Repulsive central force

1. Oct 22, 2009

gcfve

1. The problem statement, all variables and given/known data

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=.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.

2. Oct 22, 2009

Andrew Mason

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

AM

3. Oct 22, 2009

gcfve

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

4. Oct 23, 2009

Andrew Mason

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

Last edited: Oct 23, 2009
5. Oct 23, 2009

gcfve

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

6. Oct 23, 2009

Andrew Mason

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook