Newton's law of gravitation

What you'll learn

How Newton’s law of gravitation gives the force between two masses.
Why gravitational force is an inverse-square force.
How to calculate gravitational field strength using $-\frac{GM}{r^2}$ .
Why, near Earth’s surface, $g$ is treated as uniform and equals the acceleration of free fall.

The starting idea: masses attract

Any two objects with mass attract each other gravitationally. For everyday objects, this force is tiny. For planets, moons and stars, it is large enough to control orbits and motion.

The force acts along the line joining the centres of the two masses, and it is always attractive.

Two point masses attracting each other gravitationally, with centre-to-centre separation r and equal opposite forces

Definition

Point mass

A point mass is an object treated as if all its mass is concentrated at a single point. In this topic, $r$ is the distance between the two point masses, or between the centres of two spherical bodies when they are treated as point masses.

Newton’s law of gravitation

Newton’s law of gravitation says that the gravitational force between two point masses depends on both masses and on their separation.

-\frac{GMm}{r^2}

where:

$F$ is the gravitational force in newtons, N
$G$ is the gravitational constant, $6.67 \times 10^{-11}\ \text{N m}^2\text{ kg}^{-2}$
$M$ and $m$ are the two masses in kilograms, kg
$r$ is the separation between their centres in metres, m

The negative sign shows that the force is attractive. If you define the positive radial direction as “away from” the larger mass, gravity acts in the opposite direction.

Key Idea

The inverse-square law

The magnitude of the gravitational force is proportional to the product of the masses and inversely proportional to the square of the separation: $\propto \frac{Mm}{r^2}$ . Doubling $r$ makes the force one quarter as large.

In calculations, you often use the magnitude:

\frac{GMm}{r^2}

then state or show the direction separately.

Example

Calculating the gravitational force between two masses

Two masses of 8.0 kg and 12 kg are separated by 0.50 m. Calculate the magnitude of the gravitational force between them.

Identify the quantities in SI units: $8.0\ \text{kg}$ , $12\ \text{kg}$ and $0.50\ \text{m}$ .
Substitute into Newton’s law using the magnitude of the force:
$\frac{GMm}{r^2}$ $\frac{(6.67 \times 10^{-11})(8.0)(12)}{(0.50)^2}$
Calculate the denominator and evaluate the force:
$\frac{6.40 \times 10^{-9}}{0.25} = 2.56 \times 10^{-8}\ \text{N}$
Quote the answer sensibly: the gravitational force is $2.6 \times 10^{-8}\ \text{N}$ , attractive, acting along the line joining the masses.

Common Mistake

Forgetting to square the distance

The distance term is $r^2$ , not $r$ . If the separation changes, the force changes very quickly: tripling $r$ makes the force one ninth as large.

What the negative sign means

The equation in the specification is written as:

-\frac{GMm}{r^2}

This is a direction convention, not a sign that the size of the force is negative. Force is a vector quantity, so it has magnitude and direction.

If the positive direction is chosen radially outwards from mass $M$ , then the gravitational force on mass $m$ points inwards, towards $M$ . That inward direction is negative.

Tip

Magnitude first, direction second

For most numerical questions, calculate the positive magnitude using $\frac{GMm}{r^2}$ , then add the direction in words: “towards the other mass” or “towards the centre of the planet”.

Example

Using the inverse-square relationship

A satellite moves from a distance $r$ to a distance $2 r$ from Earth’s centre. Compare the gravitational force at $2 r$ with the force at $r$ .

Start from the proportionality for fixed masses: $\propto \frac{1}{r^2}$ .
Replace $r$ by $2 r$ :
$F2r∝1(2r)2=14r2F_{2r} \propto \frac{1}{(2r)^2} = \frac{1}{4r^2}$
Compare with the original force:
$F2r=14FrF_{2r} = \frac{1}{4}F_r$

So the gravitational force becomes one quarter of its original value.

Gravitational field strength

A gravitational field is a region where a mass experiences a gravitational force.

Definition

Gravitational field strength

Gravitational field strength, $g$ , is the gravitational force per unit mass on a small test mass placed in the field. Its unit is newtons per kilogram, N kg $^{-1}$ .

