Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A particle of mass m is thrown upwards from the surface of the earth with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth is

\sqrt{\frac{2GM}{R}}

\sqrt{\frac{2GM}{R^2}}

\sqrt{2gR^2}

\sqrt{\frac{2GM}{R^3}}

Step-by-Step Solution

The minimum velocity required for a particle to escape the Earth's gravitational field is called the escape velocity ( $v_e$ ). It is derived using the principle of conservation of energy. The total mechanical energy (Kinetic + Potential) at the surface must equal the total energy at infinity (which is zero for the minimum condition).

$K.E. + P.E. = 0$ $\frac{1}{2}mu^2 - \frac{GmM}{R} = 0$ $\frac{1}{2}mu^2 = \frac{GmM}{R}$ $u^2 = \frac{2GM}{R}$ $u = \sqrt{\frac{2GM}{R}}$

Alternatively, using $g = \frac{GM}{R^2}$ , the expression can be written as $\sqrt{2gR}$ . Since $\sqrt{2gR}$ is not explicitly listed (Option 3 has $R^2$ ), the correct form in terms of $G$ and $M$ is the one in Option A.

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