The period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4R is

The correct option is 8 T.

According to Kepler's Third Law of planetary motion, the square of the orbital period of a satellite (T) in a circular orbit is directly proportional to the cube of the radius of its orbit (R). This relationship is given by:
$T^2\propto R^3$
Taking the square root on both sides, we can write this relationship as:
$T\propto R^{3/2}$

Let the orbital period of the first satellite be T₁ = T and its orbital radius be R₁ = R.
Let the orbital period of the second satellite be T₂ and its orbital radius be R₂ = 4R.
By comparing the ratios of their periods and orbital radii, we get:
$\frac{T_2}{T_1}={\left(\frac{R_2}{R_1}\right)}^{3/2}$

Substitute the given values into the ratio equation:
$\frac{T_2}{T}={\left(\frac{4R}{R}\right)}^{3/2}$
Simplify the equation by canceling out the common radius term R:
$\frac{T_2}{T}=4^{3/2}$
Calculate the value of 4^3/2:
$4^{3/2}={\left(2^2\right)}^{3/2}=2^3=8$
Now, solve for the new orbital period T₂:
$T_2=8T$