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Gravitational Field Strength & Field Lines

Key idea: Define gravitational field strength g, derive g = GM/r² for a point mass, and interpret gravitational field lines (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Derive and apply gravitational field strength, including the near-surface model.
  • Derive the gravitational field strength due to a point mass from Newton's law of gravitation and the definition of field strength.

1. Definitions (Must Know)

A. Gravitational field

A gravitational field is a region of space where a mass experiences a gravitational force.

B. Gravitational field line

A gravitational field line is a model used to show the field:

  • The arrow shows the direction of the gravitational force on a small test mass.
  • Closer lines indicate a stronger field.
  • Lines never cross.
  • For a spherical mass, field lines are radial and point towards the centre.

C. Gravitational field strength, g

Gravitational field strength, g, at a point is the gravitational force per unit mass on a small test mass:

g = F/m

Unit: N kg⁻¹ (equivalent to m s⁻²).

Direction: same as the gravitational force (towards the mass creating the field).

Near Earth’s surface: g ≈ 9.81 N kg⁻¹ and is approximately constant.

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.

For a freely falling test mass, F = mg and Newton’s second law gives F = ma. Therefore its acceleration has the same magnitude and direction as the local gravitational field strength: a = g.

2. Key Ideas (What Earns Marks)

  • Link to Newton’s law: F = GMm/r² ⇒ g = F/m = GM/r²
  • r is the distance from the mass’s centre.
  • Field lines show direction; field strength decreases with distance (inverse-square).
  • Near Earth’s surface, the gravitational field can be treated as uniform (constant g, parallel field lines).
  • Equipotential surfaces cross gravitational field lines at right angles; concentric spheres are equipotentials around an isolated spherical mass.
What weightlessness really means

Astronauts in orbit are not “outside gravity”. They feel weightless because they are in free fall, so the normal reaction on them is (approximately) zero.

3. Detailed Explanations

A. Deriving g = GM/r² for a point mass

Start with Newton’s law of gravitation for a mass m in the field of a point mass M:

F = GMm/r²

By definition of gravitational field strength:

g = F/m

Substitute:

g = (1/m)(GMm/r²) = GM/r²

This is why g depends on M and r, not on the test mass m.

How g changes with distance (Earth example)

Inverse-square decrease of gravitational field strength with distance from Earth's centre.

Scroll across the graph to read all labels.

Inverse-square decrease of gravitational field strength with distance from Earth's centre.Inverse-square decrease of gravitational field strength with distance from Earth's centre.
Using g = GM/r² with GM_E ≈ 3.99 × 10¹⁴ m³ s⁻²: doubling r makes g about four times smaller.
Open full-size graph
View figure data
Values for How g changes with distance (Earth example)
Distance from Earth's centre, r (10⁷ m)g = GM/r² (Earth, scaled)
0.6379.83
13.99
20.998
4.220.224

B. Field lines: what you can infer

  • Direction of vector g is along the field line (towards the mass).
  • Where lines are closer together, | vector g| is larger.
  • For a spherical mass, lines are radial (pointing towards the centre).

C. Why the “uniform field near Earth” model works

Near Earth’s surface, r ≈ R_E and changes only a little over the heights in most questions.

So g = GM/r² is approximately constant, and the field lines are approximately parallel.

4. Common Mistakes

  • Treating r as altitude h (use r = R_E + h).
  • Using m s⁻² in one line and N kg⁻¹ in the next without realising they are equivalent.
  • Saying “no gravity in orbit” instead of “in free fall, so normal reaction ≈ 0”.
  • Treating the density of field lines as an independently measurable quantity; it is a drawing convention representing field magnitude.

5. Exam Tips

  • If the question asks “compare field strengths”, use the ratio: g₂/g₁ = (r₁/r₂)²
  • Always state the direction of vector g (towards the centre of the planet/star).
  • Near Earth, it is acceptable to take g = 9.81 m s⁻² if the question treats the field as uniform.

6. Worked Examples

Modelled example 1

Field strength at a given r

Core

Problem

Earth’s mass is 5.97 × 10²⁴ kg. Find the gravitational field strength at 1.00 × 10⁷ m from Earth’s centre, including direction.
Study the worked solution

  1. Choose the field equation

    Method

    Use g = GM/r².

    Reason

    Dividing Newton’s force law by test mass removes the test mass from the result.

    Working

    g = GM/r²

  2. Substitute

    Method

    Use the distance from Earth’s centre.

    Reason

    The spherical Earth is modelled as a point source at its centre.

    Working

    g = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)/((1.00 × 10⁷)²)

  3. Evaluate and direct

    Method

    Obtain 3.98 N kg⁻¹ towards Earth.

    Reason

    Field direction is the force direction on a small positive test mass; gravity is attractive.

