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 determine the value of \( n \) for a satellite in a circular orbit at a height of \( \frac{R}{4} \) above the surface of the Earth. The time period of revolution is given by the formula:

\[ T = 2\pi n^{3/2} \sqrt{\frac{R}{g}} \]

where \( R \) is the radius of the Earth and \( g \) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the height of the satellite above the Earth's surface**:
The satellite is at a height \( h = \frac{R}{4} \) above the surface of the Earth.

2. **Calculate the total distance from the center of the Earth**:
The radius of the orbit \( r \) is given by:
\[ r = R + h = R + \frac{R}{4} = R + 0.25R = 1.25R \]

3. **Use the formula for the time period of a satellite in orbit**:
The time period \( T \) of a satellite in a circular orbit is given by:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. We can express \( g \) as:
\[ g = \frac{GM}{R^2} \]
Thus, we can rewrite the time period as:
\[ T = 2\pi \sqrt{\frac{(1.25R)^3}{GM}} \]

4. **Substitute \( g \) into the time period formula**:
Replacing \( GM \) with \( gR^2 \):
\[ T = 2\pi \sqrt{\frac{(1.25R)^3}{gR^2}} = 2\pi \sqrt{\frac{1.25^3 R^3}{gR^2}} = 2\pi \sqrt{\frac{1.25^3 R}{g}} \]

5. **Simplify the expression for \( T \)**:
\[ T = 2\pi \cdot \frac{1.25^{3/2} R^{1/2}}{g^{1/2}} \]

6. **Equate the two expressions for \( T \)**:
From the problem statement, we have:
\[ T = 2\pi n^{3/2} \sqrt{\frac{R}{g}} \]
Setting the two expressions for \( T \) equal gives:
\[ 2\pi \cdot 1.25^{3/2} \sqrt{\frac{R}{g}} = 2\pi n^{3/2} \sqrt{\frac{R}{g}} \]

7. **Cancel common terms**:
Since \( 2\pi \sqrt{\frac{R}{g}} \) is common on both sides, we can cancel it out:
\[ 1.25^{3/2} = n^{3/2} \]

8. **Solve for \( n \)**:
Taking both sides to the power of \( \frac{2}{3} \):
\[ n = (1.25^{3/2})^{2/3} = 1.25^{1} = 1.25 \]

### Final Answer:
The value of \( n \) is \( 1.25 \).