The mass of a star is 1.490×1031 kg and it performs one rotation in 22.50 days. Find its new period (in days) if the diameter suddenly shrinks to 0.810 times its present size. Assume a uniform mass distribution before and after.
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Assuming that it shrinks in a way that doesn't change its mass or its angular momentum, those will be conserved, and,
L = I₁ω₁ = I₂ω₂
Now the anguar velocity, ω, is inversely proportional to the rotation period, T, and with the mass held constant, the moment of inertia, I, of a uniform-density sphere, is proportional to the square of its radius, r, so
r₁²/T₁ = r₂²/T₂
T₂ = (r₂/r₁)² T₁ = (0.810)²•22.50 d = 14.76 d
L = I₁ω₁ = I₂ω₂
Now the anguar velocity, ω, is inversely proportional to the rotation period, T, and with the mass held constant, the moment of inertia, I, of a uniform-density sphere, is proportional to the square of its radius, r, so
r₁²/T₁ = r₂²/T₂
T₂ = (r₂/r₁)² T₁ = (0.810)²•22.50 d = 14.76 d
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BTW, the assumptions of conserved mass and angular momentum, follow if you merely assume no interactions with outside objects during the shrinking.
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STARS DON'T HAVE VAGINAS! So they can't have periods, you dolt.