Specification: AQA A-level Physics (7408) 3.7.2.1 | Paper: 2 | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Newton’s law of gravitation: the gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation. The force is always attractive.
Point mass: a mass whose size is negligible compared with the distances involved. A uniform sphere acts as a point mass at its centre, so r is always measured centre to centre.
Inverse-square law: F ∝ 1/r², so doubling the separation quarters the force and tripling it reduces the force to one ninth.
The equation
gravitational force = (gravitational constant × mass 1 × mass 2) ÷ (separation)² F = Gm₁m₂ ÷ r²
| Quantity | Symbol | Unit |
|---|---|---|
| gravitational force (attractive) | F | newtons (N) |
| gravitational constant (6.67 × 10−11) | G | N m² kg⁻² |
| masses of the two objects | m₁, m₂ | kilograms (kg) |
| separation of the centres | r | metres (m) |
Rearranged: m₁ = Fr² ÷ (Gm₂) | r = √(Gm₁m₂ ÷ F) | G = Fr² ÷ (m₁m₂) | Measure r from centre to centre (add the radius to any height above a surface) and square it before you divide.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) Two small spheres of mass 5.0 kg and 8.0 kg are placed with their centres 0.50 m apart. Calculate the gravitational force between them, to 2 significant figures. (G = 6.67 × 10−11 N m² kg⁻²)
Show solution
F: F = Gm₁m₂ ÷ r²
I: F = 6.67 × 10−11 × 5.0 × 8.0 ÷ 0.50²
F: F is already the subject
A: 1.0672 × 10−8 N = 1.1 × 10−8 N (2 s.f.)
Q2 (Core) The mass of the Earth is 5.97 × 1024 kg and the mass of the Moon is 7.35 × 1022 kg. The mean separation of their centres is 3.84 × 108 m. Calculate the gravitational force between the Earth and the Moon, to 3 significant figures. (G = 6.67 × 10−11 N m² kg⁻²)
Show solution
F: F = Gm₁m₂ ÷ r²
I: F = 6.67 × 10−11 × 5.97 × 1024 × 7.35 × 1022 ÷ (3.84 × 108)²
F: F is already the subject
A: 1.984… × 1020 N = 1.98 × 1020 N (3 s.f.)
Watch out: forgetting to square r. Put the whole of (3.84 × 108)² in brackets on your calculator.
Q3 (Core) Two people sit on a bench with their centres of mass 60 cm apart. The gravitational force between them is 8.4 × 10−7 N. One person has a mass of 65 kg. Treating them as point masses, calculate the mass of the other person, to 2 significant figures. (G = 6.67 × 10−11 N m² kg⁻²)
Show solution
Convert first: 60 cm = 0.60 m
F: F = Gm₁m₂ ÷ r²
I: 8.4 × 10−7 = 6.67 × 10−11 × 65 × m₂ ÷ 0.60²
F: m₂ = Fr² ÷ (Gm₁) = 8.4 × 10−7 × 0.36 ÷ (6.67 × 10−11 × 65)
A: 69.7… kg = 70 kg (2 s.f.)
For comparison, the first person’s weight is 65 × 9.81 = 637.65 N, about 7.6 × 108 times bigger. This is why you never notice the gravitational pull of other people.
Q4 (Core) In a Cavendish-style experiment, a 730 g lead ball is attracted to a 158 kg lead sphere with a force of 1.5 × 10−7 N. Calculate the separation of their centres in cm, to 2 significant figures. (G = 6.67 × 10−11 N m² kg⁻²)
Show solution
Convert first: 730 g = 0.730 kg
F: F = Gm₁m₂ ÷ r²
I: 1.5 × 10−7 = 6.67 × 10−11 × 158 × 0.730 ÷ r²
F: r = √(Gm₁m₂ ÷ F) = √(7.693… × 10−9 ÷ 1.5 × 10−7) = √0.05128…
A: r = 0.2264… m = 23 cm (2 s.f.)
Watch out: stopping at r² = 0.051 m². Remember to take the square root at the end.
Q5 (Exam) A space probe at a distance r from the centre of the Earth experiences a gravitational force of 2400 N. Calculate the gravitational force on the probe when it is at a distance of (a) 2r and (b) 3r from the centre of the Earth. Give (b) to 2 significant figures.
Show solution
F: F = Gm₁m₂ ÷ r², so F ∝ 1/r² (G and both masses are constant)
I: (a) F = 2400 ÷ 2² = 2400 ÷ 4; (b) F = 2400 ÷ 3² = 2400 ÷ 9
F: new force = old force × (r_old ÷ r_new)²
A: (a) 600 N; (b) 266.6… N = 270 N (2 s.f.)
Watch out: halving the force when the distance doubles. Gravity follows an inverse-square law, not an inverse law.
Q6 (Exam) A 1200 kg satellite orbits at a height of 400 km above the Earth’s surface. Calculate the gravitational force on the satellite, to 3 significant figures. (G = 6.67 × 10−11 N m² kg⁻², mass of Earth = 5.97 × 1024 kg, radius of Earth = 6.37 × 106 m)
Show solution
Convert first: 400 km = 4.00 × 105 m; r = R + h = 6.37 × 106 + 4.00 × 105 = 6.77 × 106 m
F: F = Gm₁m₂ ÷ r²
I: F = 6.67 × 10−11 × 5.97 × 1024 × 1200 ÷ (6.77 × 106)²
F: F is already the subject
A: 10 425.6… N = 1.04 × 104 N (3 s.f.)
