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. 2E

Answer

To find the escape velocity of an object given that the magnitude of potential energy per unit mass at the surface of the Earth is \( E \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Potential Energy per Unit Mass**:
The potential energy \( U \) of an object of mass \( m \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ U = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. The potential energy per unit mass \( u \) is:
\[ u = \frac{U}{m} = -\frac{GM}{r} \]
At the surface of the Earth, \( r \) is equal to the radius of the Earth \( R \). Thus, we have:
\[ u = -\frac{GM}{R} \]

2. **Magnitude of Potential Energy per Unit Mass**:
The problem states that the magnitude of potential energy per unit mass is \( E \). Therefore:
\[ E = \frac{GM}{R} \]

3. **Escape Velocity Formula**:
The escape velocity \( v_e \) from the surface of the Earth is given by the formula:
\[ v_e = \sqrt{2gh} \]
where \( h \) is the height from which the object needs to escape. For an object at the surface, this can also be expressed in terms of gravitational potential energy:
\[ v_e = \sqrt{2 \cdot \text{(Potential Energy per unit mass)}} \]

4. **Substituting the Expression for E**:
Since we have established that the potential energy per unit mass is equal to \( E \), we can substitute:
\[ v_e = \sqrt{2E} \]

5. **Final Result**:
Therefore, the escape velocity of the object is:
\[ v_e = \sqrt{2E} \]

### Conclusion:
The escape velocity of the object is \( \sqrt{2E} \).