Use Newton's law of gravitation
Calculate the mutual attraction between point masses and apply the equivalent point-mass model outside a uniform sphere.
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
Which distance belongs in Newton's law of gravitation?
What you will learn
- Treat a point outside a uniform sphere as if its mass were at the centre
- Recall and use F = Gm₁m₂/r² for two point masses
- Distinguish centre-to-centre separation from surface separation
Two masses attract along their line of centres
Newton's law of gravitation gives the magnitude of the force between two point masses: F = Gm₁m₂/r². The forces on the two masses are equal in magnitude and opposite in direction, along the line joining their centres.
Use the universal gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻², masses in kilograms and centre-to-centre separation r in metres. The inverse-square dependence means doubling r reduces the force to one quarter.
A uniform sphere acts as a point mass from outside
At a point outside a uniform sphere, its gravitational effect can be calculated as if the sphere's entire mass were concentrated at its centre. The distance r is measured from the sphere's centre, not from its surface.
For two separated spheres, first identify the centre-to-centre distance. Add radii to a surface gap when necessary; using the gap alone makes r too small and the calculated force too large.
Pause and explain
If the separation of two point masses doubles, how does their gravitational force change?
Worked example
Two point masses of 5.0 kg are separated by 0.20 m centre to centre. Find their gravitational force.
Show the worked solution
- Use F = Gm₁m₂/r².
- Substitute 6.67 × 10⁻¹¹ × 5.0 × 5.0 / 0.20².
- Give a force magnitude in newtons.
Answer F ≈ 4.2 × 10⁻⁸ N; each mass experiences an equal force toward the other.
Try a topic test
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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.
Calculate the gravitational force between masses of 2.0 kg and 3.0 kg separated by 0.50 m.
Give me a hint
Substitute both masses and square the 0.50 m separation.
Compare my reasoning
- F = Gm₁m₂/r².
- F = 6.67 × 10⁻¹¹ × 2.0 × 3.0 / 0.50².
- Evaluate in newtons and round suitably.
F ≈ 1.6 × 10⁻⁹ N.
Look for these in your work
- Law selected
- Distance squared
- Magnitude and unit stated
A uniform spherical moon has radius 1.0 × 10⁶ m. A spacecraft is 2.0 × 10⁶ m above its surface. What r should be used to model the moon's gravity as a point mass?
Give me a hint
For an outside point, measure from the moon's centre rather than its surface.
Compare my reasoning
- Centre-to-surface radius = 1.0 × 10⁶ m.
- Add the 2.0 × 10⁶ m altitude.
- Use r = 3.0 × 10⁶ m.
Use r = 3.0 × 10⁶ m from the moon's centre.
Look for these in your work
- Outside-sphere model applied
- Radius and altitude combined
- Centre distance stated
Two satellites have the same masses and are separated by distances r and 3r. Compare the gravitational forces.
Give me a hint
Use the inverse-square dependence while keeping both masses unchanged.
Compare my reasoning
- F₁ ∝ 1/r².
- F₂ ∝ 1/(3r)² = 1/(9r²).
- The force at 3r is one ninth of the force at r.
At separation 3r, the gravitational force is one ninth as large.
Look for these in your work
- Masses held fixed
- Distance squared applied
- Force ratio correct
Common mistakes
- Using surface gap instead of centre-to-centre distance
- Forgetting to square the separation
- Treating the mutual forces as unequal
What can you explain now?
Which distance belongs in Newton's law of gravitation?
Compare with the explanation
Centre-to-centre separation
The law uses the separation between the centres of the two masses.
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Make it stick
Why must an outside observer use the centre of a uniform sphere as the point-mass location?
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.