Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is [g is acceleration due to gravity at Earth’s surface]

The correct option is 4 √2 π √(R/g).

Here is the step-by-step derivation to find the periodic time of the satellite:

1. Identify the Orbital Radius
Let $R$ be the radius of the Earth, and $h$ be the height of the satellite above the Earth's surface.
According to the problem, the satellite revolves at a height equal to the radius of the Earth:
$h=R$
The total distance of the satellite from the center of the Earth (orbital radius $r$) is:
$r=R+h=R+R=2R$

2. Relate Acceleration due to Gravity to Earth's Mass
The acceleration due to gravity $g$ at the Earth's surface is given by the formula:
$g=\frac{GM}{R^2}$
where $G$ is the universal gravitational constant and $M$ is the mass of the Earth. From this, we can express the product $GM$ as:
$GM=g{R}^2$

3. Determine the Orbital Velocity of the Satellite
The orbital velocity $v$ required for a satellite to maintain a stable circular orbit at radius $r$ is:
$v=\sqrt{\frac{GM}{r}}$
Substituting $GM=g{R}^2$ and $r=2R$ into this expression:
$v=\sqrt{\frac{g{R}^2}{2R}}=\sqrt{\frac{gR}{2}}$

4. Calculate the Time Period of Revolution
The time period $T$ is the time taken to complete one full orbit of circumference $2\pi r$:
$T=\frac{2\pi r}{v}$
Substitute the values of $r$ and $v$:
$T=\frac{2\pi \left(2R\right)}{\sqrt{\frac{gR}{2}}}$
$T=4\pi R\sqrt{\frac{2}{gR}}$
Bring the factor of $R$ inside the square root:
$T=4\pi \sqrt{\frac{2R^2}{gR}}=4\pi \sqrt{\frac{2R}{g}}$
This can be simplified and rewritten as:
$T=4\sqrt{2}\pi \sqrt{\frac{R}{g}}$