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Two satellites are monitored as they orbit the Earth; satellite X is eight (8) times as far from the Earth's center as is satellite Y. Using Kepler's 3rd Law, by what factor is the the period (or revolution) of satellite X that of satellite Y

Respuesta :

Answer:

  [tex]\frac{T_x}{T_y} = 22.63[/tex]

Explanation:

Kepler's third law is an application of Newton's second law for circular motion

          T² = K a³

let's apply this expression for each satellite

satellite X

            Tₓ² = K aₓ³

satellite Y

            [tex]T_y^2 = K a_y^3[/tex]

 the relation of the periods is

          [tex]\frac{T_x}{T_y} = \sqrt{ (\frac{a_x}{a_y} )^3 }[/tex]

they indicate us

            aₓ = 8 a_y

substitutes

           [tex]\frac{T_x}{T_y} = 8^{3/2}[/tex]

           [tex]\frac{T_x}{T_y} = 22.63[/tex]

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