Gravitational Fields and Newton's Law of Gravitation

A2 · 11 min

Gravity acts between masses that are not touching: the Sun holds the Earth in orbit across $150$ million kilometres of empty space. Physics describes this "action at a distance" with the idea of a field: a region where a mass feels a force. This note introduces the gravitational field, how to draw it, and Newton's law of gravitation, the inverse-square law that gives the force between any two masses. It is the foundation of the whole fields section of Paper 4, and the same pattern returns almost word for word for electric fields.

Fields of force

A field of force is a region of space in which an object experiences a force because of some property it has. For gravitational fields, that property is mass. For electric fields it is charge. Every mass creates a gravitational field around itself, and every mass placed in someone else's field feels a force.

The field idea separates two questions:

What field does a mass $M$ create at a point? (That depends only on $M$ and where the point is.)
What force does a second mass $m$ feel at that point? (Field multiplied by $m$ .)

To describe the field without worrying about what is placed in it, we measure the force per unit mass on a small test mass: small enough that its own field does not disturb the field being measured.

Definition

The gravitational field strength $g$ at a point is the gravitational force per unit mass acting on a small test mass placed at that point:

g = \frac{F}{m}

Its unit is $\text{N kg}^{-1}$ , equivalent to $\text{m s}^{-2}$ .

Gravitational field strength is a vector. Its direction is the direction of the force on a mass, which for gravity is always towards the mass creating the field. Gravity is always attractive.

The two units are the same thing: $1\ \text{N kg}^{-1} = 1\ \text{kg m s}^{-2}\ \text{kg}^{-1} = 1\ \text{m s}^{-2}$ . This is why the gravitational field strength at a point equals the acceleration of free fall there: if gravity is the only force, $a = F/m = g$ .

Representing a gravitational field by field lines

Field lines (lines of force) show the field as a picture.

The direction of a line at any point shows the direction of the force on a mass placed there (arrows point towards the attracting mass).
The spacing of the lines shows the strength: lines close together mean a strong field.
Field lines never cross, because the field at a point has only one direction.

Left: the radial field of an isolated spherical mass M; lines converge on the centre and get closer together, so the field is stronger near the surface. Right: near the Earth's surface over a small region, the field is approximately uniform: parallel, equally spaced lines.

Two patterns appear in exam questions:

Radial field around a spherical mass (planet, star): straight lines pointing towards the centre, getting closer together nearer the mass. The field gets weaker with distance.
Uniform field close to the surface of a planet over a small region: parallel, equally spaced vertical lines pointing downwards. The field strength is the same everywhere in the region. This is the field you used at AS when you took $g = 9.81\ \text{N kg}^{-1}$ as constant.

The uniform field is really a tiny patch of the radial field, so small compared with the Earth that the lines look parallel. Gravitational field strength explains when this approximation is valid.

A sphere behaves like a point mass

Newton's law of gravitation is stated for point masses. Real planets and stars are large spheres, so we need a bridge between the two.

Key result

For a point outside a uniform sphere (or a sphere made of uniform spherical shells), the sphere's mass may be considered to be a point mass at its centre.

This means:

the distance $r$ in the formulas is always measured from the centre of the planet, not from its surface;
for a satellite at height $h$ above a planet of radius $R$ , $r = R + h$ .

It is also why the field lines of a spherical planet look exactly like those of a point mass from outside.

Newton's law of gravitation

Definition

Newton's law of gravitation: any two point masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of their separation.

Key result

F = \frac{Gm_1m_2}{r^{2}}

$m_1$ , $m_2$ : the two masses (kg)
$r$ : the distance between their centres (m)
$G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ , the gravitational constant

Features to understand, not just memorise:

Inverse square law. Double the separation and the force falls to a quarter. Triple it and the force falls to a ninth.
Newton's third law. The force of $m_1$ on $m_2$ is equal in size and opposite in direction to the force of $m_2$ on $m_1$ . The Earth pulls the Moon exactly as hard as the Moon pulls the Earth; the Earth simply accelerates much less because its mass is larger.
Tiny constant. $G$ is so small that gravity between everyday objects is negligible. It only becomes important when at least one mass is astronomical.
Always attractive. There is no negative mass, so gravitational forces never repel.

The unit of $G$ follows by rearranging: $G = Fr^2/(m_1m_2)$ has units $\text{N m}^2\ \text{kg}^{-2}$ . In SI base units this is $\text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$ .

Worked examples

Gravity between two people

Two people, each of mass $70\ \text{kg}$ , stand $1.0\ \text{m}$ apart. Treating them as point masses, calculate the gravitational force between them and compare it with the weight of one of them.

Solution

F = \frac{Gm_1m_2}{r^2} = \frac{6.67 \times 10^{-11} \times 70 \times 70}{1.0^2} = 3.3 \times 10^{-7}\ \text{N}

The weight of one person is $70 \times 9.81 = 687\ \text{N}$ , about $2 \times 10^{9}$ times larger. Gravity between everyday objects is completely negligible.

