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23 June, 12:08

What is the approximate tangential speed of an object orbiting Earth with a radius of 1.8 * 108 m and a period of 2.2 * 104 s?

7.7 * 10^-4 m/s

5.1 * 10^4 m/s

7.7 * 10^4 m/s

5.1 * 10^5 m/s

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  1. 23 June, 15:56
    +1
    If the period is 2.2 x 104 s, then the rotational speed is:

    1 / 2.2x104 = 4.55x10-5 rad/s

    We can solve for the tangential speed using the formula:

    v = ωr

    where ω is the rotational speed

    v is the linear or tangential speed

    r is the radius

    So,

    v = 4.55x10-5 (1.8x108)

    v = 5.1 x 104 m/s
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