Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 320 / 2 / b

Solution

For a bound orbit let

E < 0

L

the specific angular momentum. Introduce the dimensionless radius suggested in the question,

x = - \frac{μ}{b Φ ( r )} = \frac{b + b ^{2} + r ^{2}}{b}, r^{2} = b^{2} x (x - 2) .

The radial energy equation becomes

\overset{x}{˙}^{2} = \frac{- 2 E ( x - x _{-} ) ( x _{+} - x )}{b ^{2} ( x - 1 ) ^{2}},

where

x_{-}

and

x_{+}

are the two turning points. Comparing coefficients gives

x_{-} + x_{+} = 2 - \frac{μ}{b E} .

\frac{T _{r}}{2} = \frac{b}{- 2 E} \int_{x_{-}}^{x_{+}} \frac{x - 1}{( x - x _{-} ) ( x _{+} - x )} d x = \frac{πb}{- 2 E} (\frac{x _{-} + x _{+}}{2} - 1) = \frac{π μ}{( - 2 E ) ^{3/2}} .

Thus the radial orbital period is

T_{r} = \frac{2 π GM}{( - 2 E ) ^{3/2}} .

It depends on

E

but not on

L

, which is the defining isochrone property.

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