Given radius of earth ‘R’ and length of a day ‘T’ the height of a geostationary satellite is [G – Gravitational constant, M – Mass of earth]

The correct option is (GMT²/4π²)⁰.³³ - R.

To find the height of a geostationary satellite, we need to balance the gravitational force acting on the satellite with the centripetal force required to keep it in a circular orbit around the Earth.

Let:
- $M$ be the mass of the Earth.
- $R$ be the radius of the Earth.
- $h$ be the height of the geostationary satellite above the Earth's surface.
- $r$ be the orbital radius of the satellite from the center of the Earth, such that:
$r=R+h$
- $m$ be the mass of the satellite.
- $T$ be the time period of the satellite's rotation, which is equal to the length of a day (since it is a geostationary satellite).

The gravitational force between the Earth and the satellite is given by Newton's law of gravitation:
$F_g=\frac{GMm}{r^2}$

The centripetal force required to keep the satellite in orbit is:
$F_c=m{\omega }^{2}r$

Here, the angular velocity $\omega$ is related to the time period $T$ by the relation:
$\omega =\frac{2\pi }{T}$

Equating the gravitational force to the centripetal force:
$\frac{GMm}{r^2}=m{\left(\frac{2\pi }{T}\right)}^{2}r$

Canceling the mass of the satellite $m$ from both sides and simplifying:
$\frac{GM}{r^2}=\frac{4{\pi }^2}{T^2}r$

Rearranging the equation to solve for the orbital radius $r$:
$r^3=\frac{GM{T}^2}{4{\pi }^2}$

Taking the cube root of both sides (which is equivalent to raising to the power of 1/3, or approximately 0.33):
$r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{0.33}$

Since the height of the satellite above the surface is $h=r-R$, we substitute the expression for $r$:
$h={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{0.33}-R$