Solution 5

Method: Gravitational Focusing

The planet collects all particles that graze its surface or hit it directly. We need to find the maximum impact parameter $b$ such that the distance of closest approach is $R$.

1. Conservation of Energy

A particle starts at infinity with speed $v_0$ and reaches the planet’s surface (distance $R$) with speed $v_{max}$.

$$ \frac{1}{2} m v_0^2 = \frac{1}{2} m v_{max}^2 – \frac{GMm}{R} $$ $$ v_{max}^2 = v_0^2 + \frac{2GM}{R} $$

We know escape velocity $v_e = \sqrt{\frac{2GM}{R}}$, so $v_{max}^2 = v_0^2 + v_e^2$.

2. Conservation of Angular Momentum

Angular momentum about the planet’s center is conserved. At infinity, $L = m v_0 b$. At the surface (grazing), $L = m v_{max} R$.

$$ v_0 b = v_{max} R $$ $$ b = R \frac{v_{max}}{v_0} = R \frac{\sqrt{v_0^2 + v_e^2}}{v_0} = R \sqrt{1 + \frac{v_e^2}{v_0^2}} $$

3. Mass Calculation

The effective cross-sectional area for capture is $A = \pi b^2$.

$$ A = \pi R^2 \left( 1 + \frac{v_e^2}{v_0^2} \right) $$

The volume of the dust cloud passing through this area is $Vol = A \times l$. The total mass collected is $M = \text{Density} \times Vol$.

$$ M = \rho \pi R^2 \left( 1 + \frac{v_e^2}{v_0^2} \right) l $$