The moon with orbital radius 384 000 km takes 27 days to complete the orbiting of the earth, determine the period of orbiting for a satellit

Question

The moon with orbital radius 384 000 km takes 27 days to complete the orbiting of the earth, determine the period of orbiting for a satellite at height 500 km from the earth surface.​

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Thành Đạt 6 months 2021-07-16T20:34:36+00:00 1 Answers 6 views 0

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    2021-07-16T20:36:32+00:00

    Answer:

     T = 1,096 10⁵ s

    Explanation:

    For this exercise we can use Kepler’s third law which is an application of Newton’s second law

               F = ma

    force is gravitational force

               F = G m M / r²

    centripetal acceleration

               a = v² / r

    we substitute

              G M / r² = v² / r

              v² = GM / r

    since the orbit is circular, we can use the uniform motion kinematics relations

              v = d / t

    the distance of a circle is

              d = 2π r

              v = 2π r / T

    in this case the time has the name e period t = T

             4π² r² /T² = GM / r

             T² = (4π² /GM) r³

    let’s use the initial data to find the value of the constant

               

    let’s reduce the magnitudes to the SI system

              r₁ = 384000 km (1000 m / 1 km) = 3.84 10⁶ m

              T₁ = 27 day (24 h / 1 day) (3600 s / 1h) = 2.3328 10⁶ s

    we look for the constant

              Ks = 4π² / GM

              T₁² = Ks r₁³

              Ks = T₁² / r₁³

              Ks = (2.3328 10⁶ )² / (3.84 10⁶ )³

              Ks = 9.61 10⁻⁸

    now we can determine the period of the satellite in an orbit of

    r = 500 km = 5 10⁵ m

             T² = 9.61 10⁻⁸  (5 10⁵ )³

             T = \sqrt{ 1.20125 \ 10^{10}}

             T = 1,096 10⁵ s

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