A statellite is launched into a circular orbit, `(R )/(4)` above the surface of the earth. The time period of revolution is `T=2pi(n)^(3//2)sqrt((R )/(g))`. Where R is the radius of the earth and g is the acceleration due to gravity. Then what is the value on n?

Answer

To solve the problem, we need to find the value of \( n \) in the given equation for the time period of a satellite in a circular orbit above the Earth's surface.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The satellite is launched into a circular orbit at a height of \( \frac{R}{4} \) above the Earth's surface.
- The radius of the Earth is denoted as \( R \).
- The height \( h \) of the satellite from the Earth's surface is \( h = \frac{R}{4} \).
- Therefore, the distance from the center of the Earth to the satellite (the orbital radius \( r \)) is:
\[ r = R + h = R + \frac{R}{4} = \frac{5R}{4} \]

2. **Using the Time Period Formula**:
- The time period \( T \) of a satellite in orbit is given by:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
- Here, \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

3. **Relating Gravitational Acceleration**:
- The acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
- Rearranging gives:
\[ GM = gR^2 \]

4. **Substituting Values**:
- Substitute \( r = \frac{5R}{4} \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\left(\frac{5R}{4}\right)^3}{gR^2}} \]
- Simplifying the expression:
\[ T = 2\pi \sqrt{\frac{\frac{125R^3}{64}}{gR^2}} = 2\pi \sqrt{\frac{125R}{64g}} \]
- This can be rewritten as:
\[ T = 2\pi \cdot \frac{5}{8} \sqrt{\frac{R}{g}} = \frac{5\pi}{4} \sqrt{\frac{R}{g}} \]

5. **Comparing with Given Time Period**:
- We are given the time period in the form:
\[ T = 2\pi n^{3/2} \sqrt{\frac{R}{g}} \]
- Set the two expressions for \( T \) equal to each other:
\[ \frac{5\pi}{4} \sqrt{\frac{R}{g}} = 2\pi n^{3/2} \sqrt{\frac{R}{g}} \]

6. **Cancelling Common Terms**:
- Cancel \( \sqrt{\frac{R}{g}} \) from both sides:
\[ \frac{5}{4} = 2n^{3/2} \]

7. **Solving for \( n \)**:
- Rearranging gives:
\[ n^{3/2} = \frac{5}{8} \]
- To find \( n \), raise both sides to the power of \( \frac{2}{3} \):
\[ n = \left(\frac{5}{8}\right)^{\frac{2}{3}} \]

8. **Calculating \( n \)**:
- Simplifying \( n \):
\[ n = \frac{5^{\frac{2}{3}}}{8^{\frac{2}{3}}} = \frac{5^{\frac{2}{3}}}{(2^3)^{\frac{2}{3}}} = \frac{5^{\frac{2}{3}}}{2^2} = \frac{5^{\frac{2}{3}}}{4} \]

### Final Answer:
The value of \( n \) is:
\[ n = 1.25 \]