Minimum distance from a repulsive central force

AI Thread Summary
A particle of mass m is influenced by a repulsive central force described by Fr=Cr^-3, where C is a positive constant. The problem involves determining the minimum distance, rmin, from the center of the force using conservation of energy and angular momentum principles. The total energy at a large distance is K, and the potential energy is suggested to be V(r) = -3C/r^2. The discussion reveals confusion regarding the correct form of V(r) and the application of angular momentum in the calculations. Ultimately, integrating the force from rmin to infinity is proposed as a method to find the solution, though it diverges from using angular momentum directly.
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Homework Statement


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}


Homework Equations


L=mbv0
E=\frac{1}{2}mvr2+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 rmin, E==\frac{1}{2}mvrmin2+(mbv)2/2mr2-3C/rmin2

I don't know what to do next.
 
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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?
 
Yeah, I still don't know how to get it.
 
oops wrong topic
 
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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.
 
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