Newton's law of gravitation — inverse square law — Physics

Before this

This leads to

A force that reaches across space

Drop a pen. It falls, because the Earth pulls it. Now think about the Moon. It is about 384 000 km away, with nothing in between but empty space. In the circular motion lectures you found that the Moon accelerates towards the Earth all the time, at about 0.0027 m/s². Something must pull it. The only candidate is the Earth.

So the same Earth pulls the pen with enough force to make it accelerate at 9.8 m/s², and pulls the Moon with only enough for 0.0027 m/s². Both pulls point to the Earth's centre. Is this one force that gets weaker with distance, or two different things?

Newton's answer was that it is one force, and that it acts between every pair of objects in the universe. The pen pulls the Earth. You pull the person next to you. The Sun pulls the Moon. This lecture builds the rule that tells you how strong that pull is, and explains why it weakens with distance in exactly the way it does.

What the pull depends on

Start with mass. Two pens pulled by the Earth feel twice the force of one pen. That is just weight: $W = m g$ , so doubling the mass doubles the force. If the pull really is between the pen and the Earth, the same must be true from the Earth's side. A planet with twice the Earth's mass would pull the pen twice as hard. So the force is proportional to both masses. Double one mass, and the force doubles. Double both, and the force becomes four times as large. The force follows the product $m_{1} m_{2}$ .

Now distance. The circular motion lectures used the Moon test. The Moon is about 60 times as far from the Earth's centre as a pen on the ground. Its acceleration, 0.0027 m/s², is very close to $9.8 \div 3600$ . And $6 0^{2} = 3600$ . So at 60 times the distance, the pull per kilogram is $\frac{1}{3600}$ as strong. The force falls with the square of the distance.

Put the two together, and you have Newton's law.

Newton's law of gravitation

Every two masses attract each other. The force is proportional to the product of the masses and inversely proportional to the square of the distance between their centres.

$F = \frac{G m _{1} m _{2}}{r ^{2}}$

$F$ is the force on each mass in N. $m_{1}$ and $m_{2}$ are the masses in kg. $r$ is the distance between their centres in m. $G \approx 6.67 \times 1 0^{- 11}$ N m²/kg² is the gravitational constant.

Three words in that box need care. Every: the law has no exceptions for small objects. Each: the same size of force acts on both masses, in opposite directions. Centres: for a planet or a ball, $r$ is measured from centre to centre, not from surface to surface. Newton proved that a uniform sphere pulls on things outside it exactly as if all its mass were at its centre. That is why the Earth's radius is always part of $r$ .

Why the square?

This is the part most textbooks rush. Spend time here, because it turns a rule into something you can see.

Picture the influence of a mass as lines spreading out from it in every direction, like paint sprayed from a single point. The number of lines never changes. What changes is how much space they have to spread over.

The same lines pass through one square at distance r, four equal squares at 2r and nine at 3r. Each square at 2r receives a quarter of the lines; each square at 3r receives a ninth.

At distance $r$ , the lines pass through one square of area $A$ . At twice the distance, the same lines have spread out in both directions, sideways and up-and-down. So they fill a square twice as wide and twice as tall: an area of $2 \times 2 = 4$ times $A$ . At three times the distance, they fill $3 \times 3 = 9$ times the area. The area grows with the square of the distance because space has two directions across the path of the lines, not one.

So a square of the original size at $2 r$ catches only $\frac{1}{4}$ of the lines, and at $3 r$ only $\frac{1}{9}$ . The strength of the pull goes the same way. That is it. That is the whole idea of an inverse square law: a fixed amount of influence, spread over the surface of a sphere whose area grows as $r^{2}$ .

This argument does not depend on what is spreading. Light from a lamp and the electric force from a charge obey inverse square laws for the same geometric reason. A torch beam held twice as far from a wall lights an area four times as large, and each part of it is a quarter as bright.

Use the law below. The astronaut has a mass of 70 kg. Move them away from the Earth and watch the force fall.

Explore — move an astronaut away from the Earth

Move the astronaut's height above the surface. Watch the distance from the centre and the force. Then try the presets, and find the height where the force is exactly a quarter of its value on the ground.