So:

\frac{F}{m}

For a point mass $M$ , combine this with Newton’s law:

-\frac{GMm}{r^2}

Divide by the small mass $m$ :

-\frac{GM}{r^2}

This is the gravitational field strength due to a point mass $M$ at distance $r$ from its centre.

Key Idea

Field strength does not depend on the test mass

The value of $g$ at a point depends on the source mass $M$ and the distance $r$ , not on the mass of the object placed there.

Example

Calculating gravitational field strength near Earth

Use $kgM_\text{Earth} = 5.97 \times 10^{24}\ \text{kg}$ and $\times 10^6\ \text{m}$ to estimate the gravitational field strength at Earth’s surface.

Use the point-mass equation for gravitational field strength:
$-\frac{GM}{r^2}$
For the magnitude:
$\frac{GM}{r^2}$
Substitute the values:
$\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.37 \times 10^6)^2}$
Evaluate the numerator and denominator:
$\frac{3.98 \times 10^{14}}{4.06 \times 10^{13}}$
Calculate the field strength:
$9.8\ \text{N kg}^{-1}$

So the gravitational field strength at Earth’s surface is about $kg−19.8\ \text{N kg}^{-1}$ , directed towards Earth’s centre.

Why $g$ has two equivalent units

Gravitational field strength is measured in N kg $^{-1}$ because it is force per unit mass.

But from Newton’s second law, $F = ma$ , so:

Fm=a\frac{F}{m} = a

Therefore N kg $^{-1}$ is equivalent to m s $^{-2}$ .

That is why near Earth’s surface you will see:

\approx 9.81\ \text{N kg}^{-1}

and also:

\approx 9.81\ \text{m s}^{-2}

They are numerically the same physical quantity, expressed in equivalent units.

Uniform gravitational field near Earth’s surface

A gravitational field around a planet is actually radial: field lines point towards the planet’s centre, and the magnitude decreases with distance from the centre.

However, close to Earth’s surface, a small laboratory region is tiny compared with Earth’s radius. Over this small height, the field lines are almost parallel and the value of $g$ changes very little. So we model the field as uniform.

Radial gravitational field around Earth compared with approximately uniform gravitational field near the surface

Definition

Uniform gravitational field

A uniform gravitational field has the same field strength and direction at every point in the region being considered.

Near Earth’s surface, the field is taken as:

\approx 9.81\ \text{N kg}^{-1}

directed vertically downwards.

Common Mistake

Close to the surface only

The uniform-field approximation is excellent for everyday heights near Earth’s surface, but it is not valid over distances comparable with Earth’s radius. For satellites and planets, use $-\frac{GM}{r^2}$ .

Example

Checking whether g changes much over a small height

Estimate the percentage change in $g$ between Earth’s surface and a height of 100 m above it. Take Earth’s radius as $6.37 \times 10^6\ \text{m}$ .

Since $\propto \frac{1}{r^2}$ , compare the field strengths using:
$gtopgsurface=(rsurfacertop)2\frac{g_\text{top}}{g_\text{surface}} = \left(\frac{r_\text{surface}}{r_\text{top}}\right)^2$
Substitute $mr_\text{surface} = 6.37 \times 10^6\ \text{m}$ and $mr_\text{top} = 6.37 \times 10^6 + 100\ \text{m}$ :
$gtopgsurface=(6.37×1066.3701×106)2\frac{g_\text{top}}{g_\text{surface}} = \left(\frac{6.37 \times 10^6}{6.3701 \times 10^6}\right)^2$
Evaluate the ratio:
$gtopgsurface≈0.99997\frac{g_\text{top}}{g_\text{surface}} \approx 0.99997$
Convert the difference from 1 into a percentage:
$\times 100\% \approx 0.003\%$

So over 100 m, the change in $g$ is tiny. Treating the field as uniform is reasonable.

Acceleration of free fall

An object is in free fall when the only significant force on it is gravity. Air resistance is ignored unless the question says otherwise.

Near Earth’s surface:

F = m g

But Newton’s second law gives:

F = ma

So for free fall:

ma = m g

and therefore:

a = g

This means the gravitational field strength is numerically equal to the acceleration of free fall.

Key Idea

Mass cancels in free fall

In the same gravitational field, all objects have the same acceleration of free fall if air resistance is negligible. A larger mass has a larger weight, but also a larger inertia, so the acceleration is unchanged.

