Asatellite is orbiting at a height of about 650 km above the earth's surface. the mass of the earth is 5.976 x 10^24 kg and the radius of the earth is 6.378 x 10^6 m. a) assuming a circular orbit, what is the speed of the satellite in its orbit? b) assuming a circular orbit, what is the period of the satellite's orbit?

a) v = 7533.3 m/s

b) T = 54677.0 seconds

Explanation:

let Fc be the centripetal force and Fg be the gravitational force.

m be the mass of the satelite and M be the mass of the earth.

a) A satelite in orbit experiences two forces, centripetal force and gravitation force due to the center of mass.

this forces are balanced, meaning at each point in the orbit:

Fc = Fg

m×v^2/r = G×m×M/(r^2)

because r is the same and m is the same:

v^2 = G×M/r

      = [(6.67408×10^-11)×(5.976×10^24)]/(6378×10^6 + 650×10^3)

      = 56750572.11

then v = \sqrt{56750572.11} = 7533.3 m/s.

therefore, the speed of the satelite in it's orbit will be 7533.3 m/s.

b)  for a given circular orbit, the satelite will cover a displacement of 2πr in speed of 2πr/T where T is the period of the orbit:

then.

2πr/T = \sqrt{G×M/r} = v

T = 2πr/(v)

  = 2π(6378×10^6 + 650×10^3)/(7533.3)

  = 54677.0 seconds

therefore, it will take the satelite 54677 secords to orbit earth.