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Force fields and Newton's law of gravitationAQA A-Level Physics: Revision notes

Section 1

Force fields

A force field is a region in which a body experiences a non-contact force. The force is exerted without the bodies touching, and its size and direction depend on position in the field. At A Level you meet three sources of field: mass (gravitational fields), static charge (electric fields) and moving charges (magnetic fields).

A field can be represented by a vector at each point. Its direction must be determined by inspection, for example towards a mass for gravity, or away from a positive charge for electric fields. Field lines show these directions, and lines closer together show a stronger field.

Key termsforce fieldnon-contact force
Exam tip

State the property that the field acts on: mass for gravitational, charge for electric.

Section 2

Gravity as a universal attractive force

Gravity acts between all matter and is always attractive; there is no gravitational repulsion. Every mass attracts every other mass, but the force is only noticeable when at least one body is very massive, such as a planet or star.

The forces on the two bodies are equal in magnitude and opposite in direction. They are a Newton's third law pair, so a falling apple pulls the Earth upwards with the same force as the Earth pulls the apple down.

Key termsuniversal
Common mistake

The Earth attracting the Moon and the Moon attracting the Earth are two forces of equal size, not one big and one small. The accelerations differ because the masses differ.

Section 3

Newton's law of gravitation

The magnitude of the force between two point masses is F=Gm1m2r2F = \dfrac{G m_1 m_2}{r^2}, where m1m_1 and m2m_2 are the masses, rr is the separation of their centres, and GG is the gravitational constant, 6.67×10−11 N m2 kg−26.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} (given in the data booklet).

This is an inverse square law: F∝1/r2F \propto 1/r^2, so doubling the separation divides the force by 4, and trebling it divides the force by 9. The force is proportional to each mass, so doubling one mass doubles the force.

Key termsgravitational constant, Ginverse square law

Section 4

Spheres and point masses

A uniform sphere acts, for points outside it, as if all of its mass were concentrated at its centre. So for planets and stars rr is the distance between centres, not between surfaces.

Worked example: Earth (5.97×10245.97\times10^{24} kg) and Moon (7.35×10227.35\times10^{22} kg), centres 3.84×1083.84\times10^{8} m apart. F=6.67×10−11×5.97×1024×7.35×1022(3.84×108)2=1.98×1020F = \dfrac{6.67\times10^{-11}\times5.97\times10^{24}\times7.35\times10^{22}}{(3.84\times10^{8})^2} = 1.98\times10^{20} N.

For a body between two masses the forces act in opposite directions, so subtract them. The resultant is zero at the point where the two forces are equal.

Key termspoint mass
Common mistake

Using the radius or surface-to-surface distance instead of the distance between centres is a common source of lost marks.

Section 5

Comparing gravitational and electrostatic forces

Similarities: both forces obey inverse square laws for point sources; both can be described using field lines; both have the idea of potential and equipotential surfaces; both act at a distance.

Differences: masses always attract, but charges may attract or repel (like charges repel, unlike attract). Electrostatic forces are also vastly larger than gravitational forces between subatomic particles. Moving charges additionally give rise to magnetic forces.

Key termsequipotential surface

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Force fields and Newton's law of gravitation

  1. A torsion-balance experiment measures the gravitational attraction between a large lead sphere of mass 15 kg and a small lead sphere of mass 0.015 kg. The centres of the spheres are 0.060 m apart and the spheres are uncharged.
    Calculate the magnitude of the gravitational force on the small sphere. Use G=6.67×10−11 N m2 kg−2G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}.2 marks
  2. A student compares the gravitational field around a planet with the electric field around an isolated, positively charged metal sphere. Both bodies are treated as point sources.
    Explain what is meant by a force field, and why a small mass near the planet and a small charge near the sphere are both described as being in a force field.2 marks
  3. The Earth has mass 5.97 × 10²⁴ kg and the Moon has mass 7.35 × 10²² kg. Their centres are 3.84 × 10⁸ m apart. Treat both bodies as point masses at their centres.
    Calculate the magnitude of the gravitational force of the Earth on the Moon. Use G=6.67×10−11 N m2 kg−2G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}.3 marks
See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).