Equations of Motion in Polar Basis

Question

A particle of mass m moves under a central force field

$ \mathbf{F}=-k\mathbf{r}$

where k is a constant with dimensions $ N m^{-1} $. Assuming that the particle moves in the equatorial plane ( $\theta=\frac{\pi}{2} $) write the equations of motion in the polar basis:

$ \mathbf{e}_r = (\cos \phi, \sin\phi,0) $

$ \mathbf{e}_{\phi} = (-\sin\phi, \cos\phi,0) $

[Hint: first use the variation of the basis vectors {$ \mathbf{\dot e}_r, \mathbf{\dot e}_{\phi} $} for an arbitrary function of t, then use these results to find $ \mathbf{\ddot r}. $]

Hence show that $ h=r^2 \dot\phi $ is a conserved quantity of the motion, and write down an equation of motion for $ r = |\bf r| $. Your equation should not contain $ \dot\phi $.

Show that circular orbits ( $\dot r =\ddot r=0 $) are allowed within this potential, and find their angular frequency, $\dot\phi =\omega$. Comment on your result.

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I am very lost with this question, and I am having a lot of difficulty wondering where to begin, all help is very much appreciated.

Answer 1

We have $\vec r(t)=\hat e_r(t)r(t)$. Therefore, the first velocity, $\vec v(t)\equiv \dfrac{d\vec r(t)}{dt}$ is given by $$\vec v(t)=\hat e_r(t)r'(t)+\hat e_{\phi}(t)r(t)\phi'(t)$$while the acceleration $\vec a(t)\equiv \dfrac{d^2\vec r(t)}{dt^2}$ is given by $$\vec a(t)=\hat e_r(t)\large(r''(t)-r\phi'^2(t)\large)+2\hat e_{\phi}(t)\large(2r'(t)\phi'(t)+r(t)\phi''(t)\large)\tag 1$$

Using $(1)$, we have the equation of motion of a particle of mass $m$ under a potential $U=\frac12 kr^2(t)$ is governed by the equations

$$mr''(t)-r\phi'^2(t)=-kr(t) \tag 2$$

and

$$m\large(2r'(t)\phi'(t)+r(t)\phi''(t)\large)=0 \tag 3$$

Solving $(3)$ for $\phi'(t)$ reveals that $\phi'(t)=h/r^2(t)$ for some integration constant $h$. Inasmuch as $\phi'(t)r^2(t)$ is a constant, it is a conserved quantity as was to be shown.

Now, using this result in $(2)$ exposes the ODE for $r(t)$ as

$$\bbox[5px,border:2px solid #C0A000]{r''(t)+\dfrac{k}{m}r(t)-\dfrac{h^2}{r^3}=0}$$

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SPECIAL CASE:

For the special case for which $h=0$, we have $$\phi'(t)=0\implies \phi(t)=\phi(0)$$and $$r''(t)+\dfrac{k}{m}r(t)=0$$ implying that $$\bbox[5px,border:2px solid #C0A000]{r(t)=r'(0)\sqrt{\dfrac{m}{k}}\sin \left(\sqrt{\dfrac{k}{m}}\,t\right)+r(0)\cos \left(\sqrt{\dfrac{k}{m}}\,t\right)}$$

Answer 2

Something to get you interested, expressing acceleration in arbitrary co-ordinates (component equations where dot denotes absolute derivative):

$$ \dot{v}^i = \frac {\partial v^i}{\partial t} + \Gamma^i_{jk}v^jv^k $$

Einstein summation convention is implied, and $\Gamma^i_{jk}$ are the christoffel symbols for the co-ordiante system used. I find it useful to use this aproach to deriving the equations of motion when dealing with wierd and wonderful co-ordinate systems.