Solution

Take the sphere to move in the positive

z

-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form

A cos θ / r^{2}

. The no-penetration condition in the laboratory frame is

\frac{\partial ϕ}{\partial r}_{r = a} = U cos θ .

It fixes

A = - U a^{3} /2

ϕ (r, θ) = - \frac{U a ^{3}}{2 r ^{2}} cos θ .

The laboratory-frame velocity components are

u_{r} = \frac{U a ^{3}}{r ^{3}} cos θ, u_{θ} = \frac{U a ^{3}}{2 r ^{3}} sin θ, u_{φ} = 0 .

K_{f} = \frac{ρ}{2} \int_{a}^{\infty} \int_{S^{2}} \frac{U ^{2} a ^{6}}{r ^{6}} (cos^{2} θ + \frac{1}{4} sin^{2} θ) r^{2} d Ω d r = \frac{π}{3} ρ a^{3} U^{2} = \frac{1}{2} (\frac{1}{2} ρ V) U^{2},

where

V = 4 π a^{3} /3

ρ V /2

When the sphere falls, replacing fluid of density

ρ

by material of density

ρ_{s}

lowers the gravitational potential energy at rate

\dot{P} = - (ρ_{s} - ρ) V gU .

The total kinetic energy of sphere and fluid is

K = \frac{1}{2} (ρ_{s} + \frac{ρ}{2}) V U^{2} .

Conservation of total energy gives

(ρ_{s} + \frac{ρ}{2}) V U \frac{d U}{d t} - (ρ_{s} - ρ) V gU = 0.

For

U \neq = 0

\frac{d U}{d t} = \frac{ρ _{s} - ρ}{ρ _{s} + ρ /2} g .

The same formula holds at release by continuity.

Solved by gpt-5.6-sol high.

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