A satellite orbits the earth at a distance of 1.50 × 10^{7} m above the planetʹs surface and takes 8.65 hours for each revolution about the

Question

A satellite orbits the earth at a distance of 1.50 × 10^{7} m above the planetʹs surface and takes 8.65 hours for each revolution about the earth. The earthʹs radius is 6.38 × 10^{6} m. The acceleration of this satellite is closest to

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Trúc Chi 5 months 2021-08-20T17:32:47+00:00 1 Answers 16 views 0

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    2021-08-20T17:34:10+00:00

    Answer:

    a = 0.43 m/s²

    Explanation:

    First, we need to find the velocity of the satellite:

    Velocity = V = Distance Covered/Time Taken

    here,

    Distance = 1 revolution = 2π(1.5 x 10⁷ m) = 9.43 x 10⁷ m

    Time = (8.65 hours)(3600 s/1 hour) = 31140 s

    Therefore,

    V = (9.43 x 10⁷ m)/(31140 s)

    V = 3028.26 m/s

    Now, the acceleration of the satellite will be equal to the centripetal acceleration, with the center of circular motion as the center of earth:

    a = V²/R

    where,

    R = 1.5 x 10⁷ m + 0.638 x 10⁷ m

    R = 2.138 x 10⁷ m

    Therefore,

    a = (3028.26 m/s)²/(2.138 x 10⁷ m)

    a = 0.43 m/s²

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