Question:

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):

Updated On: Mar 22, 2025
  • \( 2mgR_E \)
  • \( mgR_E \)
  • \( \frac{1}{2}mgR_E \)
  • \( 4mgR_E \)
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The Correct Option is B

Solution and Explanation

The total energy required to project a body to infinity is equal to the work needed to overcome the gravitational potential energy at the surface of the Earth.

Gravitational potential energy at the surface of Earth:

\[ \text{Potential Energy} = -\frac{GMm}{R_E}, \]

where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the body, and \( R_E \) is the radius of the Earth.

The kinetic energy required for escape velocity:

\[ K = \frac{1}{2}mv_e^2. \]

Equating this to the energy needed to overcome the gravitational pull:

\[ \frac{1}{2}mv_e^2 = \frac{GMm}{R_E}. \]

Substituting \( g = \frac{GM}{R_E^2} \), we get:

\[ \frac{GMm}{R_E} = mgR_E. \]

Thus, the required kinetic energy is:

\[ K = mgR_E. \]

Final Answer: \( mgR_E \) (Option 2)

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