So I have this question:

There is a particle moving in response to a central force per unit mass of $$F(r) = {\alpha\over r^2} + {\beta\over r^3}$$ where $\alpha$ and $\beta$ are constants. Initially the particle is at $r = \beta^2/3\alpha$, $\theta = 0$ and is moving with speed $4\alpha/\beta$ in a direction making an angle of $\pi/3$ with the radius vector pointing towards the origin. Starting from Newton's Second Law, show that, if $u = 1/r$, then $${d^2u \over d\theta^2} + \frac u4 = {3\alpha\over 4\beta^2}$$, with $$u = {3\alpha\over \beta^2}, {du \over d\theta} = {\alpha \sqrt3 \over \beta^2}$$ when $\theta = 0$.

My attempt at solution:

In polar coordinates, $$\ddot {\mathbf r}= (\ddot r - r \dot \theta)\mathbf e_r+ \left( \frac 1r {d \over dt}(r^2 \ddot \theta)\right) \mathbf e_\theta$$ where $\mathbf e_r$ is the unit vector in the direction of increasing $r$ and respectively for $\mathbf e_\theta$.

Now, if we have a central force, then $\mathbf F=-F(r)*\mathbf e_r$ because the force acts in the direction of $-r$.

Therefore, $$-F(r)\mathbf e_r = m\left[(\ddot r - r \dot \theta)\mathbf e_r+ \left( \frac 1r {d \over dt}(r^2 \ddot \theta)\right) \mathbf e_\theta\right]$$ Equating the coefficients of $\mathbf e_r$ and $\mathbf e_\theta$ gives $$-F(r) = m\left(\ddot r-r \dot \theta^2\right)$$ $$r^2 \dot \theta = h$$ where $h$ is a constant.

Then $$\dot r = {dr \over dt} = {dr \over d\theta}{d\theta \over dt} = {h \over r^2}{dr \over d\theta} =-h{d \over d\theta}\left(\frac 1r\right)$$ If we let $u = \frac 1r$, then $$\text{radial velocity} \; \dot r = -h{d \over d\theta}\left(\frac 1r\right) = -h{du \over d\theta}$$ $$\text{transverse velocity} \; r \dot \theta = \frac hr = hu$$

So $\ddot r$ becomes $-h^2u^2{d^2u \over d \theta ^2}$. Thus, $$F\left(\frac 1u\right) = mh^2u^2{d^2u \over d\theta ^2}-\frac mu h^2u^4$$.

Now the problem is, I don't really know how to derive $h$ from the initial conditions.

  • 1
    $\begingroup$ A small note on (La)TeX: when you wish to end a displayed equation $$…$$ with punctuation, put the punctuation before the final $$. $\endgroup$ Nov 28 '14 at 19:33

Hint: $r\dot\theta$ is the component of velocity orthogonal to the radius vector. You are given the initial velocity, and you know that $r^2\dot\theta=h$, so now you need to decompose the given initial velocity into orthogonal components, one radial and one not.

  • $\begingroup$ So the initial velocity is ${4\alpha \over \beta}$ so would the radial velocity be ${4\alpha \over \beta}$? and then $r\theta = {2\alpha \sqrt3 \over \beta}$ and then $\dot \theta = 2\beta \sqrt3$? $\endgroup$ Nov 28 '14 at 20:10
  • $\begingroup$ I'd think the radial velocity is $-(4\alpha/\beta)\cos(\pi/3)=-2\alpha/\beta$ (minus sign since it is pointed inwards), and the transverse, $r\dot\theta=(4\alpha/\beta)\sin(\pi/3)=2\alpha\sqrt3/\beta$. So yes, I think you got it but just forgot the dot on $\theta$? $\endgroup$ Nov 28 '14 at 20:17
  • $\begingroup$ Oh right, yes. Oh, I made another calculation error. $\dot \theta = {6\alpha ^2 \sqrt3 \over \beta ^3}$ and instead $h = r^2\dot \theta = 2\beta \sqrt3$? $\endgroup$ Nov 28 '14 at 20:29
  • $\begingroup$ It's a bit easier to just write $h=r\cdot r\dot\theta$, since you already have the two factors. I'll leave the careful checking of such details to you. 8-) $\endgroup$ Nov 28 '14 at 20:40

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