Example

Finding weight and free-fall acceleration

A 0.75 kg object is released near Earth’s surface where $9.81\ \text{N kg}^{-1}$ . Calculate its weight and its acceleration if air resistance is negligible.

Use $W = m g$ for the gravitational force, often called weight:
$W = (0.75) (9.81)$
Calculate the weight:
$7.36\ \text{N}$
The force acts downwards.
For free fall, apply $F = ma$ and $F = m g$ :
$ma = m g$
Cancel the mass to find the acceleration:
$9.81\ \text{m s}^{-2}$

So the object’s weight is $N7.36\ \text{N}$ downwards, and its acceleration is $s−29.81\ \text{m s}^{-2}$ downwards.

Common Mistake

Confusing g with weight

$g$ is gravitational field strength in N kg $^{-1}$ or acceleration in m s $^{-2}$ . Weight is a force, so it is measured in newtons, N, and is calculated using $W = m g$ .

Linking the equations together

You should be comfortable moving between these three ideas:

Newton’s law of gravitation gives the force between two masses:
$-\frac{GMm}{r^2}$
Gravitational field strength is force per unit mass:
$\frac{F}{m}$
For a point mass, this becomes:
$-\frac{GM}{r^2}$

The negative sign means the force or field acts towards the mass producing the field.

Exam technique

In the exam

Check whether the question asks for force or field strength: use $-\frac{GMm}{r^2}$ for force, but $-\frac{GM}{r^2}$ for field strength.
Use the correct distance: $r$ is measured from the centre of the mass, not from the surface, unless the question has already defined it that way.
Treat $g$ as uniform and equal to $s−29.81\ \text{m s}^{-2}$ only close to Earth’s surface; for satellites or large distances, use the inverse-square equation.

Self review

Check yourself

If the separation between two masses is doubled, what happens to the gravitational force between them?
Why does $-\frac{GM}{r^2}$ not contain the mass of the small test object?
In what situation is it reasonable to treat Earth’s gravitational field as uniform?

What you'll learn

How Newton’s law of gravitation gives the force between two masses.
Why gravitational force is an inverse-square force.
How to calculate gravitational field strength using $-\frac{GM}{r^2}$ .
Why, near Earth’s surface, $g$ is treated as uniform and equals the acceleration of free fall.

The starting idea: masses attract

Any two objects with mass attract each other gravitationally. For everyday objects, this force is tiny. For planets, moons and stars, it is large enough to control orbits and motion.

The force acts along the line joining the centres of the two masses, and it is always attractive.

Two point masses attracting each other gravitationally, with centre-to-centre separation r and equal opposite forces

Definition

Point mass

Newton’s law of gravitation

Newton’s law of gravitation says that the gravitational force between two point masses depends on both masses and on their separation.

-\frac{GMm}{r^2}

where:

$F$ is the gravitational force in newtons, N
$G$ is the gravitational constant, $6.67 \times 10^{-11}\ \text{N m}^2\text{ kg}^{-2}$
$M$ and $m$ are the two masses in kilograms, kg
$r$ is the separation between their centres in metres, m

The negative sign shows that the force is attractive. If you define the positive radial direction as “away from” the larger mass, gravity acts in the opposite direction.

Key Idea

The inverse-square law

In calculations, you often use the magnitude:

\frac{GMm}{r^2}

then state or show the direction separately.

Example

Calculating the gravitational force between two masses

Two masses of 8.0 kg and 12 kg are separated by 0.50 m. Calculate the magnitude of the gravitational force between them.

Identify the quantities in SI units: $8.0\ \text{kg}$ , $12\ \text{kg}$ and $0.50\ \text{m}$ .
Substitute into Newton’s law using the magnitude of the force:
$\frac{GMm}{r^2}$ $\frac{(6.67 \times 10^{-11})(8.0)(12)}{(0.50)^2}$
Calculate the denominator and evaluate the force:
$\frac{6.40 \times 10^{-9}}{0.25} = 2.56 \times 10^{-8}\ \text{N}$
Quote the answer sensibly: the gravitational force is $2.6 \times 10^{-8}\ \text{N}$ , attractive, acting along the line joining the masses.

Common Mistake

Forgetting to square the distance

The distance term is $r^2$ , not $r$ . If the separation changes, the force changes very quickly: tripling $r$ makes the force one ninth as large.