Watch out: using the height (4.00 × 105 m) as r gives 2.99 × 106 N, nearly 300 times too big. r is always measured from the centre of the Earth.
Q7 (Exam) The gravitational force between the Sun and the Earth is 3.52 × 1022 N. The mean separation of their centres is 1.50 × 1011 m. Calculate the mass of the Sun, to 3 significant figures. (G = 6.67 × 10−11 N m² kg⁻², mass of Earth = 5.97 × 1024 kg)
Show solution
F: F = Gm₁m₂ ÷ r²
I: 3.52 × 1022 = 6.67 × 10−11 × m₁ × 5.97 × 1024 ÷ (1.50 × 1011)²
F: m₁ = Fr² ÷ (Gm₂) = 3.52 × 1022 × 2.25 × 1022 ÷ (6.67 × 10−11 × 5.97 × 1024)
A: 1.988… × 1030 kg = 1.99 × 1030 kg (3 s.f.)
Q8 (Exam) A small 0.50 kg sphere sits on a straight line between a 6.0 kg sphere and a 2.0 kg sphere. The centre of each large sphere is 15 cm from the centre of the small sphere. Calculate the size and direction of the resultant gravitational force on the 0.50 kg sphere, to 2 significant figures. (G = 6.67 × 10−11 N m² kg⁻²)
Show solution
Convert first: 15 cm = 0.15 m
Step 1 – force from the 6.0 kg sphere
F: F = Gm₁m₂ ÷ r²
I: F = 6.67 × 10−11 × 6.0 × 0.50 ÷ 0.15²
A: 8.893… × 10−9 N towards the 6.0 kg sphere
Step 2 – force from the 2.0 kg sphere
F = 6.67 × 10−11 × 2.0 × 0.50 ÷ 0.15² = 2.964… × 10−9 N towards the 2.0 kg sphere
Step 3 – resultant
the forces act in opposite directions, so resultant = 8.893… × 10−9 − 2.964… × 10−9
A: 5.928… × 10−9 N = 5.9 × 10−9 N towards the 6.0 kg sphere (2 s.f.)
Watch out: adding the two forces. Both are attractive, so they pull the small sphere in opposite directions.
Q9 (Stretch) A spacecraft travels on the straight line joining the centres of the Earth and the Moon. Calculate the distance from the centre of the Earth at which the resultant gravitational force on the spacecraft is zero, to 3 significant figures. (Mass of Earth = 5.97 × 1024 kg, mass of Moon = 7.35 × 1022 kg, Earth–Moon separation = 3.84 × 108 m)
Show solution
Step 1 – set the two forces equal
let x = distance from the Earth’s centre, so the distance from the Moon’s centre = 3.84 × 108 − x
GMEm ÷ x² = GMMm ÷ (3.84 × 108 − x)² (G and the spacecraft’s mass m cancel)
Step 2 – take square roots
x ÷ (3.84 × 108 − x) = √(ME ÷ MM) = √(5.97 × 1024 ÷ 7.35 × 1022) = √81.22… = 9.012…
Step 3 – solve for x
x = 9.012… × (3.84 × 108 − x), so 10.012… x = 9.012… × 3.84 × 108
A: x = 3.456… × 108 m = 3.46 × 108 m from the centre of the Earth (3 s.f.), which is 3.84 × 107 m from the centre of the Moon
Watch out: the zero point is not halfway. The Moon is about 81 times less massive, so the point is about 9 times further from the Earth than from the Moon.
Q10 (Stretch) Two identical uniform solid lead spheres, each of radius 10 cm, rest with their surfaces touching. The density of lead is 11 300 kg m⁻³. (a) Calculate the gravitational force between them, to 2 significant figures. (b) Without recalculating the masses, find the force between two touching lead spheres each of radius 20 cm, to 2 significant figures. (G = 6.67 × 10−11 N m² kg⁻²)
Show solution
Convert first: 10 cm = 0.10 m; touching spheres have centres 2 × 0.10 = 0.20 m apart
Step 1 – mass of each sphere
m = ρV = 11 300 × (4/3)π × 0.10³ = 47.33… kg
Step 2 – (a) force
F: F = Gm₁m₂ ÷ r²
I: F = 6.67 × 10−11 × 47.33… × 47.33… ÷ 0.20²
A: 3.735… × 10−6 N = 3.7 × 10−6 N (2 s.f.)
Step 3 – (b) ratio
doubling the radius multiplies each mass by 2³ = 8, so m₁m₂ goes up by 64; the separation doubles, so r² goes up by 4
F goes up by 64 ÷ 4 = 16, so F = 16 × 3.735… × 10−6 = 5.977… × 10−5 N
A: 6.0 × 10−5 N (2 s.f.)
Watch out: using the 0.10 m radius as r. Each uniform sphere acts as a point mass at its centre, so r is the centre-to-centre distance.
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