Height above the surface: 0 km

The force on a 70 kg astronaut, from F = Gm₁m₂/r² with the Earth's mass 5.97 × 10²⁴ kg and radius 6370 km.

At 2 Earth radii from the centre the force is a quarter of its surface value; at 3 Earth radii it is a ninth.

Height (km)	Distance from centre (km)	Force (N)	Fraction of surface force
0	6370	687	1
400	6770	608	0.885
6370	12 740	172	0.25
35 790	42 160	16	0.023

What did you see? At the surface the force is about 687 N, which matches the weight you would calculate from $W = m g = 70 \times 9.8 = 686$ N. At twice the distance from the centre, it is a quarter of that, about 172 N. At three times, it is a ninth. The curve never reaches zero. Gravity gets weaker with distance, but it never stops.

Stop and think

Astronauts on the International Space Station, about 400 km up, float around inside it. Is that because there is no gravity there?

Show the answer

No. Use the explorer, or the law. At 400 km, $r = 6.37 \times 1 0^{6} + 0.40 \times 1 0^{6} = 6.77 \times 1 0^{6}$ m. The ratio $(\frac{6.37}{6.77})^{2} \approx 0.885$ , so the pull on a 70 kg astronaut is about 608 N, still 89% of the pull on the ground. The astronauts float because they and the station are falling together round the Earth, as the circular motion lectures showed. Gravity is what keeps them in orbit. Without it, they would fly off in a straight line.

How small is G?

The constant $G$ is extremely small: 0.0000000000667 in SI units. This is why you never notice the pull between ordinary objects. Gravity only becomes large when at least one of the masses is enormous, such as a planet or a star.

Measuring $G$ is hard for exactly that reason. In 1797–1798, Henry Cavendish hung a rod with a small lead ball (about 0.73 kg) on each end from a thin wire. He then placed two large lead balls (about 158 kg each) near the small ones. The pull between them was about $1.7 \times 1 0^{- 7}$ N, roughly the weight of a large grain of sand. It twisted the wire very slightly, and Cavendish measured the twist with telescopes. He reported the result as the density of the Earth, about 5.45 times that of water. Later physicists used his measurements to find $G$ . Today's value is about $6.674 \times 1 0^{- 11}$ N m²/kg².

Worked examples

Worked Example 1 — The Earth and the Moon

The Earth's mass is $5.97 \times 1 0^{24}$ kg, the Moon's mass is $7.35 \times 1 0^{22}$ kg, and their centres are about $3.84 \times 1 0^{8}$ m apart. Find the force between them.

F = (6.67 \times 10⁻¹¹ \times 5.97 \times 10²⁴ \times 7.35 \times 10²²) \div (3.84 \times 10⁸)²

Handle the numbers and the powers of ten separately. On top: $6.67 \times 5.97 \times 7.35 \approx 292.7$ , and $1 0^{- 11} \times 1 0^{24} \times 1 0^{22} = 1 0^{35}$ . Underneath: $3.8 4^{2} \approx 14.75$ , and $(1 0^{8})^{2} = 1 0^{16}$ .

So $F \approx \frac{292.7}{14.75} \times 1 0^{35 - 16} \approx 19.8 \times 1 0^{19} \approx 1.98 \times 1 0^{20}$ N.

Check against the Moon test. The Moon's acceleration should be this force divided by its mass: $\frac{1.98 \times 1 0 ^{20}}{7.35 \times 1 0 ^{22}} \approx 0.0027$ m/s². That is the centripetal acceleration you found from the Moon's orbit. The law of gravitation and the circular motion lectures give the same number.

Worked Example 2 — Two people one metre apart

Amara and Kenji each have a mass of 70 kg. Their centres are 1.0 m apart. Find the gravitational force between them, and compare it with Amara's weight.

$F = \frac{6.67 \times 1 0 ^{- 11} \times 70 \times 70}{1. 0 ^{2}} \approx 3.3 \times 1 0^{- 7}$ N.

Amara's weight is $70 \times 9.8 = 686$ N. So the pull from Kenji is about $686 \div (3.3 \times 1 0^{- 7}) \approx 2 \times 1 0^{9}$ times smaller: two thousand million times. The force is real, but it is far too small to feel. Friction with the floor is enormously larger.

