# 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 [closed]

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

## closed as off-topic by G. Smith, John Rennie, eranreches, Jon Custer, YashasMay 21 at 16:58

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• Readability leaves room for improvement. Please use the latex functionality to display your formulae. – my2cts May 20 at 17:32
• This is unreadable. Please render in LaTex. – Gert May 20 at 18:25
• Please enclose math expressions around dollar signs. For example $L^2$ renders like $L^2$. If you want to separate the equation from the text use double dollar signs, as in $$x = \frac{1}{2 \pi}$$ which is $$x = \frac{1}{2 \pi}$$ Finally exponents use ^ and subscripts _. See physics.stackexchange.com/help/notation – ja72 May 20 at 19:45
• what do you mean by frequency of the radial oscillation? – lakehal May 20 at 23:59
• Do you mean a radial or tangential impulse? Or a mixed one? – acarturk May 21 at 1:55

A few assumptions to begin with:

1. The impulse was given on the radial direction, i.e. $$L_{\text{prev}} = L_{\text{after}}$$.
2. 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.