A600 kg satellite is in a circular orbit around earth at a height above the earth’s surface equal to the radius of the earth. what is the satellite’s period of revolution in hours?

Explanation:

given,

mass of the satellite = 600 Kg

mass of earth = 5.98 x 10²⁴ Kg

height of the satellite

     r = 2 R_e

R_e is radius of earth = 6.38 x 10⁶ m

Assuming orbital velocity of the satellite

        v = 5.59 x 10³ m/s

    T = 14342.3 s

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

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