Solution to Question 51
System: Two particles (\(+q, -q\), mass \(m\)) separated by \(l\), with initial velocity \(u\) perpendicular to the separation.
(a) Condition for Collision
Since the particles interact via a central force, the total angular momentum is conserved. $$ L = m u l $$ Collision implies distance \(r \to 0\). For \(r\) to become zero while conserving angular momentum \(L = m v r\), \(L\) must be zero. $$ \therefore u = 0 $$
(b) Condition for Circular Orbits
For circular motion, the electrostatic attraction provides the centripetal force. The orbit radius for each mass about the center of mass is \(r = l/2\). $$ F_{elec} = \frac{k q^2}{l^2} = \frac{m u^2}{(l/2)} = \frac{2 m u^2}{l} $$ $$ u^2 = \frac{k q^2}{2 m l} \implies u = \sqrt{ \frac{q^2}{8\pi\epsilon_0 m l} } $$
(c) Condition for Closed Orbits
Orbits are closed (elliptical) if the Total Mechanical Energy is negative. $$ E = K + U = 2\left(\frac{1}{2}m u^2\right) – \frac{k q^2}{l} < 0 $$ $$ m u^2 < \frac{k q^2}{l} \implies u < \sqrt{ \frac{q^2}{4\pi\epsilon_0 m l} } $$
(d) Minimum Distance in Subsequent Motion
We solve for the apsidal distances using conservation of Energy and Angular Momentum. Let \(r\) be an apsidal distance (where velocity is perpendicular to radius).
1. Angular Momentum: \( m u l = m v r \implies v = \frac{ul}{r} \)
2. Energy:
$$ m u^2 – \frac{k q^2}{l} = m v^2 – \frac{k q^2}{r} $$
Substituting \(v\):
$$ m u^2 – \frac{k q^2}{l} = m \left( \frac{ul}{r} \right)^2 – \frac{k q^2}{r} $$
Let \(A = m u^2 – \frac{k q^2}{l}\). This leads to the quadratic:
$$ A r^2 + (k q^2) r – m u^2 l^2 = 0 $$
The product of the two roots \(r_1 r_2 = \frac{-m u^2 l^2}{A} = \frac{m u^2 l^2}{\frac{k q^2}{l} – m u^2} \).
One root is obviously the starting distance \(l\) (since velocity is initially perpendicular).
The other root \(r’\) is:
$$ r’ = \frac{m u^2 l^2}{k q^2/l – m u^2} = \frac{m u^2 l^2}{\frac{q^2}{4\pi\epsilon_0 l} – m u^2} $$
Case 1: If \(u < \sqrt{\frac{q^2}{8\pi\epsilon_0 m l}}\) (velocity less than circular speed), the particles start at the apogee (farthest point) and fall inwards.
\( \therefore r_{min} = r’ = \frac{m u^2 l^2}{\frac{q^2}{4\pi\epsilon_0} – m u^2 l} \)
Case 2: If \(u \ge \sqrt{\frac{q^2}{8\pi\epsilon_0 m l}}\), the particles start at the perigee (closest point) and move outwards.
\( \therefore r_{min} = l \)
If \( u < \sqrt{\frac{q^2}{8\pi\epsilon_0 m l}} \): \( \quad r_{min} = \frac{l}{ \frac{q^2}{4\pi\epsilon_0 m u^2 l} - 1 } \)
If \( u \ge \sqrt{\frac{q^2}{8\pi\epsilon_0 m l}} \): \( \quad r_{min} = l \)