Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A particle of mass $m$ is projected with a velocity, $v = kv_e$ ( $k < 1$ ) from the surface of the earth. The maximum height, above the surface, reached by the particle is: (Where $v_e =$ escape velocity, $R =$ the radius of the earth)

$\frac{R^2k}{1+k}$

$\frac{Rk^2}{1-k^2}$

$R\left(\frac{k}{1-k}\right)^2$

$R\left(\frac{k}{1+k}\right)^2$

Step-by-Step Solution

Conservation of Energy: The total mechanical energy of the particle is conserved. Let the maximum height reached be $h$ . At this point, the velocity of the particle is zero. $E_i = E_f$ $K_i + U_i = K_f + U_f$
Initial Energy: At the surface ( $r=R$ ), velocity $v = kv_e$ . Since escape velocity $v_e = \sqrt{\frac{2GM}{R}}$ , then $v^2 = k^2 \frac{2GM}{R}$ . $K_i = \frac{1}{2}mv^2 = \frac{1}{2}m \left( k^2 \frac{2GM}{R} \right) = \frac{k^2 GMm}{R}$ $U_i = -\frac{GMm}{R}$
Final Energy: At maximum height ( $r = R+h$ ), velocity is zero. $K_f = 0$ $U_f = -\frac{GMm}{R+h}$
Solve for h: $\frac{k^2 GMm}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h}$ $\frac{GMm}{R}(k^2 - 1) = -\frac{GMm}{R+h}$ $\frac{1-k^2}{R} = \frac{1}{R+h}$ $R+h = \frac{R}{1-k^2}$ $h = \frac{R}{1-k^2} - R = R \left( \frac{1 - (1-k^2)}{1-k^2} \right) = \frac{Rk^2}{1-k^2}$

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