A satellite 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.