WBJEE 2024 Physics & Chemistry Paper

Here's how to determine the kinetic energy (K) and total energy (E) of the satellite: 1. Angular Momentum and Velocity:The angular momentum ( J ) of a satellite in circular orbit is given by:$J = I \omega = m R^2 \omega$where: $m$ is the mass of the satellite$R$ is the orbital radius$\omega$ is the angular velocityWe can relate angular velocity $(\omega)$ to linear velocity ( $v$ ) using:$v = R \omega$ 2. Kinetic Energy:The kinetic energy of the satellite is:$K = \frac{1}{2} m v^2$Substituting $v = R \omega$ and $J = m R^2 \omega$, we get:$K = \frac{1}{2} m (R \omega)^2 = \frac{1}{2} m \left( \frac{J^2}{m^2 R^4} \right) = \frac{J^2}{2 m R^2}$ 3. Total Energy:The total energy ( E ) of the satellite is the sum of its kinetic energy $(K)$ and potential energy (U). The potential energy in a gravitational field is:$U = -\frac{G M m}{R}$where G is the gravitational constant and M is the mass of the Earth. The total energy is then:$E = K + U = \frac{J^2}{2 m R^2} - \frac{G M m}{R}$We can express the gravitational constant $(G)$ and the mass of the Earth $(M)$ in terms of the acceleration due to gravity $(g)$ at the Earth's surface:$G M = g R^2$ Substituting this into the equation for total energy:$E = \frac{J^2}{2 m R^2} - \frac{g R^2 m}{R} = \frac{J^2}{2 m R^2} - g R m$Now, using the centripetal acceleration equation, $g = \frac{v^2}{R} = \frac{(R \omega)^2}{R} = R \omega^2$ and substituting $J = m R^2 \omega$, we get:$g R m = R \omega^2 m = \frac{J^2}{m R^2}$Therefore, the total energy is:$E = \frac{J^2}{2 m R^2} - \frac{J^2}{m R^2} = -\frac{J^2}{2 m R^2}$ Conclusion:The correct answer is Option B:$K = \frac{J^2}{2 m R^2}, E = -\frac{J^2}{2 m R^2}$