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)
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).