    Working

    g ≈ 3.98 N kg⁻¹ towards Earth's centre

Guided practice 2

Finding g at altitude

About 5 min

Problem

A satellite is 400 km above Earth’s surface. Take R_E = 6.37 × 10⁶ m and surface field strength g₀ = 9.81 N kg⁻¹. Estimate the local g.

Try this before viewing the solution

Unit: N kg⁻¹

Hints

Hint 1: convert the radial distance
Use r = R_E + h, with 400 km = 4.00 × 10⁵ m.
View solution step by step

  1. Find centre distance

    Method

    Obtain r = 6.77 × 10⁶ m.

    Reason

    Altitude is measured from the surface, whereas inverse-square distance begins at the centre.

    Working

    r = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m

  2. Use the ratio

    Method

    Obtain about 8.7 N kg⁻¹.

    Reason

    Earth’s mass and G cancel when comparing with the surface.

    Working

    g = 9.81(6.37/6.77)² ≈ 8.7 N kg⁻¹

Common misconception 3

Reading field lines

Find and correct the mistake

Learner claim

A field-line diagram has closer lines near a planet and wider spacing farther away. A learner says the exact value of g can be calculated just by counting the drawn lines. Diagnose the claim.

Try this before viewing the solution

What the spacing supports

View solution step by step

  1. Interpret the spacing

    Method

    The field is stronger near the planet and weaker farther away.

    Reason

    Closer field-line spacing conventionally represents greater field magnitude.

    Working

    |g| decreases as radial distance increases.

  2. Limit the inference

    Method

    Do not infer an exact numerical value from an unscaled sketch.

    Reason

    The illustrator chooses how many lines to draw; the field equation supplies the quantitative inverse-square relation.

    Working

    For a spherical source, g = GM/r², so g ∝ 1/r².

Examiner practice 4

Weight of an astronaut in orbit (using W = mg with local g)

3 marks

Examination question

An astronaut of mass 75 kg is in low Earth orbit where g = 8.7 N kg⁻¹. Calculate the gravitational force and explain why the astronaut can nevertheless feel weightless. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate the force

    1 mark

    Method

    Obtain about 6.5 × 10² N.

    Reason

    Use the local field strength, not the surface value.

    Working

    F = mg = 75(8.7) = 6.5 × 10² N

  2. State direction

    1 mark

    Method

    The force is towards Earth’s centre.

    Reason

    Gravity supplies the inward acceleration of the orbit.

    Working

    Direction: radially inward.

  3. Explain apparent weightlessness

    1 mark

    Method

    The astronaut and spacecraft are in free fall together, so the normal reaction is approximately zero.

    Reason

    Feeling weight depends on contact force, not on gravity being absent.

    Working

    Gravity remains substantial even while apparent weight is near zero.

Challenge 5

Finding r when g is known

Minimal support

Independent transfer

At a point in space, Earth’s gravitational field strength is 2.50 N kg⁻¹. Take M_E = 5.97 × 10²⁴ kg. Estimate the distance from Earth’s centre.

Try this before viewing the solution

Hints

Hint 1: undo the inverse square
From gr² = GM, isolate r² before taking the square root.
View solution step by step

  1. Rearrange

    Method

    Use r = square root of (GM/g).

    Reason

    The required radial distance is the positive square root.

    Working

    r = square root of (GM/g)

  2. Evaluate

    Method

    Obtain 1.26 × 10⁷ m from Earth’s centre.

    Reason

    The result is a radial distance, not an altitude.

    Working

    r = square root of (((6.67 × 10⁻¹¹)(5.97 × 10²⁴))/2.50) ≈ 1.26 × 10⁷ m

7. Mind Stretchers

Mind stretcher 1: When is g one ninth of its surface value?Extension

At what distance from Earth’s centre is the gravitational field strength 1/9 of its value at Earth’s surface? (Express your answer in terms of R_E.)

Show answer

Since g ∝ 1/r²:

g/g₀ = (R_E/r)² = 1/9

So R_E/r = 1/3 and r = 3R_E.

Mind stretcher 2: Comparing surface g for same radius, different massExtension

Two planets have the same radius, but planet A has twice the mass of planet B. Compare the gravitational field strength at their surfaces.

Show answer

At the surface, g = GM/r². If r is the same but M doubles, then g doubles. So planet A has twice the surface field strength of planet B.

Mind stretcher 3: Optional (Enrichment)Extension

A. Why g varies slightly with latitude (beyond syllabus)

Real Earth is not a perfect sphere, and it rotates. The equator is further from Earth’s centre than the poles, and rotation introduces a small “centrifugal” effect that reduces the effective g at the equator.

B. “True” vs “apparent” weightlessness (extra context)

In questions, weightlessness almost always means apparent weightlessness: the normal reaction is zero because the person/object is in free fall.

8. Practice (Quiz)

Practice: A Level Gravitation Quiz

Ready for mixed practice across F, g, φ, escape speed, and orbits?

A Level Gravitation Quiz

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027