Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 4 / 11A (Dynamics)

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- Solution 11A

Solution

Write

h = r^{2} \dot{θ}

for the conserved specific angular momentum. Zero specific orbital energy in the inverse-square Newtonian gravitational field gives

\frac{1}{2} (\overset{r}{˙}^{2} + \frac{h ^{2}}{r ^{2}}) - \frac{GM}{r} = 0.

r_{0} = h^{2} / (2 GM)

, the periapsis radius for a nonradial parabolic trajectory. Then

\overset{r}{˙}^{2} = \frac{2 GM ( r - r _{0} )}{r ^{2}} .

On the outgoing branch, integrate from the periapsis time

t_{0}

and put

s = r - r_{0}

:

t - t_{0} = \frac{1}{2 GM} \int_{r_{0}}^{r} \frac{r ^{'} d r ^{'}}{r ^{'} - r _{0}} = \frac{2}{2 GM} (\frac{s ^{3}}{3} + r_{0} s) = \frac{2}{3 GM} (r + 2 r_{0}) r - r_{0} .

The incoming branch has the opposite sign. Squaring gives the radius-time relation for a parabolic Kepler orbit, a version of the Barker equation:

(r - r_{0}) (r + 2 r_{0})^{2} = \frac{9}{2} GM (t - t_{0})^{2} .

Here

r_{0}

and

t_{0}

are fixed by angular momentum and the time origin at periapsis. The relation applies on both branches, with

r \geq r_{0}

; it is not necessary to solve for the polar angle.

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