In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already

Question

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite, Charon, orbiting at 19,600 km with an orbital period of 6.39 days.

Required:
Assuming that the satellites do not affect each other, find the orbital periods T1 and T2 of the two small satellites without using the mass of Pluto.

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Trung Dũng 5 years 2021-08-11T23:25:10+00:00 1 Answers 136 views 0

Answers ( )

  1. thaiduong
    0
    2021-08-11T23:26:59+00:00
    Reply

    Answer:

    Time period for first satellites 24.46 days and for second satellites 37.67 days

    Explanation:

    Given :

    Distance of first satellites r_{sat1} = 48000 \times 10^{3} m

    Distance of second satellites r _{sat2} = 64000 \times 10^{3} m

    Distance of charon r_{c} = 19600 \times 10^{3} m

    Time period of charon T_{c} = 6.39 days

    From the kepler’s third law,

    Square of the time period is proportional to the cube of the semi major axis.

       T^{2} = r^{3}

       \frac{T}{r^{\frac{3}{2} } } = constant

    For first satellites,

      \frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }

    {T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

    T_{sat1} = 24.46 days

    For second satellites,

       \frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }

    {T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

    T_{sat2} = 37.67 days

    Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

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