Before the next example, test a common idea about the Earth and the Moon.

Predict — how hard does the Moon pull the Earth?

The Earth pulls the Moon with a force of about 2 × 10²⁰ N. The Earth's mass is about 81 times the Moon's. How hard does the Moon pull the Earth? Choose before you read on.

The Moon pulls the Earth with exactly the same size of force, about 2 × 10²⁰ N, in the opposite direction. The formula F = Gm₁m₂/r² contains both masses once each, so it gives one force for the pair. This is Newton's third law. The Earth's acceleration is 81 times smaller, because its mass is 81 times larger.

Worked Example 3 — Ratios instead of the formula

A spacecraft feels a gravitational force of 800 N on the Earth's surface. Find the force on it (a) at a distance of 3 Earth radii from the Earth's centre, and (b) at a height of 3 Earth radii above the surface.

Nothing changes except $r$ , so $F$ is proportional to $\frac{1}{r ^{2}}$ . On the surface, $r$ is 1 Earth radius.

(a) $r$ becomes 3 times as large, so $F$ becomes $\frac{1}{3 ^{2}} = \frac{1}{9}$ as large: $800 \div 9 \approx 89$ N.

(b) A height of 3 Earth radii above the surface is 4 Earth radii from the centre. So $F$ becomes $\frac{1}{4 ^{2}} = \frac{1}{16}$ as large: $800 \div 16 = 50$ N.

The two answers are very different, and the only difference is the word "height". Always convert a height to a distance from the centre before you square anything.

Worked Example 4 — Change two things at once

Two asteroids attract each other with a force $F$ . One asteroid's mass is tripled, and the distance between them is halved. What is the new force?

Treat each change as a factor. Tripling one mass multiplies the force by 3. Halving the distance multiplies it by $\frac{1}{( 1/2 ) ^{2}} = 4$ . So the new force is $3 \times 4 = 12$ times as large: $12 F$ .

Check with numbers. Take $m_{1} = m_{2} = 1$ , $r = 2$ , and ignore $G$ : the force is $\frac{1}{4}$ . After the change, $m_{1} = 3$ and $r = 1$ : the force is $3$ . And $3 \div \frac{1}{4} = 12$ . ✓

The same skeleton as the electric force

Later in physics you will meet Coulomb's law for the force between two electric charges:

F = \frac{k q _{1} q _{2}}{r ^{2}}

It has the same shape as Newton's law: a constant, a product of two "amounts", and $r^{2}$ underneath. The reason is the same spreading geometry. There are two differences. Charges can be positive or negative, so electric forces can attract or repel. Mass is always positive, so gravity only attracts. And the electric force is far stronger. In a hydrogen atom, the electric pull between the proton and the electron is about $2 \times 1 0^{39}$ times the gravitational pull between them. Gravity rules the universe only because large objects have almost no overall charge, so their electric forces cancel, while their masses all add up.

Three classic mistakes

Trap 1 — Using the height instead of the distance from the centre

For a satellite 400 km up, $r$ is $6.37 \times 1 0^{6} + 0.40 \times 1 0^{6} = 6.77 \times 1 0^{6}$ m, not $4.0 \times 1 0^{5}$ m. Using the height alone would make the force nearly 290 times too large.

Trap 2 — Forgetting to square

Doubling the distance does not halve the force. It divides it by $2^{2} = 4$ . When you use ratios, write the square explicitly every time: " $r$ × 3, so $F$ × $\frac{1}{9}$ ".

Trap 3 — Thinking the bigger mass pulls harder

The Earth does not pull the Moon harder than the Moon pulls the Earth. The law gives one force, and it acts on each mass. The Earth moves less because its mass is larger, not because it feels less force.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Use $G = 6.67 \times 1 0^{- 11}$ N m²/kg².

A satellite feels a pull of 2000 N on the launch pad. It is placed in orbit at a height of one Earth radius above the surface. What is the pull on it now?

Hint 1
How far is the satellite from the Earth's centre, in Earth radii?

Hint 2
It is 2 Earth radii from the centre, so the distance has doubled.

