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

This guide follows the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. It covers sections 3.7.1 (Newton’s gravitational law) and 3.7.2 (Gravitational field strength). Section 3.7 is marked International A-level only, so you meet this content on the A-level route and not in the International AS. Use it with the course hub and the printable checklist.

Companion pages: the revision notes and the practice questions. You need forces and weight from motion and Newton’s laws and careful unit handling from measurements and their errors. To find out what to revise first, take one of the free diagnostics.

What 3.7.1 and 3.7.2 ask you to do

Specification statement	What you must be able to do
3.7.1 Gravity as a universal attractive force acting between all matter	Explain that every mass attracts every other mass, and that the force is always attractive
3.7.1 Force between point masses, F = Gm₁m₂/r², where G is the gravitational constant	State the law and use it in calculations, including ratio and scaling problems
3.7.2 Concept of a force field as a region in which a body experiences a force	Define a force field and apply the idea to gravity
3.7.2 Representation by gravitational field lines	Draw and interpret field lines, including what their direction and spacing show
3.7.2 g as force per unit mass, g = F/m	Define g, give its unit and treat it as a vector
3.7.2 Magnitude of g in a radial field, g = GM/r²	Derive and use it, including how g changes with distance

Gravitational potential and orbits are separate specification sections (3.7.3 and 3.7.4). They are not covered here.

3.7.1 Newton’s law of gravitation

Gravity acts between all matter. Any two masses attract each other, whether they are planets, people or grains of sand. The force is always attractive. There is no gravitational repulsion.

Newton’s law of gravitation. Two point masses attract each other with a force F that is proportional to m₁m₂ and also proportional to 1/r²:

F = G m₁ m₂ / r²

• F is the magnitude of the force on each mass, in newtons (N).
• m₁ and m₂ are the masses, in kg.
• r is the separation of the two point masses, in metres.
• G is the gravitational constant. In calculations here, use G = 6.67 × 10⁻¹¹ N m² kg⁻².

Three facts follow from the equation.

1. Newton’s third law pair. Mass 1 pulls on mass 2 with a force of size F. Mass 2 pulls on mass 1 with a force of exactly the same size, in the opposite direction, even if one mass is far larger. Earth pulls on you and you pull on Earth with equal forces. You accelerate far more because your mass is far smaller.
2. Inverse square law. Double r and F falls to one quarter. Triple r and F falls to one ninth. Halve r and F rises by a factor of 4.
3. Proportional to each mass. Double one mass and F doubles. Double both and F quadruples.

Point masses and spheres. The law is stated for point masses. For a uniform sphere, you can treat all the mass as if it were at the centre, provided you are outside the sphere. So r is measured between centres, not between surfaces. This is a common source of lost marks when a question gives an altitude above a planet’s surface.

Worked example 1: force between two small masses

A sphere of mass 8.0 kg and a sphere of mass 2.5 kg have their centres 0.30 m apart. Calculate the force between them, then the force when the centres are 0.60 m apart.

F = G m₁ m₂ / r²
  = 6.67 × 10⁻¹¹ × 8.0 × 2.5 / (0.30)²
  = 1.334 × 10⁻⁹ / 0.090
  = 1.5 × 10⁻⁸ N   (2 s.f.)

Doubling the separation divides the force by 2² = 4:

F = 1.48 × 10⁻⁸ / 4 = 3.7 × 10⁻⁹ N

Direct substitution gives the same: 6.67 × 10⁻¹¹ × 20 / 0.36 = 3.7 × 10⁻⁹ N.

Worked example 2: scaling without calculating

A planet has 8 times the mass of Earth and twice its radius. By what factor does the gravitational force on a given object at the planet’s surface differ from the force on the same object at Earth’s surface?

The force is proportional to M/R². The factor is 8 / 2² = 8 / 4 = 2. The force is twice as large. If an object weighs 100 N on Earth, it would weigh 200 N on this planet.

3.7.2 Gravitational field strength

Force fields

A force field is a region where a body experiences a force. A gravitational field is the space around a mass where any other mass feels a gravitational force. The same idea is used for electric and magnetic fields later in the course.

Gravitational field lines

Field lines show the direction of the force on a mass placed in the field and how strong the field is.

• The arrow on a line points in the direction of the force on a mass placed there. For gravity, the lines point towards the mass that creates the field.
• Closer lines mean a stronger field. Lines further apart mean a weaker field.
• Lines do not cross, since each point in the field has just one direction of force.

Situation	Pattern of lines	What it shows
Isolated sphere (a planet)	Straight lines pointing radially inwards towards the centre	A radial field: g depends only on distance from the centre
Radial field	Lines get further apart as you move away from the sphere	g decreases with distance
A few metres or kilometres of a planet’s surface	Parallel, equally spaced lines pointing downwards	The field is approximately uniform: g is nearly constant

Near the surface, the lines are very slightly convergent, but over a few metres or kilometres the difference is too small to matter. That is why g is treated as a constant when you work with falling objects close to the ground.