Force between the Earth and the Moon

The mass of the Earth is $5.97 \times 10^{24}\ \text{kg}$ , the mass of the Moon is $7.35 \times 10^{22}\ \text{kg}$ and the distance between their centres is $3.84 \times 10^{8}\ \text{m}$ . Calculate the gravitational force between them, and the acceleration of each body caused by it.

Solution

F = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^{8})^{2}} = 1.98 \times 10^{20}\ \text{N}

The force is the same on both (Newton's third law), but the accelerations differ:

a_{\text{Moon}} = \frac{1.98 \times 10^{20}}{7.35 \times 10^{22}} = 2.69 \times 10^{-3}\ \text{m s}^{-2}, \qquad a_{\text{Earth}} = \frac{1.98 \times 10^{20}}{5.97 \times 10^{24}} = 3.32 \times 10^{-5}\ \text{m s}^{-2}

Scaling with distance

A space probe experiences a gravitational force of $800\ \text{N}$ from a planet when it is $2.0 \times 10^{7}\ \text{m}$ from the planet's centre. Calculate the force when the probe is $5.0 \times 10^{7}\ \text{m}$ from the centre.

Solution

$F \propto 1/r^2$ , so

\frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^{2} = \left(\frac{2.0}{5.0}\right)^{2} = 0.16 \quad\Rightarrow\quad F_2 = 0.16 \times 800 = 130\ \text{N}

Using a ratio avoids needing $G$ or the planet's mass at all.

Weighing the Earth

A mass $m$ at the Earth's surface has weight $mg$ , where $g = 9.81\ \text{N kg}^{-1}$ . The radius of the Earth is $6.37 \times 10^{6}\ \text{m}$ . Use Newton's law of gravitation to calculate the mass of the Earth.

Solution

The weight is the gravitational force of the Earth on the mass, with $r$ equal to the Earth's radius (treating the Earth as a point mass at its centre):

mg = \frac{GMm}{R^{2}} \quad\Rightarrow\quad M = \frac{gR^{2}}{G} = \frac{9.81 \times (6.37 \times 10^{6})^{2}}{6.67 \times 10^{-11}} = 5.97 \times 10^{24}\ \text{kg}

This is how the mass of the Earth was first found: once $G$ had been measured in the laboratory (by Cavendish), the Earth could be "weighed".

Where the pulls cancel

A spacecraft travels along the line joining the centres of the Earth and the Moon. Using the data in the earlier example, find the distance from the centre of the Earth at which the resultant gravitational force on the spacecraft is zero.

Solution

Let the point be a distance $x$ from the Earth's centre, so $d - x$ from the Moon's, with $d = 3.84 \times 10^{8}\ \text{m}$ . The forces are equal and opposite there:

\frac{GM_Em}{x^{2}} = \frac{GM_Mm}{(d - x)^{2}} \quad\Rightarrow\quad \frac{x}{d - x} = \sqrt{\frac{M_E}{M_M}} = \sqrt{\frac{5.97 \times 10^{24}}{7.35 \times 10^{22}}} = 9.01

So $x = 9.01(d - x)$ , giving $x = \dfrac{9.01d}{10.01} = 0.900d = 3.46 \times 10^{8}\ \text{m}$ from the centre of the Earth.

The mass $m$ of the spacecraft and $G$ both cancel. Taking the positive square root is correct because the point lies between the two bodies.

Who pulls the Moon harder?

The Sun has mass $1.99 \times 10^{30}\ \text{kg}$ and is $1.50 \times 10^{11}\ \text{m}$ from the Moon. Calculate the gravitational force of the Sun on the Moon and compare it with the force of the Earth on the Moon. Suggest why the Moon still orbits the Earth.

Solution

F_{\text{Sun}} = \frac{6.67 \times 10^{-11} \times 1.99 \times 10^{30} \times 7.35 \times 10^{22}}{(1.50 \times 10^{11})^{2}} = 4.34 \times 10^{20}\ \text{N}

This is about $2.2$ times the Earth's pull of $1.98 \times 10^{20}\ \text{N}$ . The Moon does orbit the Sun; but the Sun pulls the Earth and the Moon with almost the same acceleration (they are at nearly the same distance from it), so the Earth and Moon fall around the Sun together. Relative to the Earth, the Moon's motion is governed by the Earth's pull.

Watch out

Measuring $r$ from the surface. In $F = Gm_1m_2/r^2$ , $r$ is the distance between centres. For a satellite $400\ \text{km}$ above the Earth, $r = 6.37 \times 10^{6} + 4.00 \times 10^{5} = 6.77 \times 10^{6}\ \text{m}$ , not $4.00 \times 10^{5}\ \text{m}$ .

Watch out

Forgetting to square $r$ , or squaring only the number and not the power of ten. Use brackets on the calculator: (3.84E8)^2.

