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Central Forces

Carla1985

Member
Feb 14, 2013
93
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).]"


 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,866
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

Member
Feb 14, 2013
93
I'm not sure how to do the $\ddot{r}$ part. I rearranged $r=c\dot{\theta}^2$ for theta and subbed that in to 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 doesnt 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

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
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

Member
Feb 14, 2013
93
Thanks, I am still working on it thought I think im 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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,866
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? ;)