Changing a spacecraft’s circular orbit

8 marks · A-level · original practice case

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The situation

A spacecraft of mass 800 kg moves from a circular orbit of radius 7.00 × 10⁶ m to a circular orbit of radius 1.40 × 10⁷ m around a planet. The planet’s gravitational parameter GM is 3.99 × 10¹⁴ m³ s⁻². Treat the spacecraft mass as constant for the energy calculation and neglect drag. The proposed transfer has an available useful energy budget of 1.00 × 10¹⁰ J. This budget is a change in total orbital energy; it is not a rocket fuel-energy or manoeuvre model.

Circular orbit radii
Orbit	Radius from planet centre / m
Initial	7.00 × 10^6
Final	1.40 × 10^7

Part (a)

Calculate the spacecraft’s speed in the final orbit. [2 marks]

Give the numerical answer to 3 significant figures.

Final value / m s⁻¹

Working

Part (b)

Determine the increase in total orbital energy. [3 marks]

Give the numerical answer to 3 significant figures.

Final value / J

Working

Part (c)

Evaluate the claim that the stated budget is sufficient. [3 marks]

Your explanation

Marking rubric and worked answers

Indicative practice marks. Method marks assess valid working; accuracy marks require the associated valid method. Independent points stand on their own. Tick only points supported by your response. Do not award the same explanation twice. These ticks record your judgement, not an examiner’s result.

Part (a) marking points

Final answer: 5.34 × 10³ m s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Set GMm/r² = mv²/r. (1 method mark)
Use v = √(GM/r) and obtain 5.34 × 10³ m s⁻¹. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (b) marking points

Final answer: 1.14 × 10¹⁰ J. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use U = −GMm/r and K = GMm/(2r) for a circular orbit. (1 method mark)
Total energy is −GMm/(2r). (1 method mark)
Subtract initial total energy from final total energy: +1.14 × 10¹⁰ J. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (c) marking points

The required total energy increase exceeds 1.00 × 10¹⁰ J, so the stated budget is insufficient in the model. (1 independent mark)
A calculation using only potential-energy change overstates the required increase because kinetic energy decreases. (1 independent mark)
Energy sufficiency alone would not establish a viable transfer: a trajectory, thrust direction and propulsion/mass-change model are still needed. (1 independent mark)

Credit a conclusion consistent with a numerical arithmetic error in part (b), provided the comparison and physical limitations are explained. Do not credit a potential-energy-only model as total orbital energy.

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Related skills

Scope and reference sources

AQA 3.7.2.3–3.7.2.4 and 3.4.1.7. The orbital-energy derivation is AQA-specific; this case is not presented as OCR A specification practice. These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: orbital-transfer-budget · version 1.1.0 · seed 0. Choose another case