What is the Solution for a Particle Moving Under a Repulsive Central Force?

[gcfve]

Homework Statement

1. 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}



2. Homework Equations
L=mbv0
E=.5mvr2+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==.5mvrmin2+(mbv)2/2mr2-3C/rmin2

I don't know what to do next.

 

[Andrew Mason]

Homework Statement

1. 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}
2. Homework Equations
L=mbv0
E=.5mvr2+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==.5mvrmin2+(mbv)2/2mr2-3C/rmin2

I don't know what to do next.

Your nomenclature is confusing. Do you mean: F(r) = Cr^{-3} ??

AM

 

[gcfve]

Yeah, that's what i meant
I didnt realize it was like that when i copied it

 

[Andrew Mason]

Homework Statement

1. 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}
2. Homework Equations
L=mbv0
E=.5mvr2+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==.5mvrmin2+(mbv)2/2mr2-3C/rmin2

I don't know what to do next.

Conservation of angular momentum:

\vec L = m \vec v \times \vec r

so:

(1) |L| = mv_0rsin\theta = mv_0r(b/r) = mv_0b

and:

(2) U(r) + \frac{1}{2}mv^2 = constant = K

So far, you have this correct. [I will use U for potential energy since it is confusing to use v for speed and potential.]

Since F = -dU/dr, -U is the antiderivative of F

F = Cr^{-3} so U = \frac{1}{2}Cr^{-2}Therefore, from (2)

(3) \frac{1}{2}Cr^{-2} + \frac{1}{2}mv^2 = K

Next, you have to find v at minimum r. At that point, what is L (hint: what the angle between v and r?)? You should be able to express v at minimum r in terms of v_0, r and b and then K, r and b. Substitute that into (3) and you should get your answer.AM

 

[gcfve]

ok so the angle between r and v at rmin should be 90?
so,
|L| = mv_0r = mv_0b
sp r_min=bv/v₀?

 

[Andrew Mason]

ok so the angle between r and v at rmin should be 90?
so,
|L| = mv_0r = mv_0b
sp r_min=bv/v₀?

The angle is 90. But your expression for L is not correct. Why are you using v_0 at minimum r? L \ne mv_0r. v_0 is the speed at the beginning when U = 0.

AM