Scattering problem: Expression for angular momentum of particle

Question

I'm reading Goldstein's Classical Mechanics, the part on "Scattering" in the "Central Force" chapter.

In relation to the figure below, he says that angular momentum, $l$, is given by $$l=mv_0s$$ where $v_0$ is the velocity of the particle and $s$ is the distance from the line of the center of force(as shown in figure).

However, given that $l=\overrightarrow r \times \overrightarrow p=r.mv_0. sin (\theta)$, and $l$ is measured about the center of the sphere shown, it seems as if he has concluded that perpendicular drawn from the center of the sphere to the point of closest approach of the particle is of length $s$, which generally need not be true. To put it simply, how did he get the above expression?

Answer 1

don't try to find anything complicated in it. You wrote down the expression for the angular momentum correctly: $l=r.mv_{0}.\sin\theta$. Now just express $r.\sin\theta$ using $s$: $r$ is the distance from the center of force to the particle. $\theta$ is the angle between the radius vector of the particle and the horizontal axis. The side of the triangle opposite the angle $\theta$ is $s$. This triangle gives you $\sin\theta=\frac{s}{r}$ and you're done.