Question 5.5: Investigate the stability of the circular orbit of a celesti......

According to Eq. (5.66),

E = T + V = \frac{m}{2}\dot{r}^{2}+\frac{m}{2}r^{2}\dot{\theta}^{2}+V\left(r\right)= \frac{m}{2}\dot{r}^{2}+ \frac{p^{2}_{\theta }}{2mr^{2}}+V\left(r\right), (5.66)

V_{eff}\left(r\right)=\frac{p^{2}_{\theta }}{2mr^{2}}-\frac{A}{r}  (5.69)

with A = GmM > 0 where M is the mass of the Sun and m is the mass of the celestial body in circular orbit. The radius of the orbit is found from

0= \frac{dV_{eff}}{dr}=-\frac{p^{2}_{\theta }}{mr^{3}}-\frac{A}{r^{2}}\Longrightarrow  r=r_{0}=\frac{p^{2}_{\theta }}{mA},  (5.70)

a result already obtained in a different way in Example 5.3. On the other hand,

k^{\left(0\right)}=\left(\frac{d^{2}V_{eff}}{dr^{2}}\right)_{ r=r_{0}}= \frac{3p^{2}_{\theta }}{mr^{4}_{0}}- \frac{2A}{r^{3}_{0}}=\frac{A}{r^{3}_{0}}> 0 , (5.71)

and the circular orbits are stable under small perturbations,^{5} with

ω = \left(\frac{k^{\left(0\right)}}{\alpha ^{\left(0\right)}}\right)^{{1}/{2}}=\left(\frac{GM}{r^{3}_{0}}\right)^{{1}/{2}} (5.72)

being the frequency of the small oscillations around the circular orbit.

^{5} The present analysis does not apply to perturbations perpendicular to the plane of the orbit.