A geo-stationary satellite is orbiting the earth at a height of 6 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is

To find the time period of the second satellite, we can use Kepler's Third Law of Planetary Motion. According to this law, the square of the time period of a satellite orbiting a planet is directly proportional to the cube of the semi-major axis (or the orbital radius) of its orbit.

Mathematically, Kepler's Third Law is expressed as:
$T^2\propto r^3$
where $T$ is the time period of the satellite, and $r$ is the orbital radius measured from the center of the Earth.

Let $R$ be the radius of the Earth.
The orbital radius of a satellite orbiting at a height $h$ above the surface of the Earth is given by:
$r=R+h$

Let us write the parameters for both satellites:
1. For the first satellite (Geo-stationary satellite):
It is orbiting at a height $h_1=6R$ above the surface.
Therefore, its orbital radius is:
$r_1=R+6R=7R$
Since it is a geo-stationary satellite, its time period is:
$T_1=24\text{hr}$

2. For the second satellite:
It is orbiting at a height $h_2=2.5R$ above the surface.
Therefore, its orbital radius is:
$r_2=R+2.5R=3.5R$
Let its time period be $T_2$.

Using Kepler's Third Law, we can write the ratio of the time periods as:
${\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{r_2}{r_1}\right)}^3$

Substitute the values of $r_1$ and $r_2$ into the equation:
${\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{3.5R}{7R}\right)}^3$

Simplify the fraction inside the parenthesis:
$\frac{3.5}{7}=\frac{1}{2}$
So,
${\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{1}{2}\right)}^3=\frac{1}{8}$

Taking the square root on both sides:
$\frac{T_2}{T_1}=\sqrt{\frac{1}{8}}=\frac{1}{2\sqrt{2}}$

Now, solve for $T_2$ by substituting $T_1=24\text{hr}$:
$T_2=\frac{T_1}{2\sqrt{2}}=\frac{24}{2\sqrt{2}}=\frac{12}{\sqrt{2}}$

We can simplify $\frac{12}{\sqrt{2}}$ by multiplying the numerator and denominator by $\sqrt{2}$:
$T_2=\frac{12\cdot \sqrt{2}}{2}=6\sqrt{2}\text{hr}$

Thus, the time period of the second satellite is 6√2 hr, which corresponds to the correct option.