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 Incoming links: Constant-area capture of stationary dust

Past exam of the physics 1 paper of the University of Oxford / 2013 / 11 / e / Solution 2026-10-25

For constant-area capture of stationary dust, let

x = 0

at cloud entry,

M (0) = M_{p}

,

v (0) = v_{\infty}

k = ρ A

with fixed collecting area

A

. Every swept volume

A d x

adds stationary mass, so

\frac{d M}{d t} = k v, M = M_{p} + k x .

There is no external longitudinal force, and the incoming dust has zero momentum in the cloud frame. The variable-mass system therefore satisfies

d (M v) / d t = 0

, rather than

M \overset{v}{˙} = 0

. Using

M v = M_{p} v_{\infty}

gives

M \dot{M} = k M_{p} v_{\infty}, M^{2} = M_{p}^{2} + 2 k M_{p} v_{\infty} t .

Hence

M (t) = M_{p} 1 + \frac{2 ρ A v _{\infty} t}{M _{p}}, v (t) = \frac{v _{\infty}}{1 + 2 ρ A v _{\infty} t / M _{p}},

x (t) = \frac{M _{p}}{ρ A} [1 + \frac{2 ρ A v _{\infty} t}{M _{p}} - 1] .

The kinetic energy decreases during sticking:

d K / d t = - \frac{1}{2} \dot{M} v^{2}

. The lost energy becomes internal energy. Radiation, gravity, sputtering and changes in area are excluded by the ideal complete-collection model. The zero-density limit recovers

M = M_{p}

,

v = v_{\infty}

and

x = v_{\infty} t

.

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