OxfordAQA International AS and A-level Physics: Newton's law of gravitation and gravitational field strength (9630) – Revision Notes

These notes condense sections 3.7.1 (Newton’s gravitational law) and 3.7.2 (Gravitational field strength) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. Section 3.7 is International A-level only. For full explanations, use the study guide. Next, work the practice questions, follow progress on the printable checklist, and see the course hub. If you are unsure where to start, try the free diagnostics.

Definitions to learn word for word

• Newton’s law of gravitation. Every pair of point masses pulls together with a force that grows with the product m₁m₂ and falls with 1/r². The force is always attractive.
• Force field. A region where a body feels a force.
• Gravitational field strength, g. The force per unit mass acting on a small mass placed at that point in the field.
• Gravitational field line. A line showing the direction of the force on a mass placed in the field. Closer lines mean a stronger field.

Equations and units

Quantity	Equation	Unit
Force between point masses	F = Gm₁m₂/r²	N
Field strength (definition)	g = F/m	N kg⁻¹
Radial field strength	g = GM/r²	N kg⁻¹
Weight	W = mg	N
Gravitational constant	G = 6.67 × 10⁻¹¹	N m² kg⁻²

N kg⁻¹ is equivalent to m s⁻². Use the value of G supplied in the question. The 6.67 × 10⁻¹¹ here is the usual rounded figure.

Must-know distinctions

Idea A	Idea B
G is a universal constant. It is the same everywhere.	g is a field strength. It changes with position and depends on the source mass.
Mass is a scalar in kg. It does not change with position.	Weight is a force in N. It depends on the local value of g.
F = Gm₁m₂/r² needs two masses.	g = GM/r² needs only the source mass M.
r is the distance between centres.	Altitude h is measured from the surface: r = R + h.

Method boxes

Force between two masses.

1. Convert every quantity to kg and m.
2. Write F = Gm₁m₂/r², then substitute.
3. Square r in full before dividing.
4. Give the answer in N to the number of significant figures in the data.

Field strength at a distance from a planet.

1. Find r from the centre. Add the planet’s radius to any altitude.
2. Use g = GM/r².
3. Check it makes sense: g should be smaller than the surface value if r is greater than R.

“By what factor?” questions.

1. Write the proportionality, for example g ∝ M/r².
2. Put in the factor for each changed quantity.
3. Multiply the factors. Do not recalculate with G.

Derivation to reproduce

Start from the two results you know, then let the test mass cancel.

1. Force on a small mass m at distance r from a mass M: F = GMm/r².
2. Define field strength as force divided by mass: g = F/m.
3. Substitute: g = (GMm/r²) / m = GM/r².

This shows that g depends only on the source mass M and the distance r. It also explains why every freely falling object near a planet has the same acceleration, whatever its mass, when air resistance is ignored.

How g changes with distance

Because g ∝ 1/r², a table of ratios is often quicker than a calculation.

Distance from centre (in multiples of r)	g as a fraction of its value at r
1	1
2	1/4
3	1/9
4	1/16
1/2	4

The same inverse square pattern applies to the force between two masses. For a planet of radius R, r = R is the surface and r = 2R is a height of one radius above the surface. Mistaking one for the other is a very common slip.

Field line sketches

Situation	Lines
Single sphere	Radial, arrows pointing inwards, evenly spaced around the sphere
Moving away from a sphere	Lines spread out, so the field is weaker
Small region near the surface	Parallel, equally spaced, pointing down: approximately uniform field

Never let lines cross. Always include arrowheads.

Worked reminders

Reminder 1: force. Two lorries of mass 1500 kg and 2500 kg have centres 3.0 m apart.

F = 6.67 × 10⁻¹¹ × 1500 × 2500 / 3.0² = 2.8 × 10⁻⁵ N.

Reminder 2: surface field strength. An ice giant has mass 8.7 × 10²⁵ kg and radius 2.5 × 10⁷ m.

g = 6.67 × 10⁻¹¹ × 8.7 × 10²⁵ / (2.5 × 10⁷)² = 9.3 N kg⁻¹.

