Question

A body is projected from the earth's surface with a speed \(\sqrt{5}\) times the escape speed (\(V_e\)). The speed of the body when it escapes from the gravitational influence of the earth is

• A. \(V_e\)
• B. \(2V_e\)
• C. \(\sqrt{2}V_e\)
• D. \(4V_e\)

Hint:

A very useful shortcut formula derived from the conservation of energy is \(v_f^2 = v_i^2 - V_e^2\), where \(v_i\) is the projection speed from the surface and \(v_f\) is the final speed at infinity. Using this, \(v_f^2 = (\sqrt{5}V_e)^2 - V_e^2 = 5V_e^2 - V_e^2 = 4V_e^2\), which gives \(v_f = 2V_e\) directly.

Answer 1

The Correct Option is B

Step 1: Conservation of Mechanical Energy: Total Energy at Earth's Surface = Total Energy at Infinity. \[ K_i + U_i = K_f + U_f \] Initial Kinetic Energy \( K_i = \frac{1}{2} m v_{proj}^2 \). Initial Potential Energy \( U_i = -\frac{GMm}{R} \). Final Potential Energy \( U_f = 0 \) (at infinity). Final Kinetic Energy \( K_f = \frac{1}{2} m v_{\infty}^2 \).
Step 2: Relate Escape Velocity: We know escape velocity \( V_e = \sqrt{\frac{2GM}{R}} \). Thus, \( \frac{GMm}{R} = \frac{1}{2} m V_e^2 \). So, \( U_i = -\frac{1}{2} m V_e^2 \).
Step 3: Solve for Final Velocity: Given \( v_{proj} = \sqrt{5} V_e \). \[ \frac{1}{2} m (\sqrt{5} V_e)^2 - \frac{1}{2} m V_e^2 = \frac{1}{2} m v_{\infty}^2 + 0 \] \[ \frac{1}{2} m (5 V_e^2) - \frac{1}{2} m V_e^2 = \frac{1}{2} m v_{\infty}^2 \] Cancel \( \frac{1}{2} m \) from all terms: \[ 5 V_e^2 - V_e^2 = v_{\infty}^2 \] \[ 4 V_e^2 = v_{\infty}^2 \] \[ v_{\infty} = \sqrt{4 V_e^2} = 2 V_e \]