If a body is moving under central force in a circle then how its path will be changed if it is subjected to a impulse A particle of mass m moves under an attractive central force of $Kr^4$ with an angular momentum L
Find the frequency of the radial oscillations if the particle is given a small radial impulse and initially it was moving in a circle
I think that after taking radial impulse it will change its path to ellipse from circle  because there is both radial and transverse velocity but after it I am getting nothing to solve 
 A: A few assumptions to begin with:


*

*The impulse was given on the radial direction, i.e. $L_{\text{prev}} = L_{\text{after}}$.

*The perturbatory effect of this impulse is very small compared to the initial radius: $\Delta r \ll r_0$.


Then we can move on to the calculations. Since the system has polar symmetry, it is convenient to use polar coordinates as well, with basis vectors $\hat{e}_r$ and $\hat{e}_\theta$.
Then we will need the time derivatives of these unit vectors so that we can calculate the acceleration and write the Newton's second law in polar coordinates:
$$\frac{d}{dt}\,\hat{e}_r = \dot{\theta}\,\hat{e}_\theta,\ \ \frac{d}{dt}\hat{e}_\theta = -\dot{\theta}\hat{e}_r$$
$$\vec{r} = r\,\hat{e}_r$$
$$\vec{v} = \frac{d}{dt} (\vec{r}) = \frac{d}{dt} (r\,\hat{e}_r) = \frac{dr}{dt}\hat{e}_r+r\frac{\hat{e}_r}{dt} = \dot{r}\hat{e}_r+r\dot{\theta}\hat{e}_\theta$$
$$\vec{a} = \frac{d}{dt} (\vec{v}) = \frac{d}{dt} (\dot{r}\hat{e}_r+r\dot{\theta}\hat{e}_\theta) = \ddot{r}\hat{e}_r+\dot{r}\frac{d}{dt}\,\hat{e}_r+\dot{r}\dot{\theta}\hat{e}_\theta+r\ddot{\theta}\hat{e}_\theta+r\dot{\theta}\frac{d}{dt}\hat{e}_\theta$$
$$\vec{a} = \ddot{r}\hat{e}_r+\dot{r}\dot{\theta}\hat{e}_\theta+\dot{r}\dot{\theta}\hat{e}_\theta+r\ddot{\theta}\hat{e}_\theta-r\dot{\theta}\dot{\theta}\hat{e}_r = (\ddot{r}-r\dot{\theta}^2)\,\hat{e}_r + (r\ddot{\theta}+2\dot{r}\dot{\theta})\,\hat{e}_\theta$$
$$\vec{F} = m\vec{a}$$
$$-Kr^4\hat{e}_r = \vec{F} = m\vec{a} = m [(\ddot{r}-r\dot{\theta}^2)\,\hat{e}_r + (r\ddot{\theta}+2\dot{r}\dot{\theta})\,\hat{e}_\theta]$$
$$r\ddot{\theta}+2\dot{r}\dot{\theta} = 0,\ \ \ \ddot{r}-r\dot{\theta}^2 = -\frac{K}{m}\,r^4$$
A quick note: The equation $r\ddot{\theta}+2\dot{r}\dot{\theta} = 0$ proves the conservation on angular momentum. You can attempt to factor the differentials to include the term $r^2\dot{\theta} = L/m$.
Since the angular momentum is conserved, let's define it as $L$:
$$L:=m\,r^2\dot{\theta}$$
$$\therefore\ \dot{\theta} = \frac{L}{mr^2}$$
Let's investigate the system at its equilibrium:
At equilibrium, we expect $r$ (the distance) and $\dot{\theta}$ (the angular velocity) to be stable, i.e. $\dot{r} = \ddot{\theta} = 0$:
$$\ddot{r}-r\dot{\theta}^2 = 0-r_0\dot{\theta_0}^2 = -\frac{K}{m}\,{r_0}^4$$
Inserting this into the radial equation,
$${r_0}^2\,\frac{L^2}{m^2{r_0}^4} = \frac{L^2}{m^2{r_0}^2} = \frac{K}{m}\,{r_0}^4$$
$$\therefore\ r_0 = \sqrt[6]{\frac{L^2}{m\,K}}$$
Now including the $\dot{r}$, $\ddot{\theta}$ and higher order terms, i.e. after the impulse is given,
$$\ddot{r} - \frac{L^2}{m^2r^2} = -\frac{K}{m}\,r^4$$
We had claimed that $\Delta{r} := r(t)-r_0 \ll r_0$, then,
$$\ddot{(r(t)+r_0)} - \frac{L^2}{m^2(r(t)+r_0)^2} = -\frac{K}{m}\,(r(t)+r_0)^4$$
Using $1^\text{st}$ order terms in this approximation (we know that terms without $r(t)$ already goes to zero),
$$\ddot{r(t)} - \frac{L^2}{m^2{r_0}^2}\,\left(1-\frac{2r(t)}{r_0}\right) = -\frac{K}{m}\,{r_0}^4\left(1+\frac{4r(t)}{r_0}\right)$$
$$\ddot{r(t)} + \frac{L^2}{m^2{r_0}^2}\,\frac{2r(t)}{r_0} = -\frac{K}{m}\,{r_0}^4\frac{4r(t)}{r_0}$$
$$\ddot{r(t)} + \frac{2L^2}{m^2{r_0}^3}\,r(t) + \frac{K}{m}\,{r_0}^3\,4r(t) = 0$$
$$\ddot{r(t)} + \left(\frac{2L^2}{m^2{r_0}^3} + \frac{K}{m}\,{r_0}^3\right)\,r(t) = 0$$
$$\omega_0 := \frac{2L^2}{m^2{r_0}^3} + \frac{K}{m}\,{r_0}^3$$
$$\ddot{r}+{\omega_0}^2 = 0$$
This is the requested basic harmonic oscillator behavior. You can find this type of behavior on any kind of system with stationary equilibrium states driven by very small perturbations.