Definition of g

Gravitational field strength g is the force per unit mass experienced by a small mass placed at that point in the field:

g = F / m

• The unit is N kg⁻¹.
• g is a vector. Its direction is the direction of the force, towards the mass creating the field.
• N kg⁻¹ and m s⁻² are equivalent units, because 1 N = 1 kg m s⁻². So the field strength at a point is also the acceleration of a freely falling object there.
• Rearranged, F = mg. When F is the weight W of an object, W = mg.

The word “small” matters in the definition: the test mass should be small enough that it does not change the field it is measuring.

Radial field strength

Combine F = GMm/r² with g = F/m. The test mass m cancels:

g = F / m = (G M m / r²) / m = G M / r²

For a point mass, or outside a uniform sphere, the magnitude of g is g = GM/r², where M is the mass creating the field and r is the distance from its centre.

Key consequences:

• g does not depend on the test mass. A feather and a hammer feel the same g at the same place.
• g is proportional to M and inversely proportional to r².
• Because r is measured from the centre, a point at height h above a planet of radius R has r = R + h.

The next table uses Earth values (M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m) to show the inverse square pattern in the field.

Distance from Earth’s centre	g (N kg⁻¹)
R	9.81
2R	2.45
3R	1.09
4R	0.613

At 2R, g is a quarter of its surface value. At 3R it is a ninth. The field strength falls quickly but never reaches zero.

Worked example 3: g at the surface of Venus, and above it

Venus has mass 4.87 × 10²⁴ kg and radius 6.05 × 10⁶ m. Calculate g at its surface and at a point 2.00 × 10⁶ m above the surface.

At the surface:

g = GM/R² = 6.67 × 10⁻¹¹ × 4.87 × 10²⁴ / (6.05 × 10⁶)²
  = 3.248 × 10¹⁴ / 3.660 × 10¹³
  = 8.87 N kg⁻¹

Above the surface, the distance from the centre is r = 6.05 × 10⁶ + 2.00 × 10⁶ = 8.05 × 10⁶ m:

g = 3.248 × 10¹⁴ / (8.05 × 10⁶)² = 3.248 × 10¹⁴ / 6.480 × 10¹³ = 5.01 N kg⁻¹

Check with a ratio: (6.05 / 8.05)² = 0.565, and 8.87 × 0.565 = 5.01. The two methods agree.

Worked example 4: finding the mass of a planet from g

A planet has a surface field strength of 12.4 N kg⁻¹ and a radius of 7.2 × 10⁶ m. Calculate its mass.

Rearrange g = GM/R² to make M the subject:

M = g R² / G
  = 12.4 × (7.2 × 10⁶)² / 6.67 × 10⁻¹¹
  = 12.4 × 5.184 × 10¹³ / 6.67 × 10⁻¹¹
  = 9.6 × 10²⁴ kg   (2 s.f.)

Worked example 5: Sun’s field at Earth

Take the Sun’s mass as 1.99 × 10³⁰ kg and the Sun-Earth distance as 1.50 × 10¹¹ m. Calculate the Sun’s gravitational field strength at Earth and the force on Earth, whose mass is 5.97 × 10²⁴ kg.

g = GM/r² = 6.67 × 10⁻¹¹ × 1.99 × 10³⁰ / (1.50 × 10¹¹)²
  = 1.327 × 10²⁰ / 2.25 × 10²²
  = 5.90 × 10⁻³ N kg⁻¹

F = mg = 5.97 × 10²⁴ × 5.90 × 10⁻³ = 3.52 × 10²² N

The value of g is tiny compared with 9.81 N kg⁻¹, yet the force is huge because Earth’s mass is huge. The same force follows from F = Gm₁m₂/r², and Earth pulls on the Sun with a force of the same size.

Choosing the right method

You are given	Use
Two masses and their separation, asked for a force	F = Gm₁m₂/r²
One large mass and a distance, asked for g	g = GM/r²
g and asked for the force on a mass m	F = mg
A change in distance or mass, asked “by what factor”	Ratio of the proportionalities, no need for G

Common errors

• Using surface-to-surface distance. For a satellite 400 km above Earth, r is R + 400 km, not 400 km.
• Forgetting to square r. Square the whole of r, including its power of ten.
• Mixing up g and G. G is a constant of nature. g is a field strength that changes with position.
• Saying g depends on the mass of the object. The test mass cancels. The force depends on it, but g does not.
• Thinking the larger mass pulls harder. The two forces in a pair are equal and opposite.
• Drawing field lines that stop short or cross. Radial lines point to the centre and never cross.
• Dropping units. Check that N kg⁻¹ matches your calculation: GM/r² has units N m² kg⁻² × kg / m² = N kg⁻¹.

Next steps

Work through the practice questions and use the revision notes for the final weeks. Tick off each point on the checklist.

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).