JEE MAIN - Physics (2024 - 6th April Morning Shift - No. 8)

To project a body of mass $$m$$ from earth's surface to infinity, the required kinetic energy is (assume, the radius of earth is $$R_E, g=$$ acceleration due to gravity on the surface of earth):
$$1 / 2 m g R_E$$
$$4 m g R_E$$
$$m g R_E$$
$$2 m g R_E$$

Explanation

The kinetic energy required to project a body of mass $m$ from the Earth's surface to infinity, also known as the escape kinetic energy, can be calculated using the concept of gravitational potential energy. The escape velocity $v_e$ is the velocity a body must have to escape the gravitational field of the Earth without any further propulsion. The formula for escape velocity is:

$v_e = \sqrt{2gR_E}$

Where $g$ is the acceleration due to gravity on the surface of Earth and $R_E$ is the radius of the Earth. The kinetic energy $K$ required for this is given by:

$K = \frac{1}{2}mv_e^2$

Substituting the escape velocity formula into the kinetic energy formula gives:

$K = \frac{1}{2}m\left(2gR_E\right)$

$K = \frac{1}{2} \times 2 \times mgR_E$

$K = mgR_E$

Therefore, the required kinetic energy to project a body of mass $m$ from Earth's surface to infinity is $mgR_E$. So, the correct answer is:

Option C: $mgR_E$

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