Asatellite of mass 5500 kg orbits the earth (m circular motion and has a period of 6200 s. determiner the magnitude of the earth's gravitational force on the satellite. show your working.

The earth's gravitation force on the satelite is 41193.51 N.

Explanation:

if v is the orbital speed of the satelite and r is the radius of orbit of the satelite then, the period of a circular motion is given by:

T = 2πr/v

T^2 = [4×(π^2)×(r^2)]/(v^2)

v^2 =  [4×(π^2)×(r^2)]/(T^2)

but if G is gravitational constant and M is the mass of the earth then, the orbital speed of the satelite is given by:

v = √[(G×M)/r]

v^2 = G×M/r

then :

G×M/r = [4×(π^2)×(r^2)]/(T^2)

    r^3 = [G×M×(T^2)]/(4×(π^2))

          =  [(6.67×10^-11)×(5.972×10^24)×((6200)^2)]/(4×(π^2))

          = 3.8785×10^20

        r = 7292723.678 m

the Earth's gravitation force on the satelite is given by:

Fg = G×M×m/(r^2)

    = (6.67×10^-11)×(5.972×10^24)×(5500)/((7292723.678)^2)

    = 41193.51 N

Therefore, the earth's gravitation force on the satelite is 41193.51 N.

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop