Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon, which has b

Question

Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon, which has been named Dactyl. Assume the mass of Ida is 4.4 x 1016 kg, the mass of Dactyl is 2.6 x 1012 kg, and the distance between the center of Dactyl and Ida is 95 km. G = 6.672×10-11 N-m2/kg2 Part Description Answer Save Status A. Assuming a circular orbit, what would be the orbital speed of Dactyl? (include units with

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Thanh Thu 5 years 2021-07-20T14:51:04+00:00 1 Answers 20 views 0

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    2021-07-20T14:52:40+00:00

    Answer:

    The orbital speed of Dactyl is 5.55m/s

    Explanation:

    The orbital speed can be determined by the combination of the universal law of gravity and Newton’s second law:

    F = G\frac{M \cdot m}{r^{2}}  (1)

    Where G is gravitational constant, M is the mass of the asteroid, m is the mass of the moon and r is the distance between them

    In the other hand, Newton’s second law can be defined as:

    F = ma  (2)

    Where m is the mass and a is the acceleration

    Then, equation 2 can be replaced in equation 1

    m\cdot a  = G\frac{M \cdot m}{r^{2}}  (2)

    However, a will be the centripetal acceleration since the moon Dactyl describe a circular motion around the asteroid

    a = \frac{v^{2}}{r}  (3)

    m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}} (4)

    Therefore, v can be isolated from equation 4:

    m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r

    m \cdot v^{2} = G \frac{M \cdot m}{r}

    v^{2} = G \frac{M \cdot m}{rm}

    v^{2} = G \frac{M}{r}

    v = \sqrt{\frac{G M}{r}} (5)

    Finally, the orbital speed can be found from equation 5:

    Notice, that it is necessary to express r in units of meters.

    r = 95km \cdot \frac{1000m}{1km}95000m

    v = \sqrt{\frac{(6.672x10^{-11}N.m^{2}/kg^{2})(4.4x10^{16}kg)}{95000m}}

    v = 5.55m/s

    Hence, the orbital speed of Dactyl is 5.55m/s

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