Discovering the Nucleus — Question 7

Question 7

Finite impact parameter raises the closest-approach distance

A projectile of mass m, asymptotic speed v and impact parameter b scatters from a fixed repulsive center with potential $K/r$. Define $E=mv^2/2$ and $a=K/(2E)$.

1. Use conservation of energy and angular momentum to derive $r_{\min}$.

2. For $a=10.0\,\mathrm{fm}$ and $b=24.0\,\mathrm{fm}$, calculate $r_{\min}$ and compare with the head-on value.

Question 7 - Solution

Finite impact parameter raises the closest-approach distance

(a) Method and calculation

$$L=mvb,\quad E=\frac{L^2}{2mr_{\min}^2}+\frac K{r_{\min}}\Rightarrow r_{\min}^2-2ar_{\min}-b^2=0.$$$$\boxed{r_{\min}=a+\sqrt{a^2+b^2}}.$$

(b) Interpretation and further calculation

$$\boxed{r_{\min}=10+\sqrt{100+576}=36.0\,\mathrm{fm}}.$$$$b=0:\quad\boxed{r_{\mathrm{head}}=2a=20.0\,\mathrm{fm},\quad r_{\min}/r_{\mathrm{head}}=1.80}.$$

Physical check

Radial speed vanishes at closest approach, but tangential speed generally does not. Setting all kinetic energy to zero would be valid only for a head-on orbit.