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(D) $1$. Conservation of angular momentum about the black hole:$mv_0 R = mva \implies v = \frac{v_0 R}{a}$$2$. Conservation of mechanical energy:Initi

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) $1$. Conservation of angular momentum about the black hole:$mv_0 R = mva \implies v = \frac{v_0 R}{a}$$2$. Conservation of mechanical energy:Initial energy = Final energy at closest approach$\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}mv^2 - \frac{GMm}{a}$$3$. Substitute $v$ into the energy equation:$\frac{1}{2}v_0^2 = \frac{1}{2}\left(\frac{v_0 R}{a}\right)^2 - \frac{GM}{a}$$v_0^2 = \frac{v_0^2 R^2}{a^2} - \frac{2GM}{a}$$1 = \frac{R^2}{a^2} - \frac{2GM}{a v_0^2}$$\frac{R^2}{a^2} = 1 + \frac{2GM}{a v_0^2}$$\frac{R}{a} = \left(1 + \frac{2GM}{a v_0^2}\right)^{1/2}$$a = R\left(1 + \frac{2GM}{a v_0^2}\right)^{-1/2}$

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