= Solution
Use the largest Newtonian tidal-tensor <eigenvalue>, the radial <gravity> gradient $2GM/r^3$. Imposing the supplied gradient threshold gives
$$
\boxed{r_{\min}\simeq\left(\frac{2GM}{100\ \mathrm{s^{-2}}}\right)^{1/3}
=1.59\times10^6\ \mathrm m}.
$$
This is \b[about $1590\ \mathrm{km}$ from the centre], or $1580\ \mathrm{km}$ above the stated surface. The tangential <eigenvalues> have half the radial magnitude. Since $GM/(r_{\min}c^2)\simeq1.4\times10^{-3}$, the Newtonian tidal estimate is adequate; this calculation uses the given threshold without inferring any additional material-response model.
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