 Solution

The differential gravitational acceleration responsible for a tidal force is the difference between accelerations at separated points of a body. Relative to a freely falling reference point $\mathbf{R}$, a small offset $\mathbf{\xi }$ experiencesThe tidal tensor has units ${\mathrm{s}}^{-2}$; the vector called $\mathbf{T}(\mathbf{\xi })$ in the source is the resulting differential acceleration, in $\mathrm{m}{\mathrm{s}}^{-2}$. An actual differential force is this vector multiplied by the test mass. Measurable effects include relative motion of separated freely falling masses, radial stretching, transverse squeezing, deformation and, when self-gravity or strength is insufficient, disruption. A spatially uniform acceleration alone produces no tidal deformation.