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(B) Using the conservation of mechanical energy between the Earth's surface $(r = R_e)$ and the maximum height $(r = r_{max})$:$U_i + K_i = U_f + K_f$

Vedclass · Accepted Answer

$\sqrt{\frac{g R_e}{3}}$

Vedclass · Answer

(B) Using the conservation of mechanical energy between the Earth's surface $(r = R_e)$ and the maximum height $(r = r_{max})$:$U_i + K_i = U_f + K_f$$-\frac{G M m}{R_e} + \frac{1}{2} m v^2 = -\frac{G M m}{r_{max}} + 0$Given $v^2 = \frac{4 g R_e}{3}$ and $g = \frac{G M}{R_e^2}$,so $v^2 = \frac{4 G M}{3 R_e}$.$-\frac{G M m}{R_e} + \frac{1}{2} m (\frac{4 G M}{3 R_e}) = -\frac{G M m}{r_{max}}$$-\frac{G M m}{R_e} + \frac{2 G M m}{3 R_e} = -\frac{G M m}{r_{max}}$$-\frac{1}{3} \frac{G M m}{R_e} = -\frac{G M m}{r_{max}} \Rightarrow r_{max} = 3 R_e$.The maximum height $h_{max} = r_{max} - R_e = 2 R_e$.We need the velocity $v'$ at height $h = \frac{h_{max}}{2} = R_e$,which corresponds to $r = 2 R_e$.Applying conservation of energy again:$-\frac{G M m}{R_e} + \frac{1}{2} m v^2 = -\frac{G M m}{2 R_e} + \frac{1}{2} m (v')^2$$-\frac{G M m}{R_e} + \frac{2 G M m}{3 R_e} = -\frac{G M m}{2 R_e} + \frac{1}{2} m (v')^2$$-\frac{1}{3} \frac{G M m}{R_e} + \frac{1}{2} \frac{G M m}{R_e} = \frac{1}{2} m (v')^2$$\frac{1}{6} \frac{G M m}{R_e} = \frac{1}{2} m (v')^2$$(v')^2 = \frac{G M}{3 R_e} = \frac{g R_e}{3} \Rightarrow v' = \sqrt{\frac{g R_e}{3}}$.

$A$ particle is projected vertically up with velocity $v = \sqrt{\frac{4 g R_e}{3}}$ from the Earth's surface. The velocity of the particle at a height equal to half of the maximum height reached by it is .........

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