What the negative sign means

The equation in the specification is written as:

-\frac{GMm}{r^2}

This is a direction convention, not a sign that the size of the force is negative. Force is a vector quantity, so it has magnitude and direction.

If the positive direction is chosen radially outwards from mass $M$ , then the gravitational force on mass $m$ points inwards, towards $M$ . That inward direction is negative.

Tip

Magnitude first, direction second

For most numerical questions, calculate the positive magnitude using $\frac{GMm}{r^2}$ , then add the direction in words: “towards the other mass” or “towards the centre of the planet”.

Example

Using the inverse-square relationship

A satellite moves from a distance $r$ to a distance $2 r$ from Earth’s centre. Compare the gravitational force at $2 r$ with the force at $r$ .

Start from the proportionality for fixed masses: $\propto \frac{1}{r^2}$ .
Replace $r$ by $2 r$ :
$F2r∝1(2r)2=14r2F_{2r} \propto \frac{1}{(2r)^2} = \frac{1}{4r^2}$
Compare with the original force:
$F2r=14FrF_{2r} = \frac{1}{4}F_r$

So the gravitational force becomes one quarter of its original value.

Gravitational field strength

A gravitational field is a region where a mass experiences a gravitational force.

Definition

Gravitational field strength

Gravitational field strength, $g$ , is the gravitational force per unit mass on a small test mass placed in the field. Its unit is newtons per kilogram, N kg $^{-1}$ .

So:

\frac{F}{m}

For a point mass $M$ , combine this with Newton’s law:

-\frac{GMm}{r^2}

Divide by the small mass $m$ :

-\frac{GM}{r^2}

This is the gravitational field strength due to a point mass $M$ at distance $r$ from its centre.

Key Idea

Field strength does not depend on the test mass

The value of $g$ at a point depends on the source mass $M$ and the distance $r$ , not on the mass of the object placed there.

Example

Calculating gravitational field strength near Earth

Use $kgM_\text{Earth} = 5.97 \times 10^{24}\ \text{kg}$ and $\times 10^6\ \text{m}$ to estimate the gravitational field strength at Earth’s surface.

Use the point-mass equation for gravitational field strength:
$-\frac{GM}{r^2}$
For the magnitude:
$\frac{GM}{r^2}$
Substitute the values:
$\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.37 \times 10^6)^2}$
Evaluate the numerator and denominator:
$\frac{3.98 \times 10^{14}}{4.06 \times 10^{13}}$
Calculate the field strength:
$9.8\ \text{N kg}^{-1}$

So the gravitational field strength at Earth’s surface is about $kg−19.8\ \text{N kg}^{-1}$ , directed towards Earth’s centre.

Why $g$ has two equivalent units

Gravitational field strength is measured in N kg $^{-1}$ because it is force per unit mass.

But from Newton’s second law, $F = ma$ , so:

Fm=a\frac{F}{m} = a

Therefore N kg $^{-1}$ is equivalent to m s $^{-2}$ .

That is why near Earth’s surface you will see:

\approx 9.81\ \text{N kg}^{-1}

and also:

\approx 9.81\ \text{m s}^{-2}

They are numerically the same physical quantity, expressed in equivalent units.

Uniform gravitational field near Earth’s surface

A gravitational field around a planet is actually radial: field lines point towards the planet’s centre, and the magnitude decreases with distance from the centre.

Radial gravitational field around Earth compared with approximately uniform gravitational field near the surface

Definition

Uniform gravitational field

A uniform gravitational field has the same field strength and direction at every point in the region being considered.

Near Earth’s surface, the field is taken as:

\approx 9.81\ \text{N kg}^{-1}

directed vertically downwards.

Common Mistake

Close to the surface only

Example

Checking whether g changes much over a small height

Estimate the percentage change in $g$ between Earth’s surface and a height of 100 m above it. Take Earth’s radius as $6.37 \times 10^6\ \text{m}$ .

