Answer: radius r = 42360.7km
Height above ground = 35950.7km
The height of the satellite above the ground is about 100 times the height of the ISS above the ground.
This is a high orbit.
Explanation: a synchronous satellite of mass m, revolving around earth with angular speed w, having a radius of travel r will experience centripetal force F = m*r*w^2*
But w = 2¶/T
F = m*r*(2¶/T)^2
F = (4*m*r*¶^2)/T^2
For the same body on the surface of the earth of radius R, the force F will be F =mg
According to newton's law,
(4*m*r*¶^2)/T^2 is proportional to 1/r^2
also mg is proportional to 1/R^2
Therefore,
(4*m*r*¶^2)/T^2 = K/r^2,
mg = K/R^2
Equating the two we get
K = gR^2 = (4*r^3*¶^2)/T^2 (where K is a constant equal to the product of mass of earth M and gravitational constant.)
r^3 = (g*R^2*T^2)/(4x3.142^2)
Substituting values of g=9.81m/s2
R = 6400000m (radius of earth)
T = 60x60x24 = 86400s (synchronous orbit has period equal one day)
r = 42350775.04m = 42350.7km
Height above ground H = r - R
H = 42350.7 - 6400 = 35950.7km
Please verify with calculator. Thanks
