The distance of a geo-stationary satellite from the center of the earth (Radius R = 6400 km) is nearest to

The correct option is "7 R".

To understand why this is the correct option, let us go through the step-by-step derivation of the orbital radius of a geostationary satellite.

A geostationary satellite is one that remains stationary relative to a point on the Earth's surface. This means its orbital period, $T$, must be exactly equal to the rotational period of the Earth about its own axis:
$T=24\text{hours}=24\times 3600\text{seconds}=\mathrm{86,400}\text{s}$

For a satellite of mass $m$ orbiting the Earth of mass $M$ at a distance $r$ from the center of the Earth, the gravitational force provides the necessary centripetal force for its circular motion:
$\frac{GMm}{r^2}=m{\omega }^{2}r$

Since the angular velocity $\omega$ is related to the time period $T$ by $\omega =\frac{2\pi }{T}$, we can rewrite the equation as:
$\frac{GM}{r^2}={\left(\frac{2\pi }{T}\right)}^{2}r$
Rearranging this formula to solve for the orbital radius $r$ gives:
$r^3=\frac{GM{T}^2}{4{\pi }^2}$

We also know that the acceleration due to gravity at the Earth's surface is given by:
$g=\frac{GM}{R^2}$
This allows us to substitute $GM=g{R}^2$ into our equation:
$r^3=\frac{g{R}^2{T}^2}{4{\pi }^2}$

Now, substituting the standard values:
• Radius of the Earth, $R\approx 6.4\times {10}^6\text{m}$
• Acceleration due to gravity, $g\approx 9.8{\text{m/s}}^2$
• Period of rotation, $T\approx \mathrm{86,400}\text{s}$
• $\pi \approx 3.14$

Plugging these numbers in:
$r^3\approx \frac{9.8\times {(6.4\times {10}^6)}^2\times {\mathrm{86,400}}^2}{4\times {3.14}^2}$
Evaluating this expression yields:
$r\approx 4.2\times {10}^7\text{m}=\mathrm{42,000}\text{km}$

To express this distance in terms of the Earth's radius ($R=6400\text{km}$), we divide the orbital radius by $R$:
$\frac{r}{R}\approx \frac{\mathrm{42,000}}{6400}\approx 6.56$

The value 6.56 is closest to the integer value of 7. Therefore, the distance of a geo-stationary satellite from the center of the Earth is nearest to 7 R.