Why don’t we feel gravitational attraction to everyday objects?

G is extremely small. Two 1 kg masses 1 m apart attract with only 6.67 10^-11 N.

Why can a planet be treated as a point mass?

For a spherically symmetric body, the field outside it is exactly the same as if all the mass were at the centre.

Why are astronauts weightless on the ISS?

They and the station fall together around Earth, so there is no contact force between them, even though g is still about 8.7 N kg⁻¹.

If gravity is so weak, why does it dominate the universe?

Masses always add, while positive and negative charges cancel. Large bodies are neutral but enormously massive.

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174–177, 179FieldsCore

Newton’s Law of Gravitation

The force that makes an apple fall is the same force that holds the Moon in orbit. Newton’s great insight was that gravity is $\textbf{universal}$ : every mass attracts every other mass with a force that weakens as the square of the distance between them.

35 min Topic 12 · Gravitational Fields

What you'll be able to do

State and use Newton’s law of gravitation, $F = Gm_{1}m_{2}/r^{2}$
Define gravitational field strength and use $g = F/m$
Derive and use $g = GM/r^{2}$ for a radial field
Compare gravitational and electric fields

The law of gravitation

Newton’s law states that the gravitational force between two point masses is proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

The constant is $G = 6.67\times10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$ . Its small size is why you do not notice the pull between everyday objects.

Spherical objects such as planets behave as if all their mass were concentrated at their centre, so $r$ is measured centre to centre.

F = \dfrac{Gm_{1}m_{2}}{r^{2}}

Newton’s law of gravitation.

Example 1Find the gravitational force between Earth (

5.97\times10^{24}

kg) and the Moon (

7.35\times10^{22}

kg),

3.84\times10^{8}

m apart.

F = \dfrac{6.67\times10^{-11} \times 5.97\times10^{24} \times 7.35\times10^{22}}{(3.84\times10^{8})^{2}}

Answer

1.98\times10^{20}\ \mathrm{N}

Tip — When given altitude above a surface, add the planet’s radius to get $r$ .

Gravitational field strength

A $\textbf{gravitational field}$ is a region where a mass experiences a force. The field strength $g$ at a point is the force per unit mass on a small test mass placed there, in $\mathrm{N\,kg^{-1}}$ .

Field lines point in the direction a mass would be pulled — always $\textbf{towards}$ the attracting mass. Near Earth’s surface the field is effectively $\textbf{uniform}$ , with parallel, equally spaced lines, and $g = 9.81\ \mathrm{N\,kg^{-1}}$ .

On a large scale, the field around a planet is $\textbf{radial}$ , with lines converging on its centre.

g = \dfrac{F}{m}

Field strength, force per unit mass.

Radial fields

Dividing Newton’s law by the test mass gives the field strength at distance $r$ from the centre of a mass $M$ .

Outside the planet, $g$ falls as $1/r^{2}$ . Doubling the distance from the centre reduces it to a quarter.

Treating $g$ as constant is only valid for height changes that are small compared with the planet’s radius. A graph of $g$ against $r$ is a steep inverse square curve that flattens towards zero far from the planet.

g = \dfrac{GM}{r^{2}}

Radial field strength.

Example 1Find the gravitational field strength at the International Space Station, 410 km above Earth’s surface. Earth’s radius is

6.37\times10^{6}

r = 6.37\times10^{6} + 4.1\times10^{5} = 6.78\times10^{6}

g = \dfrac{6.67\times10^{-11} \times 5.97\times10^{24}}{(6.78\times10^{6})^{2}}

Answer

8.66\ \mathrm{N\,kg^{-1}}

— nearly 90% of its surface value.

Tip — Astronauts on the ISS are not weightless because gravity is absent. They are in free fall, accelerating at $g$ along with the station.

Gravitational and electric fields compared

Both are inverse square fields: $F = Gm_{1}m_{2}/r^{2}$ and $F = kQ_{1}Q_{2}/r^{2}$ . Both have radial fields for point sources and uniform fields in special cases.

Gravity acts on $\textbf{mass}$ and is always attractive. The electric force acts on $\textbf{charge}$ and can attract or repel.

Gravity is vastly weaker. It dominates astronomy only because large bodies carry huge mass but almost no net charge — positive and negative charges cancel, but masses always add.

Example 1Find the ratio of electric to gravitational force between two protons.

m_{p} = 1.67\times10^{-27}

kg.

1The

r^{2}

cancels: ratio

= \dfrac{ke^{2}}{Gm_{p}^{2}}

= \dfrac{8.99\times10^{9} \times (1.60\times10^{-19})^{2}}{6.67\times10^{-11} \times (1.67\times10^{-27})^{2}}

AnswerAbout

1.2\times10^{36}

Equation recap

F = Gm_{1}m_{2}/r^{2}

Newton’s law.

g = F/m

Field strength.

g = GM/r^{2}

Radial field.

Common mistakes to avoid

Using altitude as

r

r

is distance from the centre: radius plus altitude.

Using

1/r

instead of

1/r^{2}

Force and field strength follow the inverse square law.

Saying there is no gravity in orbit.

Orbiting objects are in continuous free fall.

Thinking

g

is 9.81 everywhere.

It is only that value near Earth’s surface.

Key takeaways

Every mass attracts every other: $F = Gm_{1}m_{2}/r^{2}$ .
Field strength $g = F/m$ , in N kg⁻¹.
Radial field: $g = GM/r^{2}$ , with $r$ from the centre.
Gravity is always attractive and far weaker than the electric force.

Test yourself

Answer a few questions, then review anything you missed.

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