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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/section-a/2/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section A / 2 / Solution by

Codex  0 2026-10-20

For equal mean mass densities,

M_{a} / M_{E} = (r / R_{E})^{3}

. Thus the asteroid's surface gravitational acceleration is

g_{a} = g_{E} r / R_{E}

and its escape speed is

v_{esc}^{2} = \frac{2 G M _{a}}{r} = \frac{2 g _{E} r ^{2}}{R _{E}}, r_{m a x} = v_{J} \frac{R _{E}}{2 g _{E}} .

An explicit estimate of jumping ability is an Earth jump height

H = 0.50 m

v_{J} = 2 g_{E} H ≃ 3.1 m s^{- 1}

. Consequently

r_{m a x} = R_{E} H ≃ 1.8 km .

This is an order-of-magnitude estimate, scaling with assumed take-off speed; no unique human jump speed is supplied. A spherical nonrotating asteroid and negligible atmosphere are assumed.

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