Question

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

Answer 1

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=(GMm)/r^2`.

Where `G=6.67384m^3/(kg*s^2)` is the gravitational constant, `M=5.972*10^24kg` 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 * r * omega^2`.

Where `m` and `r` are as above and

`omega = (2pi)/(24*60*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, `2pi` radians, in 24 hours.

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

`(GMm)/r^2=mromega^2`

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

`r^3=(GM)/(omega^2)`.

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

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

`h=((GM)/(omega^2))^(1/3) - R`

If the stated values of G, M, `omega` and `R` are put into the formula it gives a value of about `"35,870,000 m"`.