Solving for Central Force in $r = c\theta^2$ Orbit

[Carla1985]

Could someone please explain how I work with $-mf(r)e_r$ in this question. Usually we get given an equation (like the one for f(r)) and have to work out the orbit by getting a differential equation etc. I'm not too sure how to work it this way around.

"A particle of mass m moves under the influence of a central force $\textbf{F}(\textbf{r}) =−mf(r)e_r$, in the orbit

$r = c\theta^2$, (1)

​

where c > 0 and (r, θ) and er , eθ are the polar co-ordinates and corresponding basis vectors in the plane of motion of the particle. Show that:
\[
f(r)=-h^2(\frac{6c}{r^4}+\frac{1}{r^3})
\]
where $r^2\dot{\theta}=h$ is constant

​

[Hint: Use the substitution $u(\theta)=\frac{1}{r(\theta}$ to write the radial equation $\ddot{r}-r\dot{\theta}^2=-f(r)$ in terms of u(θ), and then determine f using this equation and (1).]"

 

[I like Serena]

Could someone please explain how I work with $-mf(r)e_r$ in this question. Usually we get given an equation (like the one for f(r)) and have to work out the orbit by getting a differential equation etc. I'm not too sure how to work it this way around.

"A particle of mass m moves under the influence of a central force $\textbf{F}(\textbf{r}) =−mf(r)e_r$, in the orbit

$r = c\theta^2$, (1)

​

where c > 0 and (r, θ) and er , eθ are the polar co-ordinates and corresponding basis vectors in the plane of motion of the particle. Show that:
\[
f(r)=-h^2(\frac{6c}{r^4}+\frac{1}{r^3})
\]
where $r^2\dot{\theta}=h$ is constant

​

[Hint: Use the substitution $u(\theta)=\frac{1}{r(\theta}$ to write the radial equation $\ddot{r}-r\dot{\theta}^2=-f(r)$ in terms of u(θ), and then determine f using this equation and (1).]"

Hi Carla!

Your equations are:

\(\displaystyle \ddot r - r \dot \theta^2=-f(r) \qquad\) central acceleration
\(\displaystyle r^2\dot \theta = h \qquad\qquad\qquad\) preservation of angular momentum in a conservative central field
\(\displaystyle r=c\theta^2 \qquad\qquad\qquad\) the given orbit

Can you solve $f(r)$ from these, expressing it using only $r$?

 

[Carla1985]

I'm not sure how to do the $\ddot{r}$ part. I rearranged $r=c\dot{\theta}^2$ for theta and subbed that into get
$f(r)=-(\ddot{r}-\frac{h^2}{r^3})$

I differentiated r and got $\dot{r}=2c\theta\dot{\theta}$
and then subbed in theta again: $\frac{2ch\theta}{r^2}$ and then did the same thing again to get $\ddot{r}=\frac{2ch}{r^2}\theta\dot{\theta}=\frac{2ch^2}{r^2}\theta$ which doesn't add up at all.

I think the hint is confusing me too, it says to do a substitution u=1/r which we always do our other questions but I'm not sure where that applies here.

 

[topsquark]

The substitution simply makes some of the problem easier to work. But it is not necessary. See if you can finish it the way I like Serena is helping you with. It will help if you if you then want to go back and use the hint.

-Dan

 

[Carla1985]

Thanks, I am still working on it thought I think I am going to take a break and get some sleep. its past midnight and the kids will have me up at 7am lol. I am so close its frustrating though.
I have $\ddot{r}=\frac{2ch^2\theta}{r^4}$ instead of $\frac{6ch^2}{r^4}$ and I cannot for the life of me see how I get rid of theta and get a 3 instead.

Maybe a fresh look in the morning will tell me where I've gone wrong though hehe.

 

[I like Serena]

Maybe a fresh look in the morning will tell me where I've gone wrong though hehe.

Well... the morning has passed.
Any new insights? ;)