At three radii from the centre, g falls by a factor of 9, to 1.0 N kg⁻¹.

Reminder 3: inverse square scaling. If the distance from a mass is tripled, g becomes (1/3)² = 1/9 of its original value. If the mass creating the field is also halved, the new value is 1/18 of the original.

Reminder 4: rearranging for mass. A small body has surface g = 0.30 N kg⁻¹ and radius 2.0 × 10⁵ m.

M = gR²/G = 0.30 × (2.0 × 10⁵)² / 6.67 × 10⁻¹¹ = 1.8 × 10²⁰ kg.

Reminder 5: weight in a different field. An object of mass 12 kg is moved from a place where g = 9.8 N kg⁻¹ to one where g = 1.6 N kg⁻¹. Its mass stays 12 kg. Its weight changes from 118 N to 19 N, and 12 × 1.6 = 19.2 N rounds to 19 N.

Quick self-test

1. State Newton’s law of gravitation in words.
2. Two masses of 4.0 × 10⁵ kg and 6.0 × 10⁵ kg are 25 m apart. Calculate the force between them.
3. The separation of two point masses is doubled and one mass is tripled. By what factor does the force change?
4. Define gravitational field strength and give its unit.
5. A moon has mass 2.0 × 10²² kg and radius 1.2 × 10⁶ m. Calculate g at its surface.
6. At what distance from the centre of Earth (mass 5.97 × 10²⁴ kg) is g equal to 1.0 N kg⁻¹?
7. A 0.80 kg mass is at a point where g = 6.4 N kg⁻¹. What is the force on it?
8. What does the spacing of gravitational field lines show?
9. Why can g be taken as constant over a few metres above the ground?
10. A student says the Sun pulls on Earth harder than Earth pulls on the Sun. Is this correct?

Answers

1. Attractive force between point masses, proportional to m₁m₂ and to 1/r².
2. F = 6.67 × 10⁻¹¹ × 4.0 × 10⁵ × 6.0 × 10⁵ / 25² = 16.0 / 625 = 2.6 × 10⁻² N.
3. Force is proportional to m₁m₂/r². Factor = 3 / 2² = 3/4, so the force falls to three quarters.
4. Force per unit mass on a small mass at that point; N kg⁻¹.
5. g = 6.67 × 10⁻¹¹ × 2.0 × 10²² / (1.2 × 10⁶)² = 1.334 × 10¹² / 1.44 × 10¹² = 0.93 N kg⁻¹.
6. r² = GM/g = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 1.0 = 3.98 × 10¹⁴, so r = 2.0 × 10⁷ m.
7. F = mg = 0.80 × 6.4 = 5.1 N, directed towards the mass creating the field.
8. The strength of the field: closer lines mean a stronger field.
9. The lines are almost parallel and equally spaced over that region, because a few metres is tiny compared with the planet’s radius, so r barely changes.
10. No. The two forces are an equal and opposite pair, both of the same magnitude.

Wording that earns credit in explanations

• For a definition of g, include all three parts: force, per unit mass, on a small mass at that point.
• For field lines, link spacing to strength and arrows to the direction of the force.
• For “explain why g is the same for all masses”, say that m cancels in g = F/m = GM/r².
• For a force pair, say the forces are equal in magnitude, opposite in direction and act on different bodies.

Where marks are usually lost

• Using the altitude instead of the distance from the centre.
• Not squaring r, or squaring only the number and not the power of ten.
• Calling g “gravity” with no unit, or giving the unit as N only.
• Saying the Earth’s pull on a falling object is larger because the object is heavier, rather than recognising that g is independent of the test mass.
• Writing a gravitational force as repulsive, or leaving out the idea that it is attractive.
• Drawing field lines without arrows, or lines that do not point towards the centre.
• Giving too many significant figures when the data is given to two.
• Describing a field as “strong” because the lines are long, when it is the spacing that shows strength.
• Confusing mass and weight when a planet’s g is different from 9.81 N kg⁻¹.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.7.1 Newton’s gravitational law and 3.7.2 Gravitational field strength (International A-level only).