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28 September, 17:11

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.

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  1. 28 September, 18:26
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    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.
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