Use a local Cartesian frame rotating at $\Omega=\sqrt{GM/r_0^3}$ about a circular orbit at $r_0$, with $x=r-r_0\ll r_0$, and neglect curvature beyond leading order. The radial gravitational-plus-centrifugal potential is
$$
\Phi_{\rm eff}(r)=-\frac{GM}{r}-\frac12\Omega^2r^2.
$$
Its first derivative vanishes at $r_0$ and its second derivative there is $-3\Omega^2$. Dropping a constant gives the <shearing-sheet tidal potential>
$$
\boxed{\Phi_t=-\frac32\Omega^2x^2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 3 / a / Solution

{title2=$\Phi_t$}

For a frame rotating about a Keplerian circular orbit, radial gravity and centrifugal acceleration combine to give $\Phi_t=-3\Omega^2x^2/2$.


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 Incoming links: Shearing-sheet tidal potential