AQA A-Level Physics Paper 2, June 2025: Question 7

1 mark · Medium difficulty · Multiple Choice

Determine the escape velocity of a moon with density 2ρ and radius R/3 in terms of the escape velocity v of a moon with density ρ and radius R.

Practise this question

Question

Question 7 asks: An object on the surface of a moon has an escape velocity v. The moon has a density ρ and radius R. Another moon X has a density 2ρ and radius R/3. What is the escape velocity for an object on the surface of X? Four multiple-choice options are given: A: 2/9 v, B: (√2)/3 v, C: √(2/3) v, D: 2/(√3) v.

Mark scheme

Show the mark scheme Mark scheme for question 7 showing the correct answer is B, corresponding to the expression (√2)/3 v, assessed under AO2.

How to answer it

Escape Velocity & Planetary Density Ratios

📋 What this question tests

This question assesses your ability to derive and apply the relationship between escape velocity, mean density (ρ), and radius (R) of a celestial body using gravitational potential energy and sphere volume formulas. It tests multi-step proportional reasoning under exam time pressure.

Question 07

Multiple Choice: Determining Escape Velocity from Density and Radius

✅ Correct Answer

B   ( (√2 / 3) v )

1 Mark awarded (AO2): Selecting option B correctly identifies how the scaling factors for radius and density combine under the square root in the escape velocity expression.

💡 Key Knowledge

  • Escape Velocity Formula: Derived by setting kinetic energy equal to gravitational potential energy:
    ½mv² = GMm / R ⇒ v = √(2GM / R)
  • Mass in terms of Density: For a sphere of uniform density:
    M = ρ × V = ρ × (4/3)πR³
  • Combined Proportionality: Substituting mass into velocity yields:
    v = √[2G(4/3 π R³ ρ) / R] = R · √(8/3 π G ρ)
    Therefore: v ∝ R√ρ

📐 Step-by-Step Derivation & Ratio Calculation

  1. Establish the fundamental relationship:
    v = √(2GM / R) and M = (4/3)πR³ρ
    Substituting M gives:
    v = √[ (8/3)πGρR² ] = R · √(ρ) · √[(8/3)πG]
    Since G and π are constants, this reduces to the direct scaling law:
    v ∝ R√ρ
  2. Identify the scaling factors for Moon X:
    • Radius of X: RX = (1/3) R
    • Density of X: ρX = 2 ρ
  3. Substitute the factors into the proportional relationship:
    vX / v = (RX / R) × √(ρX / ρ)
    vX / v = (1/3) × √2 = √2 / 3
  4. Conclusion:
    vX = (√2 / 3) v  →  Option B

🧠 Exam Technique

  • Memorise or quickly re-derive: Questions linking density to surface gravity ( g ∝ ρR ) or escape velocity ( v ∝ R√ρ ) appear frequently. Recognising v ∝ R√ρ immediately saves over a minute of derivation time.
  • Separate the constants: In multiple-choice questions, ignore constant factors like 2 , G , and 4/3 π . Focus entirely on the variables being modified ( R and ρ ).
  • Track radical signs carefully: Keep clear notes on which factor is inside the square root ( ρ ) and which ends up outside ( R ).

❌ Common Errors & Pitfalls

  • Leaving R under the root (Selecting C): Students substitute R³ and ρ but forget that R²/R leaves an R² inside the root, which simplifies to R outside. Keeping the 3 inside the radical gives √(2/3) .
  • Squaring the 3 incorrectly (Selecting A): Forgetting to take the square root of ρ , leading to 2/3² = 2/9 .
  • Inverting the radius ratio (Selecting D): Incorrectly placing the radius factor in the denominator inside the final expression, giving 2/√3 .
  • Assuming v is independent of density: Attempting to use v = √(2GM/R) directly without taking into account that changing R and ρ alters mass M .

Topics

Physics · 3.7 Fields and their consequences (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.