Inverse-square gravitation and near-Earth field

Key idea: Newton’s inverse-square law describes the gravitational interaction at any separation. Constant g is a useful near-Earth approximation, not a universal rule.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 2 of 3

Check your understanding

By the end of this lesson, you should be able to

Use Newton’s law of gravitation for two masses.
Relate gravitational field strength to free-fall acceleration.
Compare field strengths at different distances using the inverse-square dependence.

Learn the idea

Big question: When is constant g a safe approximation, and when must distance from a planet's centre be kept?

Apply the inverse-square interaction

Point masses, and spherically symmetric bodies viewed from outside, attract with F = Gm1m2/r². The separation r is centre to centre, not height above a surface.

The forces on the two masses are equal and opposite even when their masses differ. Their accelerations differ because a = F/m.

Check your understanding: Earth pulls an astronaut more strongly than the astronaut pulls Earth. True or false?

False. The forces are equal; Earth's much larger mass gives it a much smaller acceleration.

Turn force per mass into field strength

Gravitational field strength g = F/m gives g = GM/r² around a spherical mass M. A freely falling small body has acceleration equal to local g if other forces are negligible.

Near Earth's surface, small laboratory height changes barely alter r, so constant g is useful. Across planetary distances, the inverse-square variation is essential.

Check your understanding: At height R above a planet of radius R, what fraction of surface g remains?

The distance from the centre is 2R, so g is one quarter of its surface value.

Scroll diagram horizontally to read all labels.

Field lines point in the force direction on a small test mass and cross equipotentials at right angles. Near a planet's surface, a small region can be approximated as uniform.

Key ideas

Measure r from the centre of a spherical body.
Gravitational force is always attractive.
Doubling r reduces F and g to one quarter, not one half.

Relationships to know

F = Gm₁m₂/r²
g = F/m = GM/r²
near Earth, free-fall acceleration ≈ g

Follow the reasoning

Worked example

Find a planet's mass from surface field

Question: A spherical moon has radius 1.74 × 10⁶ m and surface gravitational field strength 1.62 N kg⁻¹. Find its mass. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².

Step 1: Choose the field equation
Why: Surface field uses centre distance equal to the moon's radius.
Working: g = GM/R², so M = gR²/G.
Step 2: Substitute with the squared radius
Why: The inverse-square relationship is a common source of power-of-ten mistakes.
Working: M = 1.62(1.74 × 10⁶)²/(6.67 × 10⁻¹¹).
Step 3: Calculate and check units
Why: N kg⁻¹, m² and G combine to kilograms.
Working: M = 7.35 × 10²² kg.

Answer: The moon's mass is approximately 7.35 × 10²² kg.

Check: Substitution back into GM/R² returns about 1.62 N kg⁻¹.

Now try it with support

Practise with support

Two 5.0 kg masses are 0.20 m apart centre to centre. Find their gravitational attraction using G = 6.67 × 10⁻¹¹ N m² kg⁻².

Hints

Square the separation in metres.
The force magnitudes on the two masses are equal.

View the guided answer

F = 6.67 × 10⁻¹¹(5.0)(5.0)/(0.20²) = 4.17 × 10⁻⁸ N, attractive.

Your turn

Practise independently

A moon orbits at three planetary radii from the centre. Express the local gravitational field strength as a fraction of the surface value.

Check your answer

At r = 3R, g/gₛ = R²/(3R)² = 1/9. The local field strength is one ninth of the surface value.

Common mistakes and exam guidance

Watch out for

Using height above the surface instead of distance from the planet’s centre.
Treating g as constant throughout space because it is nearly constant in a laboratory.

In an exam

A ratio method often cancels G and the planet mass cleanly.
State ‘centre-to-centre’ when defining r in Newton’s law.

Put the ideas together

Exam-style practice [6 marks]

A planet has surface field strength 12 N kg⁻¹ and radius R. Find the field strength at height 2R above its surface. A 5.0 kg probe is there: find the gravitational force and explain why using g = 12 N kg⁻¹ would be wrong.

Plan before you answer

Convert height above surface to centre distance.
Use a ratio to avoid needing G or planet mass.
Then use F = mg with the local field.

View the marking points and model answer

Marking points

Uses centre distance 3R.
Uses inverse-square ratio.
Finds g = 12/9 = 1.33 N kg⁻¹.
Finds force 6.67 N.
States surface g is only a near-surface value.
Explains the centre distance has changed substantially.

Model answer

At height 2R, the probe is 3R from the centre. Therefore g/g0 = R²/(3R)² = 1/9, so g = 12/9 = 1.33 N kg⁻¹. The force is 5.0(1.33) = 6.67 N towards the planet. Using 12 N kg⁻¹ would ignore the large change in centre-to-centre distance.

Finish from memory

Three-question recap

State Newton's law of gravitation.
Check
F = Gm1m2/r², attractive along the line joining the masses.
How is r measured for spherical bodies?
Check
Centre to centre.
When does free-fall acceleration equal g?
Check
When gravity is the only significant force.

Try this next

Set gravitational force equal to the required centripetal force for a circular orbit.

Continue to Circular and geostationary orbits Return to Circular Motion and Gravitation Review Uniform circular-motion kinematics and force

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027