Newton’s law of gravitation — OCR A-Level Physics
Test yourself on Newton’s law of gravitation with OCR A-Level practice questions.
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Newton’s law of gravitation explained
Newton’s law of gravitation states that the gravitational force F between two point masses is directly proportional to the product of their masses (M and m) and inversely proportional to the square of the distance r between their centres.
Read the full explanation
The force is attractive, so the vector form includes a negative sign: F = −GMm/r², where G is the gravitational constant, approximately 6.67 × 10⁻¹¹ N m² kg⁻². The law applies to point masses and, for spherical masses, to their centres. The force acts along the line joining the centres. In an MCQ, you may calculate F, M, m or r, or compare forces when masses or distances change.
(b) gravitational field strength g r GM 2 =- for a point mass
Gravitational field strength g at a point is the force per unit mass on a small test mass placed there. For a point mass M, or any spherically symmetric body treated as one, the field at distance r from its centre has magnitude g = GM/r², where G is the gravitational constant. The field is directed towards the mass, so the vector form carries a minus sign when r is measured outwards from the mass: g = −GM/r². Because r appears squared in the denominator, doubling the distance quarters the field strength. For example, at twice the Earth's surface radius from the centre, g falls to about one quarter of its surface value. The expression applies outside the body; inside a uniform sphere the field varies linearly with distance from the centre.
(c) gravitational field strength is uniform close to the surface of the Earth and numerically equal to the acceleration of free fall.
Close to the Earth's surface, over distances small compared with the Earth's radius, the gravitational field strength g is effectively uniform: its magnitude and direction are the same everywhere in that region. Numerically g equals the acceleration of free fall, about 9.81 m s⁻², because a freely falling object experiences only the gravitational force, so Newton's second law gives a = F/m = g. The two quantities share the same value but differ in meaning: field strength is force per unit mass at a point, while acceleration of free fall is the acceleration of an object moving under gravity alone. Uniformity fails over large heights, where g decreases as 1/r².
Your focus
- State and use Newton’s law of gravitation to calculate forces between point masses.
- Explain the inverse-square dependence on distance and the product dependence on masses.
- Apply the law to uniform spheres using centre-to-centre distance.
Show all 9 objectives
- Define gravitational field strength and give its SI unit.
- Use g = GM/r² to calculate the field strength at a point outside a spherical mass.
- Explain the meaning of the minus sign and the inverse-square dependence in g = −GM/r².
- Describe the gravitational field near the Earth's surface as uniform.
- Explain why the numerical value of g equals the acceleration of free fall.
- Distinguish between gravitational field strength and acceleration while recognising their numerical equality.
Newton’s law of gravitation exam tips
Marking Points
- Newton’s law of gravitation: F = −GMm/r² for two point masses, where the negative sign indicates attraction.
- The force is directly proportional to the product of the masses (M and m).
- The force is inversely proportional to the square of the distance r between the centres of the masses.
- G is the gravitational constant, approximately 6.67 × 10⁻¹¹ N m² kg⁻².
- The law applies to point masses and to uniform spheres, where r is the distance between centres.
- States that g is the gravitational force per unit mass at a point, with units N kg⁻¹.
- Recalls the magnitude relation g = GM/r² for a point mass or spherically symmetric mass.
- Explains that the minus sign in g = −GM/r² indicates the field points towards the mass when r is measured outwards.
- Applies the inverse-square dependence, so doubling r reduces g by a factor of 4.
- Recognises that the expression holds outside the body and that a uniform sphere behaves as a point mass at its centre for external points.
- States that near the Earth's surface the gravitational field is approximately uniform in magnitude and direction.
- Explains that g is numerically equal to the acceleration of free fall, approximately 9.81 m s⁻².
- Derives the equality from Newton's second law: for a freely falling body, mg = ma, so a = g.
- Distinguishes field strength (force per unit mass, N kg⁻¹) from acceleration (m s⁻²) while noting their numerical equivalence.
- Recognises that the uniform-field approximation breaks down at heights that are a significant fraction of the Earth's radius.
Examiner Tips
- 💡Use the equation F = GMm/r² for magnitude; include the negative sign only when direction is required.
- 💡Ensure r is the centre-to-centre distance, not surface-to-surface.
- 💡Remember G is a universal constant with value 6.67 × 10⁻¹¹ N m² kg⁻².
- 💡Check whether the question gives distance from the surface or from the centre, and convert to centre distance before substituting.
- 💡Keep G, M and r in SI units (N m² kg⁻², kg, m) so that g comes out in N kg⁻¹.
- 💡When comparing two points, use ratios such as g₂/g₁ = (r₁/r₂)² to avoid unnecessary arithmetic.
- 💡State the direction of g explicitly when a vector answer is expected, using the minus sign convention given in the specification.
- 💡Quote g as 9.81 m s⁻² unless the question specifies a different value such as 9.8 m s⁻².
- 💡When asked to justify uniformity, refer to the small change in height compared with the Earth's radius.
- 💡Show the link mg = ma explicitly when explaining why g equals the acceleration of free fall.
- 💡Use N kg⁻¹ when discussing field strength and m s⁻² when discussing acceleration, even though the values match.
Common Mistakes
- Misunderstanding: the force is repulsive. Correction: gravitational force is always attractive; the negative sign in the vector form indicates attraction.
- Misunderstanding: r is the distance between surfaces. Correction: r is the distance between the centres of the masses.
- Misunderstanding: the law applies to any shape. Correction: it applies strictly to point masses; for uniform spheres, it applies as if all mass were at the centre.
- Misunderstanding: G is the same as g. Correction: G is the universal gravitational constant; g is gravitational field strength, which depends on location.
- Writing g = GM/r instead of g = GM/r²; the field obeys an inverse-square law, so the correct denominator is r².
- Treating r as the height above the surface rather than the distance from the centre of the mass; for a planet, r must be measured from its centre.
- Ignoring the direction of the field and claiming g is a scalar with no sign; the vector form g = −GM/r² shows the field acts towards the mass.
- Assuming the formula applies inside a solid sphere in the same way; inside a uniform sphere the field is proportional to r, not to 1/r².
- Claiming g and the acceleration of free fall are the same physical quantity; they are numerically equal but conceptually distinct.
- Using g = 9.81 m s⁻² at all heights above the Earth; the field weakens with distance and is only uniform close to the surface.
- Confusing the units N kg⁻¹ and m s⁻² as different in value; they are equivalent units for the same numerical quantity.
- Forgetting that the uniform-field approximation requires the region to be small compared with the Earth's radius.