- #1

gcfve

- 13

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## Homework Statement

A particle of mass m moves under action of a repulsive central force F

_{r}=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 r

_{min}=[tex]\sqrt{b^2+ C/2K}[/tex]

## Homework Equations

L=mbv

_{0}

E=[tex]\frac{1}{2}[/tex]mv

_{r}

^{2}+L

^{2}/2mr

^{2}+V(r)

## 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/r

^{2}, but I'm not sure.

And at r

_{min}, E==[tex]\frac{1}{2}[/tex]mv

_{rmin}

^{2}+(mbv)

^{2}/2mr

^{2}-3C/r

_{min}

^{2}

I don't know what to do next.