AQA A-Level Physics Paper 2, June 2023: Question 12

1 mark · Easy difficulty · Multiple Choice

Determine the expression for the linear speed of a satellite orbiting at height h above a planet of mass M and radius R.

Practise this question

Question

Question 12 asks: 'A satellite is in a circular orbit at a height h above the surface of a planet of mass M and radius R. What is the linear speed of the satellite?' followed by four multiple-choice options: A is the square root of GM divided by (R + h); B is the square root of the quantity GM divided by (R + h); C is GM divided by the square root of (R + h); D is GM divided by (R + h).
Question text

12 A satellite is in a circular orbit at a height h above the surface of a planet of mass M

and radius R.

What is the linear speed of the satellite?

[1 mark]

GM

A

(R + h)

GM

B

(R + h)

GM

C

R + h

GM

D

(R + h)

Mark scheme

Show the mark scheme Mark scheme for Question 12 shows the correct answer is B, with the formula given as the square root of the fraction GM over (R + h).

12 B GM

(R + h)

How to answer it

Orbital Speed of a Satellite

📌 What this question tests

This question tests your understanding of circular orbital mechanics in gravitational fields. Specifically, it assesses whether you can equate gravitational force to centripetal force, correctly identify the orbital radius measured from the centre of mass (including planetary radius and altitude), and rearrange algebraically for linear orbital speed v .

Multiple Choice Question (1 Mark)

Question 12: Linear Speed of an Orbiting Satellite

A satellite is in a circular orbit at a height h above the surface of a planet of mass M and radius R. What is the linear speed of the satellite?

✅ Correct Answer

Option B: √[GM / (R + h)]

Mark Scheme Breakdown:
• 1 Mark for selecting B.

💡 Key Knowledge

  • Centripetal Force: For a body of mass m in circular motion at speed v: F = mv² / r .
  • Newton's Law of Gravitation: The attractive force between two masses is F = GMm / r² .
  • Orbital Radius: Gravitational equations always require distance from the centre of mass of the planet, which is r = R + h .

📐 Step-by-Step Derivation

  1. Define total orbital radius:
    Distance from the centre of the planet to the satellite:
    r = R + h
  2. Equate centripetal force to gravitational force:
    The gravitational pull provides the necessary centripetal force keeping the satellite in orbit:
    mv² / r = GMm / r²
  3. Cancel satellite mass m and one factor of r:
    v² = GM / r
  4. Substitute r = (R + h) and take the square root:
    v = √(GM / (R + h))

🧠 Exam Technique & Dimensional Analysis

  • Dimensional quick-check:
    Gravitational potential is proportional to GM/r (units: J kg⁻¹ = m² s⁻²). Therefore, speed (m s⁻¹) must have the dimension of the square root of potential: √(GM/r) .
  • This immediately eliminates C and D without even doing full algebraic rearrangements!
  • Comparing A and B: the square root must apply to the whole fraction, because v² = GM/r ⇒ v = √(GM/r) , not √GM / r .

❌ Common Errors & Traps

  • Distractor A: Forgetting that taking the square root of both sides means taking the root of the denominator as well, leaving √GM / (R + h) .
  • Distractor D: Forgetting to take the square root altogether, confusing orbital speed v with v² .
  • Distractor C: Taking the square root of the denominator only.
  • Overlooking 'height above surface': Always check if a question gives orbit radius as r or as altitude h . Here, all options included (R + h) , but in calculation questions, missing R is the #1 student error.

Topics

Physics · 3.6 Further mechanics and thermal physics (A-level only) · 3.7 Fields and their consequences (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.