= Solution
A <tidal force> measures the difference of <gravitational acceleration> across an extended object. A freely falling <reference frame> removes a spatially uniform <acceleration>, but not its <gradient>. For a distant point <mass> $M$ at distance $R$ and direction $\hat{\mathbf n}$ from the reference centre, a small internal <displacement> $\mathbf x$ gives
$$
\delta\mathbf g=\frac{GM}{R^3}\left[3\hat{\mathbf n}(\hat{\mathbf n}\cdot\mathbf x)-\mathbf x\right],\qquad |\mathbf x|\ll R.
$$
The <tidal tensor> has <eigenvalues> $2GM/R^3$ radially and $-GM/R^3$ in each transverse direction: radial stretching and transverse squeezing. Its trace vanishes in the source-free region, consistent with the gravitational Gauss law. An associated relative potential is $\Phi_{\rm tide}=-GM[3(\hat{\mathbf n}\cdot\mathbf x)^2-x^2]/(2R^3)$.
The scale is $GMl/R^3$ across size $l$, not the centre's absolute gravity $GM/R^2$. This distinction explains how a more distant massive object can exert a large <force> but a modest tide. Opposite-side stretching supports the elementary two-bulge picture; real ocean tides also involve rotation, basin geometry and <fluid> dynamics, so the static <gradient> alone does not predict coastal heights or timing. The same <gradient> stresses satellites and can disrupt a self-gravitating body when it competes with the body's own gravitational restoring <acceleration>.
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