The magnitude of potential energy per unit mass of the object at the surface of earth is `E`. Then escape velocity of the object is

Options

A. `sqrt(2E)`
B. `4E^(2)`
C. `sqrt(E)`
D. `sqrt(E//2)`

Answer

To solve the problem, we need to derive the escape velocity of an object from the given potential energy per unit mass at the surface of the Earth.

### Step-by-Step Solution:

1. **Understanding Potential Energy per Unit Mass**:
The potential energy (U) per unit mass (m) at the surface of the Earth is given as \( E \). This can be expressed mathematically as:
\[ E = \frac{U}{m} \]
The gravitational potential energy at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the radius of the Earth.

2. **Relating Potential Energy to Gravitational Potential**:
At the surface of the Earth, the gravitational potential (V) can be defined as:
\[ V = -\frac{GM}{r} \]
Since \( E \) is given as the magnitude of potential energy per unit mass, we can write:
\[ E = -\frac{GM}{R} \]
where \( R \) is the radius of the Earth.

3. **Escape Velocity Formula**:
The escape velocity (\( V_E \)) from the surface of the Earth is given by the formula:
\[ V_E = \sqrt{\frac{2GM}{R}} \]

4. **Substituting for \( GM/R \)**:
From the expression for \( E \), we can express \( \frac{GM}{R} \) in terms of \( E \):
\[ \frac{GM}{R} = -E \]
However, since we are interested in the magnitude, we can take the positive value:
\[ \frac{GM}{R} = E \]

5. **Final Calculation**:
Now substituting \( \frac{GM}{R} = E \) into the escape velocity formula:
\[ V_E = \sqrt{2 \cdot \frac{GM}{R}} = \sqrt{2E} \]

### Conclusion:
Thus, the escape velocity of the object is:
\[ V_E = \sqrt{2E} \]