Given radius of Earth 'R' and length of a day 'T' the height of a geostationary satellite is [G- Gravitational constant. M - Mass of Earth]

Options

A. `((4 pi^(2) GM)/(T^(2)))^(1//3)`
B. `((4 pi^(2) GM)/(R^(2)))^(1//3) - R`
C. `((GMT^(2))/(4 pi ^(2)))^(1//3) - R `
D. `((GMT^(2))/(4 pi ^(2)))^(1//3) + R `

Explanation

To find the height of a geostationary satellite above the Earth's surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of a Geostationary Satellite:**
A geostationary satellite orbits the Earth at a height where its orbital period matches the Earth's rotation period (T). This means it appears stationary relative to the Earth's surface.

2. **Use the Formula for Orbital Period:**
The formula for the orbital period (T) of a satellite is given by:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
where:
- \( r \) is the distance from the center of the Earth to the satellite,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth.

3. **Express \( r \) in Terms of \( R \) and \( h \):**
The distance \( r \) can be expressed as:
\[ r = R + h \]
where:
- \( R \) is the radius of the Earth,
- \( h \) is the height of the satellite above the Earth's surface.

4. **Substitute \( r \) into the Period Formula:**
Substitute \( r = R + h \) into the orbital period formula:
\[ T = 2\pi \sqrt{\frac{(R + h)^3}{GM}} \]

5. **Square Both Sides:**
To eliminate the square root, square both sides:
\[ T^2 = 4\pi^2 \frac{(R + h)^3}{GM} \]

6. **Rearrange the Equation:**
Rearranging gives:
\[ (R + h)^3 = \frac{GMT^2}{4\pi^2} \]

7. **Take the Cube Root:**
Taking the cube root of both sides:
\[ R + h = \left(\frac{GMT^2}{4\pi^2}\right)^{\frac{1}{3}} \]

8. **Isolate \( h \):**
Finally, isolate \( h \):
\[ h = \left(\frac{GMT^2}{4\pi^2}\right)^{\frac{1}{3}} - R \]

### Final Result:
The height \( h \) of the geostationary satellite above the Earth's surface is:
\[ h = \left(\frac{GMT^2}{4\pi^2}\right)^{\frac{1}{3}} - R \]