Newton's Law of Universal Gravitation
Key idea: Use Newton’s law of gravitation F = GMm/r², including the inverse-square relationship and the centre-to-centre distance r (A Level Physics).
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The core idea
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Learning objectives
- Apply Newton's law of gravitation to point and spherical masses.
1. Definitions (Must Know)
A. Newton’s law of gravitation (between two point masses)
For two point masses m₁ and m₂ separated by a distance r, the magnitude of the gravitational force is:
F = Gm₁m₂/r²
where G = 6.67 × 10⁻¹¹ N m² kg⁻².
The force is always attractive (towards each other).
B. Meaning of r (centre-to-centre distance)
r is the distance between the centres of mass of the two bodies.
For a satellite at altitude h above Earth’s surface:
r = R_E + h
C. When “point mass” is a valid model
- If each object is very small compared with r, treat it as a point mass.
- If an object is spherically symmetric and you are outside it, you can treat its mass as concentrated at its centre.
2. Key Ideas (What Earns Marks)
- F ∝ m₁m₂ (double either mass ⇒ double F).
- F ∝ 1/r² (double r ⇒ F becomes 1/4).
- The forces on the two masses are equal in magnitude and opposite in direction (Newton’s 3rd law).
- In calculations, use r in metres, and write the equation before substitution.
In F = Gm1m2/r², r is the centre-to-centre separation between masses. For spherical bodies, do not use surface distance unless the question explicitly defines it.
3. Detailed Explanations
A. What the inverse-square law means (fast ratio method)
If m₁ and m₂ stay the same, then:
F ∝ 1/r²
So you can compare forces without doing full calculations:
F₂/F₁ = (r₁/r₂)²
B. Direction (when signs matter)
The formula above gives magnitude. If you want a signed force, you must define your positive direction first.
Example: If outward radial direction is positive, then the gravitational force on a test mass points inward, so it is negative in that sign convention.
C. Why “point mass” appears in mark schemes
Newton’s law is written for point masses, but in A Level questions we often apply it to planets and stars because they are close to spherical.
If the object is not close to spherical (or you are very close to it), the mass distribution matters, and you cannot assume “all mass at the centre” without justification.
4. Common Mistakes
- Using the distance to the surface instead of centre-to-centre distance (r ≠ h; use r = R_E + h).
- Forgetting to square r.
- Mixing up G and g (they are completely different quantities).
- Forgetting to convert kilometres to metres before substitution.
5. Exam Tips
- If the question asks “how does the force change”, use the ratio method: F₂/F₁ = (m_(1,2)/m_(1,1))(m_(2,2)/m_(2,1))(r₁/r₂)²
- When a diagram is involved, draw forces as vectors pointing towards the other mass.
- Final answer should be in newtons (N) with sensible s.f.
6. Worked Examples
Modelled example 1
Gravitational force between two masses
Problem
Study the worked solution
Fix the separation
Method
Convert the centre-to-centre separation to metres.Reason
The SI value of G requires distance in metres.Working
r = 2000 km = 2.0 × 10⁶ mApply Newton's law
Method
Substitute both masses and square the separation.Reason
The force follows F = Gm₁m₂/r².Working
F = (6.67 × 10⁻¹¹)((2.0 × 10¹²)(3.0 × 10¹²))/((2.0 × 10⁶)²)Evaluate and state direction
Method
Obtain 1.0 × 10² N.Reason
The equation gives the common magnitude; each attractive force points towards the other asteroid.Working
F ≈ 1.0 × 10² N
Guided practice 2
Inverse-square change (no calculator)
Problem
Try this before viewing the solution
Hints
Hint 1: form a force ratio
View solution step by step
Use the inverse-square ratio
Method
Set r₂ = 3r₁.Reason
The masses and G cancel in the ratio.Working
F₂/F₁ = (r₁/3r₁)²Evaluate
Method
The new force is one ninth of the original.Reason
The factor of three in distance is squared.Working
F₂/F₁ = (1/3)² = 1/9
Common misconception 3
Force at altitude above Earth
Learner claim
Try this before viewing the solution
View solution step by step
Repair the distance
Method
Use r = R_E + h.Reason
Newton’s law uses centre-to-centre separation, not altitude above the surface.Working
r = 6.37 × 10⁶ + 3.00 × 10⁵ = 6.67 × 10⁶ mCalculate the force
Method
Obtain about 8.1 × 10³ N towards Earth.Reason
Earth is treated as a spherical source mass concentrated at its centre.Working
F = (6.67 × 10⁻¹¹)((5.97 × 10²⁴)(900))/((6.67 × 10⁶)²) ≈ 8.1 × 10³ N
Examiner practice 4
Gravitational force between Earth and Moon
Examination question
Try this before viewing the solution
View solution step by step
State the law
1 markMethod
Use F = GM_EM_M/r².Reason
The spherical bodies may be modelled as point masses at their centres.Working
F = GM_EM_M/r²Substitute and evaluate
1 markMethod
Obtain 2.0 × 10²⁰ N.Reason
The given separation is already centre-to-centre and in metres.Working
F = (6.67 × 10⁻¹¹)((5.97 × 10²⁴)(7.35 × 10²²))/((3.84 × 10⁸)²) ≈ 2.0 × 10²⁰ NState direction
1 markMethod
The force on the Moon is towards Earth.Reason
Gravitational force is attractive.Working
Direction: from the Moon towards Earth’s centre.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the law, value and direction.
Challenge 5
Finding G from a measured force
Independent transfer
Try this before viewing the solution
Hints
Hint 1: make the constant the subject
View solution step by step
Rearrange
Method
Make G the subject.Reason
The measured force and geometry determine the constant.Working
G = Fr²/m₁m₂Evaluate
Method
Obtain 6.7 × 10⁻¹¹ N m² kg⁻².Reason
Square the separation before dividing by both masses.Working
G = ((5.0 × 10⁻⁹)(0.20)²)/(1.50)(2.00) = 6.7 × 10⁻¹¹ N m² kg⁻²
7. Mind Stretchers
Mind stretcher 1: Ratio scaling with mass and distanceExtension
Two point masses are separated by a distance r and attract with force F. Mass 1 is doubled, mass 2 is tripled, and the separation becomes 2r. What is the new force in terms of F?
Show answer
F ∝ m₁m₂/r²
Scaling gives:
F'/F = (2)(3)/(2)² = 6/4 = 3/2
So F' = (3/2)F.
Mind stretcher 2: Why student–student gravity is negligibleExtension
Explain why the gravitational force between two students in a classroom is negligible compared with their weight.
Show answer
The masses are small and the separation is on the order of metres, so F = Gm₁m₂/r² is extremely small because G is very small (6.67 × 10⁻¹¹).
Each student’s weight is W = mg ∼ (50)(10) = 500 N, which is many orders of magnitude larger than the mutual gravitational attraction between the two students.
Mind stretcher 3: Optional (Enrichment)Extension
A. Inside a planet (beyond syllabus)
For points inside a planet, the “all mass at the centre” shortcut is not automatically valid. The gravitational field depends on the mass distribution (this is typically treated using the shell theorem / uniform-sphere models).
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