 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1/section-c/c11/a/solution

g (R) = - GM R / R^{3}

. Along the outward radial direction,

g_{R} = - GM / R^{2}

and

\frac{d g _{R}}{d R} = \frac{2 GM}{R ^{3}} .

Thus a small radial displacement

a = a R

has differential acceleration

T (a) ≃ \frac{2 GM}{R ^{3}} 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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