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.

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this