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 Incoming links: Virial mean escape speed

Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 44 / 1 / c / Solution Created 2026-10-03 Updated 2026-10-14

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Take the gravitational potential to vanish at infinity. At the position of star

i

, excluding its own point-mass field, the instantaneous test-particle escape velocity obeys

v_{esc, i}^{2} = - 2 Φ_{i}

. The factor of one half in the total gravitational potential energy is important:

W = \frac{1}{2} \sum_{i} m Φ_{i} .

(1)

Therefore the equal-mass average is

⟨ v_{esc}^{2} ⟩ = - \frac{2}{N} i \sum Φ_{i} = - \frac{4 W}{N m} = \frac{4 GN m}{R _{g}} = 4 ⟨ v_{eq}^{2} ⟩ .

(2)

Thus the rms escape speed is twice the equilibrium rms speed. This is the virial mean escape speed relation, using the same three-dimensional averaging convention as the preceding part. It concerns escape in the instantaneous system potential, rather than a subsequent collective rearrangement of the cluster.

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