A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometers above the earth’s surface (Rₑₐᵣₜₕ = 6400 km) will approximately be

The correct option is 2h.

To find the approximate time period of the satellite orbiting close to the Earth's surface, we can apply Kepler's Third Law of planetary motion.

Kepler's Third Law states that the square of the time period ($T$) of a satellite orbiting in a circular orbit is directly proportional to the cube of the radius ($r$) of its orbit:
$T^2���r^3$
Or equivalently:
$T\propto r^{3/2}$

Let $T_1$ and $r_1$ be the orbital period and orbital radius of the geostationary satellite.
A geostationary satellite orbits the Earth with a period equal to the Earth's rotational period:
$T_1=24\text{hours}$
Its orbital radius is given as:
$r_1=36000\text{km}$

Let $T_2$ and $r_2$ be the orbital period and orbital radius of the second satellite.
Since it orbits a few hundred kilometers above the Earth's surface, its orbital radius is approximately equal to the radius of the Earth:
$r_2\approx R_{\text{earth}}=6400\text{km}$

Using the ratio form of Kepler's Third Law, we have:
$\frac{T_2}{T_1}={(\frac{r_2}{r_1}}^{3/2}$

Substitute the given values into the equation:
$T_2=24\times {\left(\frac{6400}{36000}\right)}^{3/2}$
Simplify the fraction inside the parentheses:
$\frac{6400}{36000}=\frac{64}{360}=\frac{16}{90}\approx 0.1778$

Now, calculate the value of the term raised to the power of 1.5:
${\left(0.1778\right)}^{1.5}\approx 0.0749$

Finally, calculate the time period $T_2$:
$T_2\approx 24\times 0.0749\approx 1.8\text{hours}$

Approximating 1.8 hours to the nearest option, we get:
$T_2\approx 2\text{hours}$

Therefore, the time period of the satellite orbiting a few hundred kilometers above the Earth's surface will approximately be 2h.