A hypothetical Planet X of mass 7.5 ✕ 1023 kg rotates on its axis once every 5 days. Hint (a) Find the distance from the center of Planet X

Question

A hypothetical Planet X of mass 7.5 ✕ 1023 kg rotates on its axis once every 5 days. Hint (a) Find the distance from the center of Planet X (in m) at which a satellite would remain in one spot above the planet’s surface. m (b) Find the speed of the satellite (in m/s). m/s

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Huyền Thanh 3 months 2021-08-06T09:55:19+00:00 1 Answers 14 views 0

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    2021-08-06T09:56:48+00:00

    Answer:

    (A) The distance of the center of Planet X is 6.18 \times 10^{7} m

    (B) The speed of the satellite is 898.38 \frac{m}{s}

    Explanation:

    Given:

    Mass of the planet m = 7.5 \times 10^{23} kg

    Time period T = 5 days = 432000 sec

    (A)

    From kepler’s third law

     T^{2}  = \frac{4\pi ^{2}r^{3}  }{Gm}

    Where G = 6.67 \times 10^{-11}

    Now for finding the distance of planet X,

      r^{3} = \frac{Gm T^{2} }{4\pi ^{2} }

      r^{3} = \frac{6.67 \times 10^{-11} \times 7.5 \times 10^{23 } \times (432000)  ^{2} }{4 (3.14)^{2} }

      r^{3} = 236 \times 10 ^{21}

       r = (236 \times 10^{21} )^{\frac{1}{3} }

       r = 6.18 \times 10^{7} m

    (B)

    Speed of the satellite is given by,

       v = \frac{2\pi r}{T}

       v = \frac{6.28 \times 6.18 \times 10^{7} }{432000}

       v = 898.38 \frac{m}{s}

    Therefore, the speed of planet is 898.38 \frac{m}{s}

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