Specification: AQA A-level Physics (7408) 3.7.2.3 | Paper: 2 | Data and formulae booklet: given in the exam
🛒 All the equations in one place: the AQA A-level Every Key Equation PowerPoint has 127 equations, 381 practice questions and 15 derivations, for £12.
📌 KEY DEFINITIONS – learn them word for word
Gravitational potential, V: the work done per unit mass to move a small mass from infinity to that point. V is zero at infinity and negative everywhere else, because gravity is attractive: work is done by the field as a mass moves in, so the mass has less energy than at infinity.
Equipotential: a surface on which the gravitational potential is the same everywhere. No work is done moving a mass along an equipotential (ΔV = 0). Equipotentials are always perpendicular to the field lines.
Potential gradient: g = −ΔV ÷ Δr. The field strength is minus the gradient of a V–r graph; the minus sign shows that g points towards decreasing r, the direction in which V decreases.
The equation
gravitational potential = −(gravitational constant × mass) ÷ distance from centre V = −GM ÷ r with work done = mass × change in potential ΔW = mΔV and g = −ΔV ÷ Δr
| Quantity | Symbol | Unit |
|---|---|---|
| gravitational potential | V | joules per kilogram (J kg⁻¹) |
| gravitational constant | G | 6.67 × 10−11 N m² kg⁻² |
| mass of the body creating the field | M | kilograms (kg) |
| distance from the centre of the body | r | metres (m) |
| work done (change in gravitational potential energy) | ΔW | joules (J) |
| mass being moved | m | kilograms (kg) |
| gravitational field strength | g | newtons per kilogram (N kg⁻¹) |
Rearranged: r = −GM ÷ V | M = −Vr ÷ G | ΔV = ΔW ÷ m | m = ΔW ÷ ΔV | r is measured from the centre of the planet, so r = R + h. Keep the minus sign: V is always negative, and moving away from a planet makes V less negative (ΔV positive, so work must be done).
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) Calculate the gravitational potential at the surface of the Earth, 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
F: V = −GM ÷ r
I: V = −(6.67 × 10−11 × 5.97 × 1024) ÷ 6.37 × 106 = −3.98199 × 1014 ÷ 6.37 × 106
F: V is already the subject
A: −6.251… × 107 J kg⁻¹ = −6.25 × 107 J kg⁻¹ (3 s.f.)
Q2 (Core) The Moon has a mass of 7.35 × 1022 kg and a radius of 1.74 × 106 m. Calculate the gravitational potential at the Moon’s surface, to 3 significant figures. (G = 6.67 × 10−11 N m² kg⁻²)
Show solution
F: V = −GM ÷ r
I: V = −(6.67 × 10−11 × 7.35 × 1022) ÷ 1.74 × 106
F: V is already the subject
A: −2.8175 × 106 J kg⁻¹ = −2.82 × 106 J kg⁻¹ (3 s.f.)
Watch out: the answer must be negative. A positive gravitational potential loses the mark, because V is zero at infinity and gravity is attractive.
Q3 (Core) A space station orbits 400 km above the Earth’s surface. Calculate the gravitational potential at this height, 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, so r = R + h = 6.37 × 106 + 4.00 × 105 = 6.77 × 106 m
F: V = −GM ÷ r
I: V = −3.98199 × 1014 ÷ 6.77 × 106
F: V is already the subject
A: −5.881… × 107 J kg⁻¹ = −5.88 × 107 J kg⁻¹ (3 s.f.)
Watch out: examiners regularly see r = 400 km used. Distances in V = −GM/r are always from the centre of the planet.
Q4 (Core) At a point above the Earth the gravitational potential is −1.60 × 107 J kg⁻¹. Calculate the distance of this point from the centre of the Earth and its height above the surface, both 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
F: V = −GM ÷ r
I: −1.60 × 107 = −3.98199 × 1014 ÷ r
F: r = 3.98199 × 1014 ÷ 1.60 × 107
A: r = 2.488… × 107 m = 2.49 × 107 m (3 s.f.)
height = r − R = 2.488… × 107 − 6.37 × 106 = 1.851… × 107 m = 1.85 × 107 m (3 s.f.)
