= Solution
For a <point mass>, $\mathbf g(\mathbf R)=-GM\mathbf R/R^3$. Along the outward radial direction, $g_R=-GM/R^2$ and
$$
\frac{dg_R}{dR}=\frac{2GM}{R^3}.
$$
Thus a small radial displacement $\mathbf a=a\widehat{\mathbf R}$ has differential acceleration
$$
\boxed{\mathbf T(\mathbf a)\simeq\frac{2GM}{R^3}\mathbf a.}
$$
The farther point falls inward less strongly, so its acceleration relative to the reference point points outward. The formula is first order in $|a|/R$.
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