Calculate gravitational field strength around a point mass
Derive and use g = GM/r², then explain why field strength changes little over small height differences near Earth's surface.
Section 13 · Gravitational fields
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 13. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use Newton's laws, circular motion, vectors, scientific notation and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why does the test mass disappear when deriving g for a point source?
What you will learn
- Derive g = GM/r² from Newton's law and field-strength definition
- Recall and use g = GM/r²
- Explain the near-surface constant-g approximation
Derive field strength from gravitational force
A point mass M attracts a test mass m at distance r with force F = GMm/r². Field strength is force per unit test mass, g = F/m. Dividing by m gives g = GM/r².
The test mass cancels because field strength describes the source mass and the location, not the test object. Use the distance from the source's centre and give g in N kg⁻¹ or m s⁻².
Near Earth's surface, a small height change barely changes r
At height h above a spherical Earth of radius R, g = GM/(R + h)². If h is small compared with R, then R + h is close to R, so g changes only slightly and can be treated as approximately constant.
This approximation supports familiar near-surface calculations such as weight mg and gravitational potential-energy change mgΔh. It is not suitable for a height comparable with Earth's radius, where the inverse-square change matters.
Pause and explain
For a point mass M at radius r, which expression gives gravitational field strength?
Worked example
Estimate g at a distance 6.4 × 10⁶ m from the centre of a planet of mass 6.0 × 10²⁴ kg.
Show the worked solution
- Use g = GM/r².
- Substitute 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / (6.4 × 10⁶)².
- Give N kg⁻¹.
Answer g ≈ 9.8 N kg⁻¹.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A planet has mass 4.0 × 10²⁴ kg. Calculate g at radius 5.0 × 10⁶ m from its centre.
Give me a hint
Substitute into g = GM/r² and square the radius in scientific notation.
Compare my reasoning
- g = GM/r².
- g = (6.67 × 10⁻¹¹)(4.0 × 10²⁴)/(5.0 × 10⁶)².
- Evaluate and give N kg⁻¹.
g ≈ 10.7 N kg⁻¹.
Look for these in your work
- Field equation selected
- Radius squared
- Unit stated
Earth's radius is 6.4 × 10⁶ m. Compare g at the surface with g 1.0 km above it using a ratio, and explain why the difference is small.
Give me a hint
Use g(h)/g(0) = [R/(R + h)]² and note that 1000 m is small compared with R.
Compare my reasoning
- Use h = 1.0 × 10³ m and R = 6.4 × 10⁶ m.
- g(h)/g(0) = [6.4 × 10⁶/(6.401 × 10⁶)]² ≈ 0.9997.
- The ratio is very close to 1 because h ≪ R.
The field is about 0.9997 of its surface value, a change of roughly 0.03%; the near-surface constant-g approximation is reasonable.
Look for these in your work
- Centre distance ratio used
- Numerical ratio sensible
- Small-height assumption explained
A learner uses g = 9.8 N kg⁻¹ at an altitude comparable with a planet's radius. Explain why this may be inaccurate.
Give me a hint
Compare the altitude with the radius and decide whether R + h is still close to R.
Compare my reasoning
- At large altitude, h is not small compared with R.
- The distance from the centre becomes R + h.
- Since g = GM/(R + h)², the field is materially weaker than at the surface.
The approximation requires h ≪ R; at a comparable altitude the inverse-square variation cannot be ignored.
Look for these in your work
- Assumption stated
- Centre distance updated
- Inverse-square consequence explained
Common mistakes
- Leaving the test mass in g after dividing force by mass
- Using height above the surface instead of distance from centre
- Applying constant g over a height comparable with the planet radius
What can you explain now?
Why does the test mass disappear when deriving g for a point source?
Compare with the explanation
g is F divided by the test mass
The m in GMm/r² cancels with the m in F/m.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
What assumption lets g be treated as constant near Earth's surface, and when does it fail?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.