Q5 (Exam) A 1200 kg satellite is lifted from the Earth’s surface to a height of 400 km. Using ΔW = mΔV, calculate the minimum work done against gravity, 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
Step 1 – potentials
at the surface: V₁ = −3.98199 × 1014 ÷ 6.37 × 106 = −6.251… × 107 J kg⁻¹
at 400 km: r = 6.77 × 106 m, V₂ = −3.98199 × 1014 ÷ 6.77 × 106 = −5.881… × 107 J kg⁻¹
ΔV = V₂ − V₁ = −5.881… × 107 − (−6.251… × 107) = +3.693… × 106 J kg⁻¹
Step 2 – work done
F: ΔW = mΔV
I: ΔW = 1200 × 3.693… × 106
F: ΔW is already the subject
A: 4.432… × 109 J = 4.43 × 109 J (3 s.f.)
Watch out: mgΔh with g = 9.81 N kg⁻¹ gives 4.71 × 109 J, which is too big. g falls with height, so mgΔh only works for small height changes near the surface.
Q6 (Exam) A tangent is drawn to a graph of gravitational potential V against distance r from the centre of the Earth, at r = 8.0 × 106 m. The tangent passes through the points (4.0 × 106 m, −7.47 × 107 J kg⁻¹) and (1.2 × 107 m, −2.49 × 107 J kg⁻¹). Calculate the gravitational field strength at r = 8.0 × 106 m, to 2 significant figures, and state its direction.
Show solution
F: g = −ΔV ÷ Δr
I: ΔV = −2.49 × 107 − (−7.47 × 107) = 4.98 × 107 J kg⁻¹; Δr = 1.2 × 107 − 4.0 × 106 = 8.0 × 106 m
F: g = −(4.98 × 107 ÷ 8.0 × 106) = −6.225 N kg⁻¹
A: magnitude 6.2 N kg⁻¹ (2 s.f.), directed towards the centre of the Earth (the minus sign means towards decreasing r)
Check: GM ÷ r² = 3.98199 × 1014 ÷ (8.0 × 106)² = 6.22 N kg⁻¹, which agrees.
Watch out: use the gradient of the tangent at the point, not V ÷ r from a single point on the curve.
Q7 (Exam) Two equipotential surfaces around the Earth have potentials of −5.0 × 107 J kg⁻¹ and −4.0 × 107 J kg⁻¹. (a) Calculate the work done moving a 500 kg satellite from the inner surface to the outer one, to 2 significant figures. (b) State the work done moving it all the way around one equipotential. (c) Calculate the separation of the two surfaces, 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
(a) F: ΔW = mΔV
I: ΔW = 500 × (−4.0 × 107 − (−5.0 × 107)) = 500 × 1.0 × 107
F: ΔW is already the subject
A: 5.0 × 109 J (2 s.f.)
(b) ΔV = 0 along an equipotential, so the work done is zero.
(c) r = GM ÷ |V|: r₁ = 3.98199 × 1014 ÷ 5.0 × 107 = 7.963… × 106 m; r₂ = 3.98199 × 1014 ÷ 4.0 × 107 = 9.954… × 106 m
separation = 9.954… × 106 − 7.963… × 106 = 1.990… × 106 m = 1.99 × 106 m (3 s.f.)
Note: equipotentials with equal steps of ΔV get further apart as r increases, because the field gets weaker.
