Finding the Central Force Given the Orbit

[phil ess]

Homework Statement

Homework Equations

on image

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, I am 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!

 

[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

 

[phil ess]

Thanks for the reply! But I am 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 don't see how this helps me, since the equation for the force doesn't 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!

 

[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

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

As you derived already

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

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:

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

Not long ago, there was a thread with similar topics: https://www.physicsforums.com/showthread.php?t=392282

ehild

 

[phil ess]

Thanks so much for the help... this is really frustrating. Maybe I'm just making arithmetic errors, bu I just can't 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

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

 

[D H]

Maybe I'm just making arithmetic errors

You are.

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

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?

 

[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