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20 November, 11:07

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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  1. 20 November, 12:30
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    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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