# Are the value of velocity and height constant for geostationary satellite?

We know that generally for geostationary station T=24 hours, h=3.6x10⁴km, R=6400 km, and velocity is 3.1 km/s. I think this values are stationary for geostationary satellite.

Let we want to place a geostationary satellite for a height greater or less than h= 3.6x10⁴ km. Is it possible to place a geostationary satellite at a height less or more than 'h'? If it possible, how it can be done? Which steps should be taken?

The main condition for a geostationary satellite is that its time period should be the same as that of Earth. Thus, the circle that it makes around Earth (considering its orbit to be perfectly circular) should be covered in 24 hours.

Thus, we get the relation: $$24 hours = \frac{2\pi(R + r)}{v}$$

Where $R$ is the radius of the Earth, $r$ is the height of the satellite, and $v$ is the velocity of the satellite.

So now we have two variables to solve for.

Now consider the fact that the satellite is in rotational motion around the Earth. Thus, it must experience a centrifugal force, which is pushing it outwards. This centrifugal force must be balanced by the inward gravitational pull of the Earth.

The centrifugal force experienced is given by: $mv^2\over{R+r}$

The gravitational force experienced is given by: $GmM\over{R+r}^2$

Where $m$ represents the mass of satellite, $M$ represents mass of Earth, and $G$ represents the universal gravitational constant. Equating the two, we get:$$v^2 = \frac{GM}{R+r}$$

This along with the other equation can be used to solve for $r$ and $v$. Upon solving, we only get the values that you have given. Thus, it is not possible for a geostationary satellite to be placed at any other height or with any other velocity.