 Solution

ID: past-exam-of-the-relativity-course-of-the-university-of-cambridge/2014/paper-1/1/c/solution

Use the largest Newtonian tidal-tensor eigenvalue, the radial gravity gradient

2 GM / r^{3}

. Imposing the supplied gradient threshold gives

r_{m i n} ≃ (\frac{2 GM}{100 s ^{- 2}})^{1/3} = 1.59 \times 1 0^{6} m .

This is about $1590 km$ from the centre, or

1580 km

above the stated surface. The tangential eigenvalues have half the radial magnitude. Since

GM / (r_{m i n} c^{2}) ≃ 1.4 \times 1 0^{- 3}

, the Newtonian tidal estimate is adequate; this calculation uses the given threshold without inferring any additional material-response model.

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