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5 April, 22:39

Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Suppose that, at the end of its life, the Sun collapses inward to form a uniform dwarf star that is approximately the same size as Earth. Use the average radius of Earth in your calculations.

Part A

What will the period of the dwarf's rotation be?

T = ?

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  1. 5 April, 22:41
    0
    T = 184 seconds

    Explanation:

    First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

    T = 2π/w

    Where w: angular speed (in rad/s)

    So, let's calculate first the innitial angular speed:

    w = 2π/T

    Converting days to seconds:

    25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

    Then the angular speed:

    w = 2π / 2,194,560 = 2.863x10^-6 rad/s

    Now, the innitial angular momentum is:

    I = (2/5) Mr² replacing dа ta:

    I = 2/5 * (6.96x10^8) ² * M = 1.94x10^17m² * M

    so the initial angular momentum would be:

    L = Iω = 2.863x10^-6 * 1.94x10^17 M

    L = 5.55x10^11 m²/s * M = final angular momentum

    Now the final I = 2/5Mr²

    Final I = 2/5 * (6.37x10^6) ² * M = 1.62x10^13m² * M

    Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

    ω = 3.42x10^-2 rad/s

    Then the new period

    T = 2π/ω = 2*3.14 / 3.42x10^-2

    T = 184 seconds
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