 Solution

ID: 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

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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