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15 January, 03:02

Consider a neutron star of radius 10 km that spins with a period of 0.8 seconds. Imagine a person is standing at the equator of this neutron star. Calculate the centripetal acceleration of this person

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  1. 15 January, 06:55
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    a = 616850.28 m/s²

    Explanation:

    Given that,

    The radius of the neutron star, r = 10 Km

    = 10,000 m

    The time period of the neutron star, T = 0.8 s

    The centripetal acceleration is given by the formula,

    a = v²/r

    The linear velocity is given by the relation,

    v = rω

    The time taken to complete one complete rotation is given by the relation

    T = 2π / ω

    Where,

    ω = 2π / T

    Substituting v and ω into the equation for centripetal acceleration. It becomes

    a = 4π²r/T²

    Substituting the given values in the above equation

    a = 4π² x 10000 / 0.8²

    = 616850.28 m/s²

    Hence, the centripetal acceleration of this person is, a = 616850.28 m/s²
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