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

[Luke Cohen]

Homework Statement

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!

Homework Equations

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

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?

 

[andrewkirk]

F~ = −kr(rˆ)

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.

dr/dt = 0

THis contradicts the statement that the orbit is non-circular.

 

[Luke Cohen]

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

 

[andrewkirk]

I tried to do the above and solve the r^4 + ...r^2 + ... = 0 quadratic, but I can't.

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.

 

[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.

 

[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?

 

[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

 

[andrewkirk]

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

 

[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?

 

[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.

 

[Luke Cohen]

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

 

[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.

 

[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.

 

[andrewkirk]

then square my answer to get r

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.

 

[Luke Cohen]

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