Q8 (Exam) An 850 kg satellite is moved from a circular orbit of radius 7.0 × 106 m to a geostationary orbit of radius 4.22 × 107 m. Calculate the increase in its gravitational potential energy, 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
F: ΔW = mΔV, with ΔV = (−GM ÷ r₂) − (−GM ÷ r₁) = GM(1 ÷ r₁ − 1 ÷ r₂)
I: V₁ = −3.98199 × 1014 ÷ 7.0 × 106 = −5.688… × 107 J kg⁻¹; V₂ = −3.98199 × 1014 ÷ 4.22 × 107 = −9.435… × 106 J kg⁻¹
F: ΔV = −9.435… × 106 − (−5.688… × 107) = 4.744… × 107 J kg⁻¹, so ΔW = 850 × 4.744… × 107
A: 4.033… × 1010 J = 4.03 × 1010 J (3 s.f.)
Watch out: this is the change in potential energy only. The satellite moves more slowly in the higher orbit, so the total energy that must be supplied is less than this (see the orbits page).
Q9 (Stretch) (a) Show that the escape velocity from the surface of a planet of mass M and radius R is v = √(2GM ÷ R). (b) Calculate the escape velocity from the Earth, to 3 significant figures. (c) Calculate the escape velocity from the Moon (mass 7.35 × 1022 kg, radius 1.74 × 106 m), 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) Ignore air resistance.
Show solution
Step 1 – (a) derivation
to escape, the object must reach infinity (V = 0), so the kinetic energy at launch must equal the work needed: ½mv² = mΔV = m × (0 − (−GM ÷ R)) = GMm ÷ R
the m cancels: v² = 2GM ÷ R, so v = √(2GM ÷ R)
Step 2 – (b) Earth
F: v = √(2GM ÷ R)
I: v = √(2 × 3.98199 × 1014 ÷ 6.37 × 106) = √(1.250… × 108)
F: v is already the subject
A: 1.118… × 104 m s⁻¹ = 1.12 × 104 m s⁻¹ (3 s.f.), about 11.2 km s⁻¹
Step 3 – (c) Moon
v = √(2 × 6.67 × 10−11 × 7.35 × 1022 ÷ 1.74 × 106) = 2373.8… m s⁻¹ = 2.37 × 103 m s⁻¹ (3 s.f.)
Watch out: escape velocity does not depend on the mass of the object being launched, because m cancels.
Q10 (Stretch) A probe is launched vertically upwards from the Earth’s surface at 5.00 km s⁻¹. Ignoring air resistance and the Earth’s rotation, use energy conservation with V = −GM/r to calculate the maximum height the probe reaches above the surface, 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: 5.00 km s⁻¹ = 5000 m s⁻¹
Step 1 – energy per kilogram
F: loss of kinetic energy per kg = gain in potential: ½v² = Vtop − Vsurface = −GM ÷ r + GM ÷ R
I: ½ × 5000² = 1.25 × 107 J kg⁻¹, so 1 ÷ r = 1 ÷ R − v² ÷ 2GM
Step 2 – solve for r
v² ÷ 2GM = 2.5 × 107 ÷ (2 × 3.98199 × 1014) = 3.139… × 10−8 m⁻¹; 1 ÷ R = 1 ÷ 6.37 × 106 = 1.569… × 10−7 m⁻¹
F: 1 ÷ r = 1.569… × 10−7 − 3.139… × 10−8 = 1.255… × 10−7 m⁻¹, so r = 7.962… × 106 m
Step 3 – height
A: h = r − R = 7.962… × 106 − 6.37 × 106 = 1.592… × 106 m = 1.59 × 106 m (3 s.f.), about 1590 km
Watch out: v² = 2gh with g = 9.81 N kg⁻¹ gives only 1.27 × 106 m. Over large heights g is not constant, so you must use the change in potential.
🧮 Want every A-level equation in one place?
The AQA A-level Every Key Equation PowerPoint has 127 equations, 381 practice questions and 15 derivations, covering the core and all five options, for £12. For the rest of the specification, the AQA A-level Physics Focus Questions turn every point of the spec into a question with the answer underneath, for £24.