A mass is taken to a height R from the surface of earth and then is given horizontal velocity v. The minimum value of v so that mass escapes to infinity is `sqrt((GM)/(nR))`. Find the value of n.

Answer

To solve the problem, we need to determine the minimum horizontal velocity \( v \) that a mass must have when taken to a height \( R \) above the Earth's surface in order to escape the Earth's gravitational field. The formula given for this minimum velocity is \( v = \sqrt{\frac{GM}{nR}} \), and we need to find the value of \( n \).

### Step-by-Step Solution:

1. **Understand the Situation**:
- The mass is taken to a height \( R \) above the Earth's surface.
- The distance from the center of the Earth to the mass at this height is \( R + R = 2R \).

2. **Use Conservation of Energy**:
- The total mechanical energy at the height \( R \) must equal the total mechanical energy at infinity.
- At infinity, the potential energy (U) is zero, and we want the kinetic energy (K) to also be zero for the minimum escape velocity.

3. **Write the Energy Conservation Equation**:
\[ K_i + U_i = K_f + U_f \]
- Where:
- \( K_i = \frac{1}{2} mv^2 \) (initial kinetic energy)
- \( U_i = -\frac{GMm}{2R} \) (potential energy at height \( R \))
- \( K_f = 0 \) (final kinetic energy at infinity)
- \( U_f = 0 \) (potential energy at infinity)

Thus, the equation becomes:
\[ \frac{1}{2} mv^2 - \frac{GMm}{2R} = 0 \]

4. **Simplify the Equation**:
- Rearranging gives:
\[ \frac{1}{2} mv^2 = \frac{GMm}{2R} \]
- Dividing both sides by \( m \) (mass of the object):
\[ \frac{1}{2} v^2 = \frac{GM}{2R} \]

5. **Solve for \( v^2 \)**:
- Multiply both sides by 2:
\[ v^2 = \frac{GM}{R} \]

6. **Relate to the Given Formula**:
- We want to express \( v \) in the form \( v = \sqrt{\frac{GM}{nR}} \).
- From our derived formula, we have:
\[ v^2 = \frac{GM}{R} \]
- Comparing this with \( v^2 = \frac{GM}{nR} \), we can see that:
\[ n = 1 \]

### Conclusion:
The value of \( n \) is \( 1 \).