Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 1 / a

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- Solution a

Solution

For the Paczyński-Wiita potential, circular balance gives

r Ω^{2} = \frac{d Φ}{d r} = \frac{GM}{( r - r _{S} ) ^{2}} .

Therefore

Ω (r) = \frac{GM}{r ^{1/2} ( r - r _{S} )}, h (r) = r^{2} Ω = \frac{GM r ^{3}}{r - r _{S}} .

The squared radial epicyclic frequency is

κ_{r}^{2} = \frac{1}{r ^{3}} \frac{d ( h ^{2} )}{d r} = \frac{GM ( r - 3 r _{S} )}{r ( r - r _{S} ) ^{3}} .

Circular orbits change from stable to unstable where

κ_{r}^{2}

vanishes, so

r_{ISCO} = 3 r_{S} .

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