Watch out

Defining field strength as "force on a mass". It is force per unit mass. A definition without "per unit mass" scores zero.

Exam tip

"Define gravitational field strength" (1 mark): force per unit mass (on a small test mass). Do not write "the acceleration due to gravity": that is a consequence, not the definition.
"State Newton's law of gravitation" (2 marks): the force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation. Each bold phrase is a marking point.
"Explain why the planet can be treated as a point mass": it is a uniform sphere and the point considered is outside it.
When sketching field lines, use a ruler, arrows on every line pointing towards the mass, and radial lines that would meet at the centre if extended.

Summary

A gravitational field is a field of force: a region where a mass experiences a force.
Gravitational field strength $g = F/m$ is the force per unit mass; unit $\text{N kg}^{-1}$ $= \text{m s}^{-2}$ ; it is a vector pointing towards the mass.
Field lines show direction and (by spacing) strength. Radial for a sphere; parallel and equally spaced (uniform) near the Earth's surface.
For points outside a uniform sphere, the sphere acts as a point mass at its centre; $r$ is measured from the centre.
Newton's law of gravitation: $F = Gm_1m_2/r^2$ , attractive, inverse square, equal and opposite on the two masses.
$G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ ; gravity is only significant when a mass is very large.

Practice questions

Question

Define gravitational field strength and show that its unit is equivalent to $\text{m s}^{-2}$ .
Calculate the gravitational force between a $1.0 \times 10^{3}\ \text{kg}$ satellite and the Earth when the satellite is $2.0 \times 10^{7}\ \text{m}$ from the Earth's centre. ( $M_E = 5.97 \times 10^{24}\ \text{kg}$ )
The force between two point masses is $F$ . Both masses are doubled and the separation is tripled. Find the new force in terms of $F$ .
Express the unit of $G$ in SI base units.
Sketch the gravitational field lines (a) around an isolated spherical planet, (b) in a room on the Earth's surface. State the difference between the two fields.
The weight of an astronaut is $780\ \text{N}$ on the Earth's surface. Calculate her weight at a height above the surface equal to the Earth's radius.
Two spheres of mass $4.0\ \text{kg}$ and $9.0\ \text{kg}$ have their centres $0.50\ \text{m}$ apart. Find the distance from the $4.0\ \text{kg}$ sphere, on the line joining them, where a small mass would feel no resultant gravitational force.
Jupiter has mass $1.90 \times 10^{27}\ \text{kg}$ . Its moon Europa has mass $4.80 \times 10^{22}\ \text{kg}$ and orbits at $6.71 \times 10^{8}\ \text{m}$ from Jupiter's centre. (a) Calculate the gravitational force between them. (b) Calculate Europa's centripetal acceleration and hence its orbital period in days.

Answers

Gravitational field strength is the force per unit mass on a small test mass. $\text{N kg}^{-1} = (\text{kg m s}^{-2})/\text{kg} = \text{m s}^{-2}$ .
$F = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1.0 \times 10^{3}/(2.0 \times 10^{7})^2 = 996\ \text{N} \approx 1.0 \times 10^{3}\ \text{N}$ .
$F_{\text{new}} = F \times (2 \times 2)/3^2 = 4F/9$ .
$G = Fr^2/(m_1m_2)$ : $\text{kg m s}^{-2} \times \text{m}^2/\text{kg}^2 = \text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$ .
(a) Radial lines with arrows pointing inwards to the centre, closer together near the surface. (b) Parallel, equally spaced vertical lines pointing down. The radial field weakens with distance; the field in the room is uniform (same strength and direction everywhere).
$r$ doubles from $R$ to $2R$ , so weight falls by a factor of $4$ : $780/4 = 195\ \text{N}$ .
$\dfrac{4.0}{x^2} = \dfrac{9.0}{(0.50 - x)^2} \Rightarrow \dfrac{0.50 - x}{x} = \dfrac{3}{2} \Rightarrow 2(0.50 - x) = 3x \Rightarrow x = 0.20\ \text{m}$ from the $4.0\ \text{kg}$ sphere.
(a) $F = 6.67 \times 10^{-11} \times 1.90 \times 10^{27} \times 4.80 \times 10^{22}/(6.71 \times 10^{8})^2 = 1.35 \times 10^{22}\ \text{N}$ . (b) $a = F/m = 1.35 \times 10^{22}/4.80 \times 10^{22} = 0.281\ \text{m s}^{-2}$ . $a = r\omega^2 \Rightarrow \omega = \sqrt{0.281/6.71 \times 10^{8}} = 2.05 \times 10^{-5}\ \text{rad s}^{-1}$ , $T = 2\pi/\omega = 3.07 \times 10^{5}\ \text{s} = 3.55$ days.

Fields of force

Representing a gravitational field by field lines

A sphere behaves like a point mass

Newton's law of gravitation

Worked examples

Practice questions

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