# Homework Help: Minimum distance from a repulsive central force

1. Oct 21, 2009

### gcfve

1. The problem statement, all variables and given/known data
A particle of mass m moves under action of a repulsive central force Fr=Cr-3 with constant C greater than 0. At a very large distance from the centre of the force, the partcle has kinetic energy K and its impact parameter is b. Use conservation of energy and angular momentum to show that the closest m comes to the centre of force is rmin=$$\sqrt{b^2+ C/2K}$$

2. Relevant equations
L=mbv0
E=$$\frac{1}{2}$$mvr2+L2/2mr2+V(r)

3. The attempt at a solution
So far, I said that since the object is far away its total energy is K
I think V(r) = -3C/r2, but I'm not sure.
And at rmin, E==$$\frac{1}{2}$$mvrmin2+(mbv)2/2mr2-3C/rmin2

I don't know what to do next.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 22, 2009

### MiguelSanchez

PHY2333...Assignment is due tomorrow... Still can't get this either. Where do you get -3C/r^2 for V(r)? Would it not be -0.5C/r^2?

3. Oct 22, 2009

### gcfve

Yeah, I still don't know how to get it.

4. Oct 22, 2009

### gcfve

oops wrong topic

Last edited: Oct 22, 2009
5. Oct 22, 2009

### MiguelSanchez

Take Force integrate from rmin to b, as when r is infinity, it will approximately equal to be, because it is essentially uneffected by the force anyway. this integration will give you the answer. I know its not using angular momentum, but I don't care at this point.