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 Incoming links: Hollow-sphere gravitational transit

Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Section A / 5 / Solution 2026-10-18

The tunnel is taken thin enough that removal of its material does not significantly disturb spherical symmetry. For

a \leq r \leq 2 a

, the enclosed mass is

M (r) = \frac{4 π ρ}{3} (r^{3} - a^{3}) .

By the spherical shell theorem, the inward gravitational acceleration has magnitude

g (r) = \frac{GM ( r )}{r ^{2}} = \frac{4 π Gρ}{3} (r - \frac{a ^{3}}{r ^{2}}) .

The released gravitational potential energy is the positive work done by gravity while the mass falls from radius

2 a

to

a

:

U_{A} - U_{B} = m \int_{a}^{2 a} g (r) d r = \frac{4 π Gρ m}{3} [\frac{r ^{2}}{2} + \frac{a ^{3}}{r}]_{a}^{2 a} = \frac{4 π Gρ m a ^{2}}{3} .

By conservation of mechanical energy,

\frac{1}{2} m v_{B}^{2} = U_{A} - U_{B}, v_{B} = a \frac{8 π Gρ}{3} .

Inside the cavity, the spherical shell theorem gives zero gravitational field, so the velocity stays constant. The original figure places B and C at opposite ends of a cavity diameter, a distance

2 a

apart. Therefore

t_{BC} = \frac{2 a}{v _{B}} = \frac{3}{2 π Gρ} .

Gρ

has units

s^{- 2}

, so this is a time. There are no numerical values for

a

or

ρ

in the paper. The transit time is independent of

a

and the falling mass

m

. The cavity's gravitational potential is constant, not necessarily zero; only its gradient vanishes. The calculation is a hollow-sphere gravitational transit.

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