^{-11}}$M=GgR2​=6.67×10−111.62×(1.74×106)2​

Numerator: $(1.74)^{2} = 3.03$(1.74)2=3.03, so $(1.74 \times 10^{6})^{2} = 3.03 \times 10^{12}$(1.74×106)2=3.03×1012 m²; then $1.62 \times 3.03 \times 10^{12} = 4.90 \times 10^{12}$1.62×3.03×1012=4.90×1012.

$M = \frac{4.90 \times 10^{12}}{6.67 \times 10^{-11}} = 7.35 \times 10^{22} \text{ kg}$M=6.67×10−114.90×1012​=7.35×1022 kg

The accepted lunar mass is $7.35 \times 10^{22}$7.35×1022 kg — to 3 sf the calculation is exact. The same method, applied to any astronomical body with a measurable surface $g$g, gives its mass. Apollo-era surface gravimeters used this to refine the lunar mass; modern equivalents (orbiter Doppler tracking, GRAIL twin-spacecraft mission) give the Moon's mass to better than 1 part in $10^{8}$108.

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Worked Example 3 — Gravitational Force Between Two Students

Two students, each of mass $m = 60$m=60 kg, stand $r = 0.50$r=0.50 m apart. Calculate the gravitational force of attraction between them.

$F = \frac{Gm^{2}}{r^{2}} = \frac{6.67 \times 10^{-11} \times 60 \times 60}{(0.50)^{2}} = \frac{6.67 \times 10^{-11} \times 3600}{0.25}$F=r2Gm2​=(0.50)26.67×10−11×60×60​=0.256.67×10−11×3600​

Numerator: $6.67 \times 3600 = 2.40 \times 10^{4}$6.67×3600=2.40×104, so $6.67 \times 10^{-11} \times 3600 = 2.40 \times 10^{-7}$6.67×10−11×3600=2.40×10−7 N m².

$F = \frac{2.40 \times 10^{-7}}{0.25} = 9.60 \times 10^{-7} \text{ N}$F=0.252.40×10−7​=9.60×10−7 N

About one micronewton — utterly negligible compared with friction, drag, or even the static-electricity force from rubbing hair on a jumper. This is why mutual gravitational attraction between ordinary objects is never directly noticeable: a single 60 kg human exerts the same force on you as the weight of about $10^{-7}$10−7 kg, equivalent to about a hundredth the weight of a grain of sand. Gravity wins astronomically only because large masses are involved.

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Worked Example 4 — Earth-Moon Null Point

The Earth ($M_{E} = 5.97 \times 10^{24}$ME​=5.97×1024 kg) and the Moon ($M_{M} = 7.35 \times 10^{22}$MM​=7.35×1022 kg) are separated by $d = 3.84 \times 10^{8}$d=3.84×108 m centre-to-centre. At what point along the Earth-Moon line does a test mass experience zero net gravitational field?

Let $x$x be the distance from the Earth's centre to the null point; the distance from the Moon is then $d - x$d−x. At the null point the two field magnitudes balance:

$\frac{GM_{E}}{x^{2}} = \frac{GM_{M}}{(d-x)^{2}}$x2GME​​=(d−x)2GMM​​

Dividing and taking the square root:

$\frac{(d-x)^{2}}{x^{2}} = \frac{M_{M}}{M_{E}} = \frac{7.35 \times 10^{22}}{5.97 \times 10^{24}} = 0.01231$x2(d−x)2​=ME​MM​​=5.97×10247.35×1022​=0.01231

$\frac{d-x}{x} = \sqrt{0.01231} = 0.1109$xd−x​=0.01231​=0.1109

Rearranging:

$d - x = 0.1109 \, x \quad \Rightarrow \quad d = 1.1109 \, x \quad \Rightarrow \quad x = \frac{d}{1.1109} = 3.46 \times 10^{8} \text{ m}$d−x=0.1109x⇒d=1.1109x⇒x=1.1109d​=3.46×108 m

The null point sits about $346\,000$346000 km from the Earth's centre and only $38\,000$38000 km from the Moon's centre — much closer to the lighter body, as you would expect (the weaker field strength has to be "boosted" by reducing $r$r). This is not a Lagrange point; it is purely the location where the gravitational fields cancel, ignoring the centrifugal contribution from co-rotation. The actual Earth-Moon L1 (a true equilibrium of gravity + centrifugal force in the rotating frame) is at a slightly different location.

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$g$g Inside and Outside a Uniform-Density Planet

A standard OCR graph asks for $g$g as a function of $r$r both inside and outside a spherical planet of mass $M$M and radius $R$R.

Outside ($r \ge R$r≥R): by the shell theorem, $g = GM/r^{2}$g=GM/r2 — the $1/r^{2}$1/r2 tail.

Inside ($r \le R$r≤R): by the inside shell theorem, only the mass enclosed within radius $r$r contributes; shells outside that radius cancel. For a uniform-density sphere, the enclosed mass scales as $r^{3}$r3:

$M_{\text{enc}}(r) = M \times \frac{(4/3)\pi r^{3}}{(4/3)\pi R^{3}} = M \frac{r^{3}}{R^{3}}$Menc​(r)=M×(4/3)πR3(4/3)πr3​=MR3r3​

so

$g(r) = \frac{G M_{\text{enc}}}{r^{2}} = \frac{GM r^{3}/R^{3}}{r^{2}} = \frac{GM}{R^{3}} r$g(r)=r2GMenc​​=r2GMr3/R3​=R3GM​r

— a linear rise from $g = 0$g=0 at the centre to $g_{s} = GM/R^{2}$gs​=GM/R2 at the surface. Sketch:

Three features for OCR:

1. $g = 0$g=0 at the centre ($r = 0$r=0): a marble at the centre of a uniform planet experiences no net gravitational force.
2. $g$g peaks at the surface ($r = R$r=R): this is where the gravitational field is strongest.
3. The curve has a cusp at $r = R$r=R: the gradient changes discontinuously from $+\text{const}$+const (linear inside) to $-2g_{s}/R$−2gs​/R (the slope of the $1/r^{2}$1/r2 tail) at the surface. Real planets have density gradients (denser cores), so the inside curve is not a perfect straight line, but the qualitative shape is unchanged.

A common misconception is "the deeper you dig, the stronger gravity gets" — wrong! Inside a uniform-density planet, $g$g falls linearly from its surface maximum down to zero at the centre. The shell of mass outside your radius cancels by the inside shell theorem and stops contributing.

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Why the Moon Does Not Fall on Us

Newton's most famous conceptual insight was the recognition that the Moon is falling towards the Earth — continuously — but its tangential velocity is large enough that its path curves inward at exactly the rate needed to maintain a circular (well, nearly circular) orbit. The acceleration at the Moon's orbital radius $r = 3.84 \times 10^{8}$r=3.84×108 m is

$g_{\text{at Moon}} = \frac{GM_{E}}{r^{2}} = \frac{3.98 \times 10^{14}}{(3.84 \times 10^{8})^{2}} = \frac{3.98 \times 10^{14}}{1.47 \times 10^{17}} = 2.70 \times 10^{-3} \text{ m s}^{-2}$gat Moon​=r2GME​​=(3.84×108)23.98×1014​=1.47×10173.98×1014​=2.70×10−3 m s−2