What are the radius and height above the ground of a circular geosynchronous orbit around the Earth (in m and Earth radii)? Is this a high or low orbit? How does its height compare with the height of the orbit of the International Space Station (about 360 km)? [HINT: You'll need to derive an algebraic relation between orbit radius and period, instead of just radius and speed.] (b)

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

answer: pseudoscience