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: Formula for escape velocity
The escape velocity from a planet is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) is the gravitational constant 
- \( M \) is the mass of the planet 
- \( R \) is the radius of the planet 

Step 2: Original planet’s escape velocity 
Let the original planet have mass \( M \) and radius \( M \) and radius \( R \). Its escape velocity is: \[ v_e = \sqrt{\frac{2GM}{R}} \] 
Step 2: New planet’s parameters 
The new planet has: - Mass = \( M’ = 2M \) 
- Radius = \( R’ = 2R \) 

Step 3: Calculate new escape velocity 
Substitute \( M’ = M’’ = 2M \) and \( R’ = R’ = 2R \) into the escape velocity formula: \[ v_e’ = v_e’ = \sqrt{\frac{2G \cdot 2M}{2R}} \] \[ = \sqrt{\frac{4GM}{2R}} \] \[ = \sqrt{\frac{4GM}{2R}} = \sqrt{2 \cdot \frac{2GM}{R}} \] \[ = \sqrt{\frac{2GM}{R}} = v_e \] The new escape velocity \( v_e’ = v_e \), so it remains unchanged. 

Step 4: Alternative approach 
Notice that escape velocity depends on 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 is unchanged, the escape velocity remains \( v_e \). Thus, the correct answer is option (1) \( v_e \).