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Altitude and speed of a geostationary satellite

작성자 Uploader : alice 작성일 Upload Date: 2023-05-18변경일 Update Date: 2023-05-18조회수 View : 1771

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The gravitational force and the centrifugal force acting on a geostationary satellite are the same.

mg = mv^2/r

m : satellite mass (kg)
g : acceleration of gravity (m/s^2)
v : satellite speed (m/s)
r : distance from the center of the earth to the satellite (m)

Substituting g = G*M/r^2 into the above equation,

G*M/r^2 = v^2/r

G : gravitational constant, = 6.673*10^(-11) (m^3/(kg*s^2))
M : mass of the earth, = 5.972*10^24 (kg)

and summed up as follows:

r = G*M/v^2

v = 2πr/t

t : sidereal time, the time it takes for a satellite to orbit the earth once, 86164 (s)

v^2 = 4*π^2*r^2 / t^2

r = G*M*t^2/(4*π^2*r^2)

r^3 = G*M*t^2/(4*π^2)

r = (G*M*t^2/(4*π^2))^(1/3)

If the coefficients of G and M are Gn and Mn, respectively,

r = (Gn*Mn*10^(13)*t^2/(4*π^2))^(1/3)

Subtracting the radius of Earth R = 6378 km gives the altitude h (km) of the satellite.

h = r/1000 - R

The speed of satellite is

v = 2*π*r/t
▶ distance from the center of the earth to the geostationary satellite (m)




▶ altitude of the satellite (km)


▶ speed of satellite (km/s)


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