A mass m,travelling at speed V_0 in a straigh | vedclass

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(D) At the point of closest approach,the velocity $v$ is perpendicular to the position vector $a$. By conservation of angular momentum about the black

Vedclass · Accepted Answer

$a = R{\left( {1 + \frac{{2GM}}{{aV_0^2}}} \right)^{ - 1/2}}$

Vedclass · Answer

(D) At the point of closest approach,the velocity $v$ is perpendicular to the position vector $a$. By conservation of angular momentum about the black hole:$m V_0 R = m v a \implies v = \frac{V_0 R}{a} \quad ...(i)$By conservation of mechanical energy:$\frac{1}{2} m V_0^2 = \frac{1}{2} m v^2 - \frac{G M m}{a} \quad ...(ii)$Substitute $v$ from $(i)$ into $(ii)$:$\frac{1}{2} m V_0^2 = \frac{1}{2} m \left( \frac{V_0 R}{a} \right)^2 - \frac{G M m}{a}$Divide by $\frac{1}{2} m$:$V_0^2 = \frac{V_0^2 R^2}{a^2} - \frac{2GM}{a}$$V_0^2 = \frac{V_0^2 R^2}{a^2} \left( 1 - \frac{2GM}{a V_0^2} \right)$$a^2 = R^2 \left( 1 - \frac{2GM}{a V_0^2} \right)$$a = R \left( 1 - \frac{2GM}{a V_0^2} \right)^{1/2}$Note: The potential energy is negative,so the energy equation is $\frac{1}{2} m V_0^2 = \frac{1}{2} m v^2 - \frac{G M m}{a}$. Rearranging gives $v^2 = V_0^2 + \frac{2GM}{a}$. Substituting into angular momentum conservation $v = \frac{V_0 R}{a}$ leads to $a = R \left( 1 + \frac{2GM}{a V_0^2} \right)^{-1/2}$.

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