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

Question

Josiah Adams · Accepted Answer

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

The following formula determines the gravitational force acting on a satellite: #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 formula provides the centripetal force needed to maintain the satellite in orbit. #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. The gravitational and centripetal forces must be equal for a satellite to be in orbit, giving rise to 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#. Thus, the following formula yields the height of geostationary orbit h: #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"#.

Ariana Acosta · Answer

undefined The distance to geosynchronous orbit above the Earth's surface is approximately 35,786 kilometers.