Full answer
$r$ goes from 1 to 2 Earth radii, so the force is multiplied by $\frac{1}{2 ^{2}} = \frac{1}{4}$ : $2000 \div 4 = 500$ N. Check: 500 × 4 = 2000 ✓.
Two cargo ships each have a mass of $5.0 \times 1 0^{7}$ kg. Their centres are 100 m apart. Find the gravitational force between them. Is it enough to pull them together?

Hint 1
Use the formula directly: $F = \frac{G m _{1} m _{2}}{r ^{2}}$ .

Hint 2
$m_{1} m_{2} = 2.5 \times 1 0^{15}$ kg², and $r^{2} = 1 0^{4}$ m².

Full answer
$F = \frac{6.67 \times 1 0 ^{- 11} \times 2.5 \times 1 0 ^{15}}{1 0 ^{4}} \approx 17$ N. That is about the weight of a 1.7 kg bag of rice. Each ship weighs about $5.0 \times 1 0^{7} \times 9.8 \approx 4.9 \times 1 0^{8}$ N, so 17 N is tiny by comparison. The resistance of the water, the wind and the current are all far larger. The ships will not drift together.
Two masses attract with a force of 36 N. Both masses are doubled, and the distance between them is tripled. Find the new force.

Hint 1
Write one factor for each change, then multiply them.

Hint 2
Doubling both masses gives × 4. Tripling the distance gives × $\frac{1}{9}$ .

Full answer
The force is multiplied by $2 \times 2 \times \frac{1}{9} = \frac{4}{9}$ . So $F = 36 \times \frac{4}{9} = 16$ N. Check with $m_{1} = m_{2} = 1$ , $r = 1$ (force 1) and then $m_{1} = m_{2} = 2$ , $r = 3$ (force $\frac{4}{9}$ ) ✓.
The Sun's mass is about $1.99 \times 1 0^{30}$ kg, and the Moon is about $1.50 \times 1 0^{11}$ m from it. (a) Find the Sun's pull on the Moon and compare it with the Earth's pull, $1.98 \times 1 0^{20}$ N, from Worked Example 1. (b) If the Sun pulls harder, why does the Moon go round the Earth?

Hint 1
For (a), use the formula with the Sun's mass and the Moon's mass, $7.35 \times 1 0^{22}$ kg.

Hint 2
For (b), think about the Earth too. The Sun pulls on the Earth as well, and the Earth is at almost the same distance from the Sun.

Full answer
(a) $F = \frac{6.67 \times 1 0 ^{- 11} \times 1.99 \times 1 0 ^{30} \times 7.35 \times 1 0 ^{22}}{( 1.50 \times 1 0 ^{11} ) ^{2}} \approx 4.3 \times 1 0^{20}$ N. That is about 2.2 times the Earth's pull. (b) The Sun gives the Earth and the Moon almost the same acceleration, because they are at almost the same distance from it. So the Sun carries both of them round its orbit together. The Earth's pull is what makes the Moon move round the Earth while the pair travels. This surprising result is also why the Moon's path around the Sun always curves towards the Sun.

The bigger idea

Newton published the law in 1687, in the Principia. Before it, many people thought that the heavens and the Earth followed different rules. The law of gravitation says they do not. The pull that makes a pen fall is the same pull that bends the Moon's path, at a strength set only by the masses and the distance. One equation connected a dropped object with the motion of the planets.

The law describes a force between two masses. The next lecture looks at it from the point of view of one mass alone: the pull it would exert on each kilogram placed near it. That is the gravitational field strength, and it explains why $g$ is 9.8 N/kg here, and less on a mountain or on the Moon.

Key concepts in this lesson

Every two masses attract each other with a gravitational force $F = \frac{G m _{1} m _{2}}{r ^{2}}$ .
The force is proportional to the product of the masses and inversely proportional to the square of the distance between their centres.
The square comes from geometry: the same influence spreads over a sphere whose area grows as $r^{2}$ .
The distance $r$ is measured centre to centre, so a height above the surface must have the planet's radius added.
$G \approx 6.67 \times 1 0^{- 11}$ N m²/kg² is so small that gravity between everyday objects is too weak to notice.
The two masses feel forces of the same size in opposite directions, as Newton's third law requires.