# Find the min/max r of particle with a central force

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1. Feb 18, 2016

### Luke Cohen

1. The problem statement, all variables and given/known data
A force field in 2-d F~ = −kr(rˆ) with U(r) = k(r^2)/2 acts on a particle of mass m.
The particle is now in a non-circular orbit. In terms of the particle’s angular momentum L and energy E,
d) What is its closest approach to the origin? e) What is its furthest distance from the origin?
e) What is its furthest distance from the origin?

I am confused on how to start this problem. I can't switch my variable so that I get an equation for phi.... Can someone help me out starting this off?

Thanks!!!

2. Relevant equations
E = T + U
T = L^2/(2r^2) b/c dr/dt = 0
U = Kr^2/2

3. The attempt at a solution
I tried to do the above and solve the r^4 + ....r^2 + .... = 0 quadratic, but I can't. Can someone please help me?

2. Feb 18, 2016

### andrewkirk

What does this mean? It looks like you're trying to write the formula for the force, but this collection of symbols makes no sense.
THis contradicts the statement that the orbit is non-circular.

3. Feb 18, 2016

### Luke Cohen

For the first part, it's an r hat. And ****, you're right about the second part. I am going to change that. How should I go from there?

4. Feb 18, 2016

### andrewkirk

Do you mean quartic?
Write out in full the equation that needs to be solved. If you've derived the equation correctly, it should be easy to see how to solve it once it's written down.

5. Feb 18, 2016

### Luke Cohen

I don't think I did, as I was going under the assumption that dr/dt = 0, which I can see now only applies for circular motion. Is there a particular direction you think I should tackle this problem from? It says I need to answer in terms of E and L, and I am really stuck on how that is possible.

6. Feb 18, 2016

### andrewkirk

It is not true in general that dr/dt=0, but there may be points in the orbit where that condition momentarily holds. What points would they be and what energy equation can you write that applies at such a point?

7. Feb 18, 2016

### Luke Cohen

ah, okay so my quartic equation was correct.

I said E = L^2/(2mr^2) + kr^2/2 giving me a quartic equation of r^4 - 2Er^2/k + L^2/(mk) = 0

8. Feb 18, 2016

### andrewkirk

Can you think of a way of converting it into a quadratic?

9. Feb 18, 2016

### Luke Cohen

I had been working on that for a while and couldn't figure anything out. Can you point me in the right direciton?

10. Feb 18, 2016

### andrewkirk

Try substitution - in the same meaning of 'substitution' as applies in integration by substitution. We are not integrating here. But the same trick may work.

11. Feb 18, 2016

### Luke Cohen

substituting what though? the only common term is 1/k, but that wouldn't help

12. Feb 18, 2016

### andrewkirk

In integration by substitution you substitute a new variable, call it u, for some function of the variable you are trying to integrate over.
So here, try substituting a new variable for some function of the variable you are trying to solve for.

13. Feb 18, 2016

### Luke Cohen

Okay, so I get u^2 - 2Eu/k + L^2/mk = 0 using the substitution u = r^2.
After that, I use the quadratic formula, then square my answer to get r. That would give me my Rmin and Rmax? Is the physics I performed correct in the above equation? Thanks for the help thus far.

14. Feb 18, 2016

### andrewkirk

Square root, not square.
The physics is fine. It's just the conservation of energy and angular momentum.
A physical system that would act like this would be a heavy ball attached to a fixed metal rod by a spring that can rotate around the rod.

15. Feb 18, 2016

### Luke Cohen

Awesome. you are a huge help. Thanks so much!