The universal law of gravitation

Newton's inverse-square law of gravity, adding gravitational forces, and where the value of g comes from.

Follows MIT 8.01SC lesson 5MIT 8.01SC Lesson 5: Gravity (5.1–5.3)On MIT OpenCourseWare: the lecture and its readings

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Full reading path (5 pages)

Vectors (8.01)
Position and velocity in one dimension (8.01)
Acceleration in one dimension (8.01)
Motion in two dimensions and projectiles (8.01)
Newton's laws of motion (8.01)

On this page · 6 sections

Newton's law of gravity
Adding gravitational forces
Gravity at Earth's surface: the value of g
Why everything falls the same way
Try it
Summary

The same force that pulls an apple to the ground keeps the Moon in its orbit, holds galaxies together and, on the largest scales, slows the expansion of the universe. Newton's great step was to see that one law covers all of these.

Newton's law of gravity#

Every two point masses attract each other. The force acts along the line joining them, grows with both masses, and falls off with the square of the distance:

Universal law of gravitation

\vec F^{\,G}_{1,2} = -G\,\frac{m_1 m_2}{r_{1,2}^2}\,\hat r_{1,2}, \qquad G = 6.67\times10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2

In words

vector F to the power G sub 1,2 equals minus G m sub 1 m sub 2 over r sub 1,2 squared r hat sub 1,2 , G equals 6.67 times 10 to the power minus 11 N m squared / kg squared

Source: 8.01SC Course Notes, Chapter 8: Applications of Newton's Second Law, Section 8.2 (8.2.1–8.2.3), Sec. 8.2.1, Eq. (8.2.1)

As in Newton's laws, $\vec F_{1,2}$ is the force on body 2 due to body 1. The unit vector $\hat r_{1,2}$ points from 1 to 2, and the minus sign turns the force around so that it points back toward body 1: gravity attracts.

Swap the labels and only the unit vector changes sign, so $\vec F_{2,1} = -\vec F_{1,2}$ . Gravity automatically obeys Newton's third law.

The inverse-square dependence is the heart of the law: double the distance and the force drops to a quarter.

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m_1 m_2 \text{ (relative)}

How the force falls with distance. Doubling r cuts the inverse-square force to a quarter; a 1/r force would only halve.

Adding gravitational forces#

When several bodies pull on one object, the forces add as vectors. This superposition is what MIT's worked example 5.2 practises.

Worked example

Where Earth and Moon pull equally

A spacecraft travels in a straight line from the Earth to the Moon. At what distance from the Earth's centre do the two gravitational pulls cancel? Use $M_E = 5.97\times10^{24}$ kg, $M_M = 7.35\times10^{22}$ kg and an Earth–Moon distance $D = 3.84\times10^{8}$ m.

Write both forces

At distance $x$ from the Earth, the Earth pulls back toward it with $GM_E m/x^2$ , and the Moon pulls forward with $GM_M m/(D - x)^2$ .

Set the forces equal

$\dfrac{M_E}{x^2} = \dfrac{M_M}{(D - x)^2}$ , so $\dfrac{D - x}{x} = \sqrt{\dfrac{M_M}{M_E}} = 0.111$ . The spacecraft's own mass $m$ and $G$ cancel.

Solve

$x = \dfrac{D}{1 + 0.111} = 3.46\times10^{8}$ m, about 90% of the way to the Moon.

Sanity check

The balance point should be much closer to the Moon, because the Moon is about 81 times less massive. It is: the distance ratio is $\sqrt{81} = 9$ , and $3.46 : 0.38 \approx 9 : 1$ .

Gravity at Earth's surface: the value of g#

A sphere with its mass spread evenly in layers attracts an outside object as if all its mass sat at its centre. So a body of mass $m$ on the surface of the Earth (mass $M_E$ , radius $R_E$ ) feels a force of size $mg$ , with

The value of g

F = mg, \qquad g = \frac{G M_E}{R_E^2} \approx 9.8\ \text{m/s}^2

In words

F equals mg, g equals G M sub E over R sub E squared is approximately 9.8 m/s squared

Source: MIT 8.01SC segment 5.3: Gravity at the Surface of the Earth – The Value of g

With $M_E = 5.97\times10^{24}$ kg and $R_E = 6.37\times10^{6}$ m this gives $9.8\ \text{m/s}^2$ , the constant used for projectiles.

