Solution 1

Derivation

We analyze the motion in the relative frame (reduced mass $\mu = \frac{M \cdot M}{M+M} = \frac{M}{2}$).

Let relative position be $\vec{r} = \vec{r}_{12}$ and relative velocity be $\vec{v} = \vec{v}_2 – \vec{v}_1$.

1. Conservation of Energy

The total energy $E$ is conserved. At the closest approach distance $r_{min}$, the radial velocity is zero, and velocity is purely tangential ($v_{tan}$).

$$ E_{initial} = \frac{1}{2}\mu v^2 – \frac{G(M)(M)}{r} = \frac{1}{2}\mu v_{tan}^2 – \frac{G M^2}{r_{min}} $$

2. Angular Momentum

Angular momentum $L$ is conserved:

$$ L = |\vec{r} \times \mu \vec{v}| = \mu |\vec{r} \times \vec{v}| $$

At closest approach: $L = r_{min} (\mu v_{tan}) \implies v_{tan} = \frac{L}{\mu r_{min}} = \frac{|\vec{r} \times \vec{v}|}{r_{min}}$.

Substitute $v_{tan}$ back into the energy equation:

$$ \frac{1}{2}\mu v^2 – \frac{GM^2}{r} = \frac{1}{2}\mu \left( \frac{|\vec{r} \times \vec{v}|}{r_{min}} \right)^2 – \frac{GM^2}{r_{min}} $$

3. Condition for Collision

Collision occurs if $r_{min} \le 2R$. The limiting case is $r_{min} = 2R$. Substituting $r_{min} = 2R$ and $\mu = M/2$:

$$ \frac{1}{2}\left(\frac{M}{2}\right) v^2 – \frac{GM^2}{r} = \frac{1}{2}\left(\frac{M}{2}\right) \frac{|\vec{r} \times \vec{v}|^2}{(2R)^2} – \frac{GM^2}{2R} $$

Cancel $M$ from all terms and simplify fractions:

$$ \frac{v^2}{4} – \frac{GM}{r} = \frac{|\vec{r} \times \vec{v}|^2}{16R^2} – \frac{GM}{2R} $$

Rearranging for the cross-product term:

$$ \frac{|\vec{r} \times \vec{v}|^2}{16R^2} = \frac{v^2}{4} + \frac{GM}{2R} – \frac{GM}{r} $$

For collision, the impact parameter must be small enough, satisfying the inequality: