Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 4 / 9B / a / i

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- Solution i

Solution

The central force has no transverse component. Its angular equation is therefore

r \ddot{θ} + 2 \overset{r}{˙} \dot{θ} = 0 ⟹ \frac{d}{d t} (r^{2} \dot{θ}) = 0.

Thus the specific angular momentum is the constant

l = r^{2} \dot{θ} .

The radial equation is

\overset{r}{¨} - r \dot{θ}^{2} = - V^{'} (r) .

Substitute

\dot{θ} = l / r^{2}

:

\overset{r}{¨} = \frac{l ^{2}}{r ^{3}} - V^{'} (r) = - \frac{d}{d r} [V (r) + \frac{l ^{2}}{2 r ^{2}}] .

Consequently the radial motion for

r > 0

is exactly a unit-mass one-dimensional motion in the effective potential

V_{eff} (r) = V (r) + \frac{l ^{2}}{2 r ^{2}} .

Its conserved energy is

\frac{1}{2} \overset{r}{˙}^{2} + V_{eff} (r)

; the second term in the effective potential accounts for the angular kinetic energy.

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