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 Incoming links: Axial gravitational field of a uniform ring

Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / A6 / Solution 2026-10-20

The total ring mass is

M = 2 π r λ_{m} = 1.257 \times 1 0^{7} kg

. For the axial gravitational field of a uniform ring, transverse force components cancel by symmetry; each axial component has projection factor

z / r^{2} + z^{2}

. Thus

F = - \frac{GM m z}{( r ^{2} + z ^{2} ) ^{3/2}} \hat{z}, ∣ F ∣ = 3.04 \times 1 0^{- 10} N .

The force points towards the ring's centre along its axis.

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