Solution to Question 33
The particle moves in a central potential field $U(r) = -k/r^4$. We apply the principles of conservation of mechanical energy and angular momentum.
Energy Conservation:
At infinity, potential energy is zero and velocity is $u$. At any distance $r$, let the velocity be $v$.
$$ E = \frac{1}{2} m u^2 = \frac{1}{2} m v^2 – \frac{k}{r^4} $$
Angular Momentum Conservation:
Angular momentum about the origin is conserved throughout the motion.
$$ L = m u b = m v_{\perp} r $$
Note that at the distance of closest approach ($r_{min}$), the velocity is purely perpendicular to the radius vector ($v = v_{\perp}$).
At the closest approach, let distance be $r$. Substituting $v = \frac{ub}{r}$ (from angular momentum) into the energy equation: $$ \frac{1}{2} m u^2 = \frac{1}{2} m \left( \frac{ub}{r} \right)^2 – \frac{k}{r^4} $$ Multiply the entire equation by $2/m$: $$ u^2 = \frac{u^2 b^2}{r^2} – \frac{2k}{m r^4} $$
We rearrange the equation to form a quadratic in terms of $r^2$. Multiply by $r^4 / u^2$: $$ r^4 = b^2 r^2 – \frac{2k}{m u^2} $$ $$ r^4 – b^2 r^2 + \frac{2k}{m u^2} = 0 $$
Let $X = r^2$. The equation becomes: $$ X^2 – b^2 X + \frac{2k}{m u^2} = 0 $$ For the particle not to hit the source, there must be a valid, real solution for the distance of closest approach (i.e., a real turning point exists where $\dot{r}=0$). This requires the quadratic equation for $X$ to have real roots.
The condition for real roots is that the discriminant $D \ge 0$: $$ D = (coefficient\_of\_X)^2 – 4(1)(constant\_term) \ge 0 $$ $$ (-b^2)^2 – 4 \left( \frac{2k}{m u^2} \right) \ge 0 $$ $$ b^4 – \frac{8k}{m u^2} \ge 0 $$
From the inequality: $$ b^4 \ge \frac{8k}{m u^2} $$ Taking the fourth root on both sides: $$ b \ge \left( \frac{8k}{m u^2} \right)^{1/4} $$