Gravitational force and field strength

Key idea: H2 Physics lessons on inverse-square gravitation, field strength, gravitational potential, escape speed and satellite orbits.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: How does a mass create a force field around it?

Two point masses attract with F = GMm/r². Gravitational field strength is force per unit test mass, so a spherical mass produces g = GM/r² directed inward outside the mass. Superpose field vectors when more than one source acts, and distinguish the source mass from the test mass.

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.

Radial and approximately uniform gravitational fieldsA spherical mass has radial field lines pointing inward and concentric equipotential surfaces. A small region near a large spherical surface is represented by parallel downward field lines and horizontal equipotentials.Radial fieldMdashed circles: equipotentialsNear-surface modelgparallel lines: approximately constant g
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 to keep

  • Distance r is measured centre to centre.
  • Field strength exists whether or not a test mass is placed there.
  • Gravitational force is attractive, so direction matters even when the equation uses magnitudes.

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.

  1. 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.

  2. 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).

  3. 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.

Question

A planet's field strength is 4.00 N kg⁻¹ at 3.00 × 10⁶ m from its centre. Estimate its mass using G = 6.67 × 10⁻¹¹ N m² kg⁻².

Check the worked solution

From g = GM/r², M = gr²/G = 4.00(3.00 × 10⁶)²/(6.67 × 10⁻¹¹) = 5.40 × 10²³ kg.

Practise with support

Try this

A spherical planet has mass 6.00 × 10²⁴ kg and radius 6.50 × 10⁶ m. Find g at altitude 5.00 × 10⁵ m.

Hint: The inverse-square distance is measured from the planet's centre.

Check your answer

r = 6.50 × 10⁶ + 5.00 × 10⁵ = 7.00 × 10⁶ m. Thus g = GM/r² = 6.67 × 10⁻¹¹(6.00 × 10²⁴)/(7.00 × 10⁶)² = 8.17 N kg⁻¹ towards the centre.

Practise independently

Your turn

At a planet's surface, g = 12.0 N kg⁻¹ and radius R = 8.00 × 10⁶ m. Find g at altitude R above the surface.

Check your answer

The new centre distance is r = R + R = 2R. Since g ∝ 1/r², g = 12.0(R/2R)² = 3.00 N kg⁻¹.

Common mistakes

Common mistake

Gravitational force is inversely proportional to separation.

What is wrong with this reasoning?

Show better thinking

It is inversely proportional to the square of centre-to-centre separation: doubling r reduces force to one quarter.

Common mistake

The r in GM/r² is altitude above a planet's surface.

What is wrong with this reasoning?

Show better thinking

For a spherical planet, r is measured from its centre. At altitude h, use r = R + h.

Common mistake

A heavier test mass experiences a stronger gravitational field strength.

What is wrong with this reasoning?

Show better thinking

It experiences a larger force, but g = F/m = GM/r² is a property of the source and position, not the test mass.

Exam guidance

Sketch the field direction first and keep source mass M separate from test mass m.

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.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

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.

Check what stayed with you

Recall question 1

State Newton's law of gravitation.

Check the answer

F = Gm1m2/r², attractive along the line joining the masses.

Recall question 2

How is r measured for spherical bodies?

Check the answer

Centre to centre.

Recall question 3

When does free-fall acceleration equal g?

Check the answer

When gravity is the only significant force.

Try this next

Continue to the next lesson in this topic.

Potential, potential energy and field gradient

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. The official topic states no explicit exclusions. Models use centre-to-centre distance, zero potential at infinity, point masses or spherical bodies viewed externally, and ideal circular orbits unless a question states otherwise.

  • GCE A-Level H2 PhysicsTopic 8(a) / Topic 8(b) / Topic 8(c) / Topic 8(d) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027