Question

The ratio of the escape velocity to the orbital velocity of the earth is ________.

• A. 2
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$
• E. $\sqrt{3}$

Hint:

Escape velocity is always $\sqrt{2}$ times the orbital velocity for the same planet.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This question asks for the ratio of two important velocities in orbital mechanics for an object near the Earth's surface.
- Orbital Velocity (\(v_o\)): The speed required for an object to maintain a stable circular orbit just above the Earth's surface.
- Escape Velocity (\(v_e\)): The minimum speed required for an object to escape Earth's gravitational field completely and not fall back.
Step 2: Key Formula or Approach
Let M be the mass of the Earth and R be its radius.
1. Orbital Velocity: For a circular orbit, the gravitational force provides the necessary centripetal force.
\[ \frac{GMm}{R^2} = \frac{mv_o^2}{R} \implies v_o^2 = \frac{GM}{R} \implies v_o = \sqrt{\frac{GM}{R}} \] 2. Escape Velocity: By conservation of energy, the initial total energy (kinetic + potential) must be zero for the object to just reach infinity with zero speed.
\[ \frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \implies v_e^2 = \frac{2GM}{R} \implies v_e = \sqrt{\frac{2GM}{R}} \] 3. Ratio: We need to find the ratio \(\frac{v_e}{v_o}\).
Step 3: Detailed Explanation
1. Write the formulas for escape and orbital velocities.
\[ v_e = \sqrt{\frac{2GM}{R}} \] \[ v_o = \sqrt{\frac{GM}{R}} \] 2. Calculate the ratio.
\[ \frac{v_e}{v_o} = \frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{GM}{R}}} \] We can combine the square roots:
\[ \frac{v_e}{v_o} = \sqrt{\frac{2GM/R}{GM/R}} \] The term \(\frac{GM}{R}\) cancels out.
\[ \frac{v_e}{v_o} = \sqrt{2} \] Step 4: Final Answer
The ratio of the escape velocity to the orbital velocity is \(\sqrt{2}\).