The escape velocity of a particle of mass `m` varies as

Options

A. `m^(2)`
B. `m`
C. `m^(0)`
D. `m^(-1)`

Explanation

To solve the question regarding how the escape velocity of a particle of mass `m` varies, we can follow these steps:

### Step 1: Understand Escape Velocity
Escape velocity is the minimum velocity an object must have to break free from the gravitational attraction of a celestial body without any additional propulsion.

### Step 2: Formula for Escape Velocity
The formula for escape velocity (V) is given by:
\[ V = \sqrt{\frac{2GM}{R}} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the celestial body (e.g., planet),
- \( R \) is the radius of the celestial body.

### Step 3: Analyze the Dependence on Mass
From the formula, we can see that escape velocity depends on the mass of the celestial body \( M \) and the radius \( R \), but it does not include the mass \( m \) of the escaping particle. This indicates that the escape velocity is independent of the mass of the object trying to escape.

### Step 4: Conclusion
Since escape velocity does not depend on the mass of the particle \( m \), we can conclude that:
- Escape velocity varies as \( M^{1/2} \) and \( R^{-1/2} \), but not with respect to \( m \).
- Therefore, we can say that escape velocity varies as \( m^0 \) (which means it is independent of \( m \)).

### Final Answer
The escape velocity of a particle of mass `m` varies as \( m^0 \) (independent of mass).

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