Through the use of re-arranging the above equation, we can come to the equation:

r³ = G (m2) T² / 4π²

We know that (m2) is the mass of the earth at 5.98×10^24 kg, T is the time period and G the universal gravitation constant at 6.67 x10^-11 kg^-2 .

## Radius Of A Geostationary Orbit

We know every bit of information in the above equation to work out the radius of a geostationary orbit. The time period will be 24 hours which is 86400 seconds. Therefore, for a geostationary orbit,

r = 4.23×10^7 metres.

However this is the radius to from the **center of the Earth**. Therefore, we will need to deduct the radius of the Earth from this number: the height of the satellite from Earth = r – r(E) where r is the distance of the satellite from the center of the Earth and r(E) is the radius of the Earth.

From this, the height of a geostationary orbit above the earth is 3.6×10^7 meters, with a radius of orbit of 4.23×10^7 m (from the center of the earth).

## Summary

- A geostationary orbit is an orbit which is fixed in respect to a position on the Earth. Therefore, the time period will always be 24 hours.
- From combining the centripetal force, gravitational force and basic velocity force equations, we can deduce that the radius required for a geostationary orbit is 3.6×10^7 meters from the surface of the earth.

Hi Will,

I really like the clear explanations on your website. As a Physics teacher however, I would argue that the *radius* of orbit is the first number you calculate: 4.23×10^7 m whereas 3.6×10^7 m is the height of the satellite above the Earth’s surface. I don’t think this can be called a radius as it is not the distance to the centre of the circle that its motion forms.

I hope this will help students avoid some confusion in the future.

Chris

I enjoyed your article very much but I feel I should point out that your result for the radius (42,300 km) is in error. This is because for the time “T” you used the solar day (24 hrs) instead of the sidereal day which is 23 hrs. 56 min. 4 sec. The sidereal day is the actual time it takes the earth to rotate once on its axis and is therefore the period “T” that the geostationary orbit must match. The solar dday adds almost 4 minutes to compensate for the earth’s movement around the sun. This is so that the sun reaches its highest pt. In the sky at exactly noon every day. Plug in 86,164 sec. For “T” and that should give you the correct value for “r” which is 42,164 km.

Hi Alfonso – good spot! With the article, I rounded to make the numbers nide. However, for more significant figures your answer is better!