In fact $g$ varies a little over the Earth's surface. The MIT notes give about $9.78\ \text{m/s}^2$ at the equator and $9.83\ \text{m/s}^2$ at the poles, mostly because the Earth spins and bulges at the equator. Geologists map tiny local changes in $g$ to find dense rock, oil and water underground.

Higher up, $g$ shrinks as $1/r^2$ :

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Gravitational acceleration versus height, from g = GM/r². At the International Space Station it is still about 89% of its surface value.

Why everything falls the same way#

The mass $m$ in the law of gravity, the gravitational mass, measures how strongly a body is pulled. The mass in $\vec F = m\vec a$ , the inertial mass, measures how hard it is to accelerate. Nothing in Newton's theory says these must be equal, but every experiment finds that they are. This is the principle of equivalence:

Principle of equivalence

m_\text{grav} = m_\text{inertial} \quad\Longrightarrow\quad a = \frac{F}{m} = \frac{G M}{r^2}\ \ \text{for every body}

In words

m sub grav equals m sub inertial which gives a equals F over m equals G M over r squared for every body

Source: 8.01SC Course Notes, Chapter 8: Applications of Newton's Second Law, Section 8.2 (8.2.1–8.2.3), Sec. 8.2.2

Einstein later made this coincidence the foundation of general relativity, where gravity is no longer a force at all but the curvature of spacetime.

Try it#

Switch gravity off and on, change the masses, and watch the orbits of the Earth and Moon respond.

Gravity and Orbits. Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC BY-NC 4.0 (https://phet.colorado.edu). Open full screen

Why cosmology needs this

In 8.286, the first model of an expanding universe uses this page's law plus one more fact about spheres. Picture a huge, uniform ball of matter whose every particle moves away from the centre according to Hubble's law. Newton's gravity pulls each particle back. Matter in shells outside a particle exerts no net force on it (the shell theorem), so only the mass inside counts, and the deceleration rate $\ddot a/a$ depends only on the density $\rho$ :

Newtonian deceleration of the expansion

\frac{d^2 a}{dt^2} = -\frac{4\pi}{3}\,G\rho\,a

In words

d squared a over dt squared equals minus 4 pi over 3 G rho a

Source: MIT 8.286 The Early Universe (Fall 2013), Lecture 6: The Dynamics of Homogeneous Expansion, Part II, lecture transcript

Whether the expansion stops and reverses, or goes on forever, then depends on whether gravity can win — just as a ball thrown upward either falls back or escapes. That question, and the open, closed and flat universes it leads to, is the subject of Lecture 7.

Summary#

Gravity between point masses: $F = Gm_1m_2/r^2$ , attractive, along the line joining them.
Gravitational forces add as vectors (superposition) and obey Newton's third law.
At Earth's surface $g = GM_E/R_E^2 \approx 9.8\ \text{m/s}^2$ ; it decreases as $1/r^2$ with height.
Gravitational and inertial mass are equal (principle of equivalence), so all bodies fall alike.

Symbols on this page

Show the 10 symbols used here

$\pi$ pi: The ratio of a circle's circumference to its diameter, π ≈ 3.14159. Say it: “pi”.
$\rho$ rho: Density: mass (or energy divided by c²) per unit volume. ρ_c is the critical density, ρ̄ the mean density. Say it: “rho”.
$\approx$ approximately equal: Is approximately equal to. Say it: “is approximately”.
$\propto$ proportional to: Is proportional to: doubling one doubles the other. Say it: “is proportional to”.
$\Rightarrow$ implies: Implies: what follows is a consequence of what came before. Say it: “which gives”.
$\times$ times: Times (multiplication). Between two vectors: the cross product. Say it: “times”.
$\vec v$ vector arrow: A vector: a quantity with a size and a direction. Say it: “vector”.
$\hat\imath$ hat: A unit vector: a vector of length 1 that only gives a direction (î, r̂). Say it: “hat”.
$\ddot a$ double dot: Second rate of change with time: ä = d²a/dt². Say it: “double dot”.
$G$ G: Newton's gravitational constant, G ≈ 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻². Say it: “G”.

Tap or click a symbol in a formula to see the same explanation. All symbols: symbols and notation.

Where this leads

In cosmology

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