Dynamics | Part IA, 2002

Derive the equation

d2udθ2+u=f(u)mh2u2\frac{d^{2} u}{d \theta^{2}}+u=\frac{f(u)}{m h^{2} u^{2}}

for the orbit r1=u(θ)r^{-1}=u(\theta) of a particle of mass mm and angular momentum hmh m moving under a central force f(u)f(u) directed towards a fixed point OO. Give an interpretation of hh in terms of the area swept out by a radius vector.

If the orbits are found to be circles passing through OO, then deduce that the force varies inversely as the fifth power of the distance, f=cu5f=c u^{5}, where cc is a constant. Is the force attractive or repulsive?

Show that, for fixed mass, the radius RR of the circle varies inversely as the angular momentum of the particle, and hence that the time taken to traverse a complete circle is proportional to R3R^{3}.

[You may assume, if you wish, the expressions for radial and transverse acceleration in the forms r¨rθ˙2,2r˙θ˙+rθ¨\ddot{r}-r \dot{\theta}^{2}, 2 \dot{r} \dot{\theta}+r \ddot{\theta}.]

Typos? Please submit corrections to this page on GitHub.