# Homework Help: Finding the Central Force Given the Orbit

1. Apr 9, 2010

### phil ess

1. The problem statement, all variables and given/known data

2. Relevant equations

on image

3. The attempt at a solution

The first thing I did was use the second equation to get

Theta dot = L/mr2

Which I then subbed into the first equation to eliminate Theta dot.

Now I need everything in terms of r, and this is where I get stuck.

If I just find the second derivative of r (r double dot), then sub it back in, Im left with all kinds of Thetas, but I need f as a function of r.

How do I proceed here? Any help is greatly appreciated!! Thanks!

2. Apr 9, 2010

### ehild

Derive the second equation: from that you get the second derivative of theta in terms of r and the first derivatives of both r and theta.
The first derivative of theta is just L/(mr^2).
And you have the equation of the orbit to find theta in terms of r.

ehild

3. Apr 9, 2010

### phil ess

Thanks for the reply! But Im not sure what youre getting at.

When you say derive the second equation, I assume you mean differentiate it?

In that case, is this correct?

$$\dot{\theta} = L / mr\stackrel{2}{}$$

$$\ddot{\theta} = -2L\dot{r} / mr\stackrel{3}{}$$

Im a little unsure as to how to do the derivative..

In any case, I dont see how this helps me, since the equation for the force doesnt contain any Theta double dot terms.

What I then tried to do was just find r double dot frm the equation for r, and using the equation for r, cancel out the thetas that appear.

Then i get:

$$\ddot{r} = \frac{-A}{r\stackrel{2}{}}\left(1-\frac{r\stackrel{2}{}}{A\stackrel{4}{}}\right)\stackrel{-1/2}{}$$

Which is a nice enough result I guess, and now I have r double dot and theta dot to use in equation 1, but I get a very wrong answer upon subbing them in..

If you can offer any more guidance I'd be very grateful!

4. Apr 10, 2010

### ehild

I can not check your work if you do not show it in detail. Do you know what result you should get?

This is quite a weird orbit as it has sense only between -pi/4 and pi/4, otherwise r^2 would be negative.

Anyway, the derivation for a central force
is a bit easier if you use the derivatives of r with respect to theta, (r', r''):

$$\dot r = r' \dot \theta$$

and

$$\ddot r = r'' \dot \theta^2+r' \ddot \theta$$

$$\ddot{\theta} =\frac {-2L\dot r} {mr^3}$$

this is equivalent to

$$\ddot{\theta} =\frac {-2L r' \dot \theta} {mr^3}=-\frac{2 r'}{r} \dot \theta^2$$

therefore

$$\ddot r = r'' \dot \theta^2- \frac{2 r'^2}{r} \dot \theta^2\rightarrow \ddot r =(r''-\frac{2r'^2}{r})\dot \theta^2$$

and

$$f/m= (r''-\frac{2r'^2}{r}-r)\dot \theta^2$$

If you substitute r=1/y, you get an even simpler formula for y:

$$f= -(y''+y)r^2 m \dot \theta^2=-(y''+y) L^2/(mr^2)$$

ehild

Last edited: Apr 10, 2010
5. Apr 10, 2010

### phil ess

Thanks so much for the help... this is really frustrating. Maybe I'm just making arithmetic errors, bu I just cant get the right answer. Using your method:

$$r=A cos^{1/2}(2\theta)$$

$$r'=A cos^{-1/2}(2\theta)sin(2\theta)$$

$$r'^{2}=A^{2}cos^{-1}(2\theta)sin^{2}(2\theta) = A^{2}cos^{-1}(2\theta)(1-cos^{2}(2\theta)) = \frac{A^{4}}{r^{3}} - r$$

after substituting back in from the first line to get rid of theta

$$r" = \frac{1}{2}A cos^{-3/2}(2\theta)(-2sin(2\theta))sin(2\theta) - A cos^{1/2}(2\theta)(2 cos(2\theta))$$

Then after a few lines of simplifying, just like for r'

$$r" = \frac{-A^{4}}{r^{4}}+1-2r$$

Subbing into your equation for r double dot:

$$\ddot{r}=\frac{-LA^{4}}{mr^{6}}(1+r)$$

Which gives

$$f(r) = \frac{-LA^{4}(1+r)-L^{2}r^{3}}{mr^{6}}$$

which again is wrong. the correct answer should be

$$f(r) = \frac{-3A^{4}L^{2}}{mr^{7}}$$

6. Apr 10, 2010

### D H

Staff Emeritus
You are.

A quick test of the units shows this cannot be correct. r' is the derivative of r (units=length) with respect to angle (unitless). r' therefore has units of length, so r'2 has units of length2. Your result has units of length, so it cannot be correct. This invalid result makes everything else you did invalid.

Try differentiating $r^2 = A^2\cos2\theta$ to give you $rr'$. What does squaring that result yield?

7. Apr 10, 2010

### ehild

I have found some errors at the beginning. The following is correct.

$$r'=-A cos^{-1/2}(2\theta)sin(2\theta)$$

$$r'^{2}=A^{2}cos^{-1}(2\theta)sin^{2}(2\theta) = A^{2}cos^{-1}(2\theta)(1-cos^{2}(2\theta)) = \frac{A^{4}}{r^{2}} - r^2$$

ehild