# How do you calculate the distance above the surface of the earth to geosynchronous orbit?

Jan 9, 2016

The height of a geostationary orbit is calculated as the distance required to have an orbital period of 24 hours.

#### Explanation:

The force of gravity acting on a satellite is given by the formula

$F = \frac{G M m}{r} ^ 2$.

Where $G = 6.67384 {m}^{3} / \left(k g \cdot {s}^{2}\right)$ is the gravitational constant, $M = 5.972 \cdot {10}^{24} k g$ is the mass of the Earth, $m$ is the mass of the satellite and $r$ is the distance from the centre if the Earth to the satellite.

The centripetal force required to keep the satellite in orbit is given by the formula

$F = m \cdot r \cdot {\omega}^{2}$.

Where $m$ and $r$ are as above and

$\omega = \frac{2 \pi}{24 \cdot 60 \cdot 60}$

is the angular velocity of the satellite in radians per second. The value of $\omega$ given is the angular velocity required to complete a full orbit, $2 \pi$ radians, in 24 hours.

When a satellite is in orbit the gravitational force must equal the centripetal force which gives the formula

$\frac{G M m}{r} ^ 2 = m r {\omega}^{2}$

The $m$ cancels out and the formula can be rewritten as

${r}^{3} = \frac{G M}{{\omega}^{2}}$.

The distance is from the centre of the Earth so we need to subtract the radius of the Earth $R = 6 , 371 , 000 m$.

So the height of geostationary orbit h is given by the formula:

$h = {\left(\frac{G M}{{\omega}^{2}}\right)}^{\frac{1}{3}} - R$

If the stated values of G, M, $\omega$ and $R$ are put into the formula it gives a value of about $\text{35,870,000 m}$.