Question

The escape velocity from the surface of a planet is \(v_e\). What will be the escape velocity from a planet whose mass and radius are twice that of the original planet?

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

Hint:

Escape velocity is proportional to \( \sqrt{\frac{M}{R}} \). If both mass and radius double, the ratio remains unchanged.

Answer 1

The Correct Option is A

Step 1: Escape Velocity Formula
The escape velocity \( v_e \) from a celestial body is calculated using the formula: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) represents the gravitational constant
- \( M \) is the mass of the body 
- \( R \) is the radius of the body 

Step 2: Original Planet's Escape Velocity 
Consider an initial planet with mass \( M \) and radius \( R \). Its escape velocity is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] 
Step 2: New Planet's Parameters 
A new planet is introduced with the following parameters: - Mass: \( M’ = 2M \) 
- Radius: \( R’ = 2R \) 

Step 3: Calculate New Escape Velocity 
Substitute the new parameters into the escape velocity formula: \[ v_e’ = \sqrt{\frac{2G \cdot 2M}{2R}} \] \[ = \sqrt{\frac{4GM}{2R}} \] \[ = \sqrt{2 \cdot \frac{2GM}{R}} \] \[ = \sqrt{2} \cdot \sqrt{\frac{2GM}{R}} \] The new escape velocity is \( v_e’ = \sqrt{2} v_e \), indicating it is increased by a factor of \( \sqrt{2} \). 

Step 4: Alternative Approach 
Observe that escape velocity is proportional to the square root of the ratio \( \frac{M}{R} \): \[ v_e \propto \sqrt{\frac{M}{R}} \] For the new planet: \[ \frac{M’}{R’} = \frac{2M}{2R} = \frac{M}{R} \] Since the ratio \( \frac{M}{R} \) remains constant, the escape velocity is also unchanged. Thus, the correct answer is option (1) \( v_e \).