Question

Maximum height reached by a rocket fired with a speed equal to 50\% of the escape speed from the surface of the earth is (R – Radius of the earth)

• A. \( \frac{R}{2} \)
• B. \( \frac{16R}{9} \)
• C. \( \frac{R}{3} \)
• D. \( \frac{R}{8} \)

Hint:

For problems involving escape velocity and maximum height, use the conservation of mechanical energy between the surface and the maximum height to derive the formula.

Answer 1

The Correct Option is C

Step 1: The escape velocity from the surface of the earth is given by the formula: \[ v_{\text{escape}} = \sqrt{\frac{2GM}{R}} \] where \( G \) is the gravitational constant, \( M \) is the mass of the earth, and \( R \) is the radius of the earth. When the rocket is fired with 50\% of the escape velocity, the speed \( v \) is: \[ v = \frac{1}{2} v_{\text{escape}} = \frac{1}{2} \sqrt{\frac{2GM}{R}} \] Step 2: The maximum height \( h \) reached by the rocket can be found using the energy conservation method. The total mechanical energy at the surface is the sum of kinetic and potential energy: \[ E_{\text{total}} = \frac{1}{2} m v^2 - \frac{GMm}{R} \] At the maximum height, the kinetic energy is zero, and the total energy is just the gravitational potential energy at that height. So, we have: \[ \frac{1}{2} m v^2 - \frac{GMm}{R} = - \frac{GMm}{R + h} \] Substitute \( v = \frac{1}{2} \sqrt{\frac{2GM}{R}} \) into the equation: \[ \frac{1}{2} m \left( \frac{1}{2} \sqrt{\frac{2GM}{R}} \right)^2 - \frac{GMm}{R} = - \frac{GMm}{R + h} \] Simplifying this: \[ \frac{1}{2} m \cdot \frac{GM}{2R} - \frac{GMm}{R} = - \frac{GMm}{R + h} \] \[ \frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R + h} \] \[ \frac{GMm}{R} \left( \frac{1}{4} - 1 \right) = - \frac{GMm}{R + h} \] \[ \frac{-3GMm}{4R} = - \frac{GMm}{R + h} \] \[ \frac{3}{4} = \frac{R}{R + h} \] Solving for \( h \): \[ R + h = \frac{4}{3} R \] \[ h = \frac{4}{3} R - R = \frac{R}{3} \] Thus, the maximum height reached by the rocket is \( \frac{R}{3} \).