Newton's law of gravitation and field strength
Every mass attracts every other mass. Newton's law gives the size of that force, and dividing by the mass of the object being pulled gives the field strength, which tells you how strong gravity is at any point.
Part 1 of 3: Learn it
In short
- F = Gm₁m₂ ÷ r², with G = 6.67 × 10⁻¹¹ N m² kg⁻².
- Gravitational field strength g = F ÷ m; near a planet, g = GM ÷ r².
- Doubling the distance between centres makes the force a quarter as big.
Where this is in your specification
Spec points: AQA 7408 3.7.2.1 and 3.7.2.2, OCR A H556 5.4.1 and 5.4.2, Edexcel 9PH0 Topic 12
| Board | Topic: Gravitational and electric fields |
|---|---|
| AQA 7408 | 3.7.1 to 3.7.3 |
| Edexcel 9PH0 | Topics 7 and 12 |
| OCR H556 | 5.4 and 6.2 |
| Higher C857 76 | Our dynamic Universe: gravitation; Particles and waves: forces on charged particles |
The law
r is measured between the centres of the masses. A planet or a sphere of even density behaves as if all its mass were at its centre. The force is always attractive and acts along the line joining the centres.
Field strength
g is the force per unit mass on a small test mass, in N kg⁻¹. Outside a planet it falls with the square of the distance from the centre. Near the surface, over small changes in height, it is almost constant, so the field is treated as uniform.
Field lines
Around a planet, field lines are radial, pointing inwards to the centre, and spread out with distance, showing the field getting weaker. Close to the surface they look parallel and evenly spaced, showing a uniform field.
The distance between two masses is tripled. What happens to the force between them?
Show the answer
It becomes one ninth as big.
Part 2 of 3: See it worked
Worked examples
Example 1
Find the gravitational force between the Earth (5.97 × 10²⁴ kg) and the Moon (7.35 × 10²² kg) when their centres are 3.84 × 10⁸ m apart.
- F = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²² ÷ (3.84 × 10⁸)²
- = 2.93 × 10³⁷ ÷ 1.47 × 10¹⁷
Answer: 2.0 × 10²⁰ N.
Example 2
Find g at a height of 400 km above the Earth's surface. (Earth: radius 6.37 × 10⁶ m, mass 5.97 × 10²⁴ kg)
- r = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m
- g = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ ÷ (6.77 × 10⁶)²
Answer: 8.7 N kg⁻¹, so the International Space Station is not beyond gravity: it is in free fall.
Common mistakes
- Using the height above the surface as r.
- Forgetting to square r.
- Saying astronauts in orbit float because there is no gravity.
- Mixing up G (a universal constant) and g (field strength at a place).
What are the units of gravitational field strength?
Show the answer
N kg⁻¹ (equivalent to m s⁻²).
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
The distance between two masses is tripled. What happens to the force between them?
It becomes one ninth as big.
What are the units of gravitational field strength?
N kg⁻¹ (equivalent to m s⁻²).
Describe the field lines around a planet.
Radial lines pointing towards the centre, further apart further from the planet.
Jobs that use this
- Astronomer (se abre en otra pestaña)
- Aerospace engineer (se abre en otra pestaña)
- Physicist (se abre en otra pestaña)
Each link opens the job profile on the National Careers Service (England). In the rest of the UK: My World of Work (Scotland), Careers Wales, nidirect careers (Northern Ireland).
Estos apuntes están en inglés porque siguen los programas de examen del Reino Unido.
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