Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A rocket is fired vertically upward with a speed of $v = \frac{v_e}{\sqrt{2}}$ from the Earth's surface, where $v_e$ is escape velocity on the surface of Earth. The distance from the surface of Earth upto which the rocket can go before returning to the Earth is: (given, the radius of Earth $R = 6400$ km)

1600 km

3200 km

6400 km

12800 km

Step-by-Step Solution

Principle of Conservation of Energy: The total mechanical energy of the rocket is conserved. Let $m$ be the mass of the rocket and $M$ be the mass of the Earth. The initial point is the surface ( $r=R$ ) and the final point is the maximum height ( $r=R+h$ ) where velocity becomes zero. $K_i + U_i = K_f + U_f$ $\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$
Substitute Velocity: Given $v = \frac{v_e}{\sqrt{2}}$ . We know escape velocity $v_e = \sqrt{\frac{2GM}{R}}$ . Therefore, $v^2 = \frac{v_e^2}{2} = \frac{1}{2} \left( \frac{2GM}{R} \right) = \frac{GM}{R}$ .
Solve for h: Substitute $v^2$ into the energy equation: $\frac{1}{2}m \left( \frac{GM}{R} \right) - \frac{GMm}{R} = -\frac{GMm}{R+h}$ $\frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{R+h}$ $\frac{GMm}{R} \left( \frac{1}{2} - 1 \right) = -\frac{GMm}{R+h}$ $-\frac{GMm}{2R} = -\frac{GMm}{R+h}$ $\frac{1}{2R} = \frac{1}{R+h}$ $2R = R + h \implies h = R$
Calculation: Given $R = 6400$ km, the height reached is $h = 6400$ km.

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