Since $\propto \frac{1}{r^2}$ , compare the field strengths using:
$gtopgsurface=(rsurfacertop)2\frac{g_\text{top}}{g_\text{surface}} = \left(\frac{r_\text{surface}}{r_\text{top}}\right)^2$
Substitute $mr_\text{surface} = 6.37 \times 10^6\ \text{m}$ and $mr_\text{top} = 6.37 \times 10^6 + 100\ \text{m}$ :
$gtopgsurface=(6.37×1066.3701×106)2\frac{g_\text{top}}{g_\text{surface}} = \left(\frac{6.37 \times 10^6}{6.3701 \times 10^6}\right)^2$
Evaluate the ratio:
$gtopgsurface≈0.99997\frac{g_\text{top}}{g_\text{surface}} \approx 0.99997$
Convert the difference from 1 into a percentage:
$\times 100\% \approx 0.003\%$

So over 100 m, the change in $g$ is tiny. Treating the field as uniform is reasonable.

Acceleration of free fall

An object is in free fall when the only significant force on it is gravity. Air resistance is ignored unless the question says otherwise.

Near Earth’s surface:

F = m g

But Newton’s second law gives:

F = ma

So for free fall:

ma = m g

and therefore:

a = g

This means the gravitational field strength is numerically equal to the acceleration of free fall.

Key Idea

Mass cancels in free fall

Example

Finding weight and free-fall acceleration

A 0.75 kg object is released near Earth’s surface where $9.81\ \text{N kg}^{-1}$ . Calculate its weight and its acceleration if air resistance is negligible.

Use $W = m g$ for the gravitational force, often called weight:
$W = (0.75) (9.81)$
Calculate the weight:
$7.36\ \text{N}$
The force acts downwards.
For free fall, apply $F = ma$ and $F = m g$ :
$ma = m g$
Cancel the mass to find the acceleration:
$9.81\ \text{m s}^{-2}$

So the object’s weight is $N7.36\ \text{N}$ downwards, and its acceleration is $s−29.81\ \text{m s}^{-2}$ downwards.

Common Mistake

Confusing g with weight

$g$ is gravitational field strength in N kg $^{-1}$ or acceleration in m s $^{-2}$ . Weight is a force, so it is measured in newtons, N, and is calculated using $W = m g$ .

Linking the equations together

You should be comfortable moving between these three ideas:

Newton’s law of gravitation gives the force between two masses:
$-\frac{GMm}{r^2}$
Gravitational field strength is force per unit mass:
$\frac{F}{m}$
For a point mass, this becomes:
$-\frac{GM}{r^2}$

The negative sign means the force or field acts towards the mass producing the field.

Exam technique

In the exam

Check whether the question asks for force or field strength: use $-\frac{GMm}{r^2}$ for force, but $-\frac{GM}{r^2}$ for field strength.
Use the correct distance: $r$ is measured from the centre of the mass, not from the surface, unless the question has already defined it that way.
Treat $g$ as uniform and equal to $s−29.81\ \text{m s}^{-2}$ only close to Earth’s surface; for satellites or large distances, use the inverse-square equation.

Self review

Check yourself

If the separation between two masses is doubled, what happens to the gravitational force between them?
Why does $-\frac{GM}{r^2}$ not contain the mass of the small test object?
In what situation is it reasonable to treat Earth’s gravitational field as uniform?

What you'll learn

The starting idea: masses attract

Newton’s law of gravitation

What the negative sign means

Gravitational field strength

Why $g$ has two equivalent units

Uniform gravitational field near Earth’s surface

Acceleration of free fall

Linking the equations together

Review Newton's law of gravitation by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Newton's law of gravitation

What you'll learn

The starting idea: masses attract

Newton’s law of gravitation

What the negative sign means

Gravitational field strength

Why $g$ has two equivalent units

Uniform gravitational field near Earth’s surface

Acceleration of free fall

Linking the equations together

Review Newton's law of gravitation by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Newton's law of gravitation

What you'll learn

The starting idea: masses attract

Newton’s law of gravitation

What the negative sign means

Gravitational field strength

Why ggg has two equivalent units

Uniform gravitational field near Earth’s surface

Acceleration of free fall

Linking the equations together

Review Newton's law of gravitation by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Newton's law of gravitation Revision Guide

Newton's law of gravitation

What you'll learn

The starting idea: masses attract

Newton’s law of gravitation

What the negative sign means

Gravitational field strength

Why ggg has two equivalent units

Uniform gravitational field near Earth’s surface

Acceleration of free fall

Linking the equations together

Review Newton's law of gravitation by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Newton's law of gravitation Revision Guide

Why $g$ has two equivalent units

Why $g$ has two equivalent units