If earth contracts to half its radius what would be the length of the day at equator?
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Answered by
240
conservation of angular momentum
Iw =constant for solid
2/5 mR²w= constant
wR²= constant
2πfR² = constant
R²/T = constant
so, T is directly proportional to square of radius
Thus if radius is decreased to half then,
T'= T/4
which is 24/4
time period will be 6 hours.
Iw =constant for solid
2/5 mR²w= constant
wR²= constant
2πfR² = constant
R²/T = constant
so, T is directly proportional to square of radius
Thus if radius is decreased to half then,
T'= T/4
which is 24/4
time period will be 6 hours.
Answered by
58
It depends on whether or not Earth’s mass shrinks with its radius. Let’s assume that the mass stays the same, so that the Earth just compresses as its radius shrinks. We’ll also make the simplifying assumption that Earth is a sphere of uniform density.
Angular momentum is given by:
L=I×ω
where I is a quantity called moment of inertia(for some scientific reason ) , and ω is angular velocity (approximately 2π radians per 24 hours for planet Earth).
Angular momentum is conserved, so that:
Li=Lf
Ii×ωi=If×ωf
For a solid sphere, I∝r2
r2×ωi=(r2)2×ωf
4 ωi=ωf
So planet Earth would rotate about four times in 24 hours, reducing our day length to six hours.
Angular momentum is given by:
L=I×ω
where I is a quantity called moment of inertia(for some scientific reason ) , and ω is angular velocity (approximately 2π radians per 24 hours for planet Earth).
Angular momentum is conserved, so that:
Li=Lf
Ii×ωi=If×ωf
For a solid sphere, I∝r2
r2×ωi=(r2)2×ωf
4 ωi=ωf
So planet Earth would rotate about four times in 24 hours, reducing our day length to six hours.
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