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: Understanding the Concept:
This problem can be solved using the principle of conservation of total mechanical energy. The total energy (kinetic + potential) of the body remains constant throughout its journey. We will compare the total energy at the Earth's surface with the total energy at an infinite distance (where it has "escaped" the gravitational influence).
Step 2: Key Formulae:
Let M and R be the mass and radius of the Earth.
Kinetic Energy: \(KE = \frac{1}{2}mv^2\)
Gravitational Potential Energy: \(PE = -\frac{GMm}{r}\)
Escape Speed: \(V_e = \sqrt{\frac{2GM}{R}}\), which implies \(\frac{GM}{R} = \frac{1}{2}V_e^2\).
Conservation of Energy: \(E_i = E_f \implies KE_i + PE_i = KE_f + PE_f\).
Step 3: Detailed Explanation:
Let the initial projection speed be \(v_i\). We are given \(v_i = \sqrt{5} V_e\).
Let the final speed at infinity be \(v_f\).
Initial Energy (at Earth's surface, r=R):
\[ E_i = KE_i + PE_i = \frac{1}{2}mv_i^2 - \frac{GMm}{R} \] Final Energy (at infinity, r=\(\infty\)):
At an infinite distance, the gravitational potential energy is zero.
\[ E_f = KE_f + PE_f = \frac{1}{2}mv_f^2 + 0 \] Applying Conservation of Energy:
\[ \frac{1}{2}mv_i^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2 \] Divide the entire equation by \(m\):
\[ \frac{1}{2}v_i^2 - \frac{GM}{R} = \frac{1}{2}v_f^2 \] Now, substitute \(v_i = \sqrt{5} V_e\) and \(\frac{GM}{R} = \frac{1}{2}V_e^2\):
\[ \frac{1}{2}(\sqrt{5} V_e)^2 - \frac{1}{2}V_e^2 = \frac{1}{2}v_f^2 \] \[ \frac{1}{2}(5 V_e^2) - \frac{1}{2}V_e^2 = \frac{1}{2}v_f^2 \] Multiply the entire equation by 2:
\[ 5V_e^2 - V_e^2 = v_f^2 \] \[ 4V_e^2 = v_f^2 \] Taking the square root of both sides:
\[ v_f = \sqrt{4V_e^2} = 2V_e \] Step 4: Final Answer:
The speed of the body when it escapes the Earth's gravitational influence is \(2V_e\). Option (B) is correct.