Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The escape velocity of a body on the earth's surface is $11.2 \text{ km/s}$ . If the same body is projected upward with a velocity $22.4 \text{ km/s}$ , the velocity of this body at an infinite distance from the centre of the earth will be:

$11.2\sqrt{2} \text{ km/s}$

zero

$11.2 \text{ km/s}$

$11.2\sqrt{3} \text{ km/s}$

Step-by-Step Solution

Conservation of Energy: According to the principle of conservation of energy, the total energy of the body at the surface equals its total energy at infinity. Let $v$ be the projection velocity, $v_e$ be the escape velocity, and $v_f$ be the final velocity at infinity. $K_i + U_i = K_f + U_f$ $\frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2 + 0$
Relating to Escape Velocity: We know that the potential energy at the surface can be expressed in terms of escape velocity using the relation $\frac{1}{2}mv_e^2 = \frac{GMm}{R}$ . Substituting this into the energy equation: $\frac{1}{2}mv^2 - \frac{1}{2}mv_e^2 = \frac{1}{2}mv_f^2$ $v_f = \sqrt{v^2 - v_e^2}$
Calculation: Given $v_e = 11.2 \text{ km/s}$ and $v = 22.4 \text{ km/s} = 2v_e$ . $v_f = \sqrt{(2v_e)^2 - v_e^2} = \sqrt{4v_e^2 - v_e^2} = \sqrt{3v_e^2} = v_e\sqrt{3}$ $v_f = 11.2\sqrt{3} \text{ km/s}$

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