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

1 Answer

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"#.