Minimum distance from a repulsive central force

[gcfve]

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=$$\sqrt{b^2+ C/2K}$$

Homework Equations

L=mbv₀
E=$$\frac{1}{2}$$mv_r²+L²/2mr²+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/r², but I'm not sure.
And at r_min, E==$$\frac{1}{2}$$mv_rmin²+(mbv)²/2mr²-3C/r_min²

I don't know what to do next.

 

[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?

 

[gcfve]

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

 

[gcfve]

oops wrong topic

 

[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.