Shrinking it
If Earth somehow contracted to 8 1 of its current volume instantly (through some magical internal forces), while keeping its shape and mass, then how long would a day be?
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2 solutions
Why would the Earth spin faster ? I don't understand
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It's because angular momentum is conserved, so the smaller Earth had the same angular momentum as the larger Earth. With mass held constant, a smaller radius leads to a smaller moment of inertia, so it has to spin faster for angular momentum to be conserved.
Similar to how a skater spinning on ice with his arms out spins faster if he pulls his arms close to him.
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True. They never said it would be spinning faster as by "magic" ...
kinetic energy is not conserved?
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No, as the work done in contracting the earth contributes to an increase in rotational kinetic energy.
This work has to be done by an external agent otherwise the kinetic energy of the earth would decrease
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Its similar to how you use the energy inside of you to increase your kinetic energy. The earth does work to contract itself, so that work should be put to use to something, so the kinetic energy increases. If work is done by an external agent neither angular momentum nor K.E would be conserved.
Why won't kinetic energy be conserved?
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There is work done on the system by the internal forces displacing the mass across half the radius, as when a rotating skater draws her arms in to her body and rotates faster.
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Thanks Bob. Now it makes sense.
I'm afraid it does not make sense to me: "work done by internal forces" sounds like "internal work" wich, by definition, does not change total energy. Is there some potencial energy here somewhere? I can't "see" it.
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@Rui Veríssimo – As I understand it, there would be "some" other not-mechanical energy inside the Earth. We could imagine that that force is nuclear energy that converts (by some not-understood-so-far process) some of it into work to bring the mass more compressed than it is currently. Another possible source of energy: there would be some gravitational force (not developed so far) that potentially and eventually could "crush" the ground to a new more compressed state. The "work" would be made by transforming this potential (=gravitatory) energy, that was not mechanical, into mechanical.
This implies angular kinetic energy is four times larger. No one said we were adding energy to the system. Magic?
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Some work is done in contracting the earth which leads to an increase in kinetic energy, however the forces that cause this work to be done are completely internal, thus there is no net torque on the system, conserving angular momentum.
Can we get answer by Kepler's 3rd law?
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Keplers 3rd law deals with time periods of revolutions , not rotations. So I don't think that should help
What if the earth contracted to half of its current volume instantly ......then what will be the answer??
How much would the gravity increase on earth due to the increased rotation?
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4 times (mass stays same, half radius squared) due to shrinkage. Rotation doesn’t affect gravity
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Rotation doesn’t affect gravity, right, but there would be some anti-gravity effect on the objects on Earth due to that increased angular speed, even if that anti-gravity effect is negligible against the 4 times increment of gravity.
Is thre an intuitive way of knowing that angular momentum is proportional to radius squared? Rather than just stating it...
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Linear momentum was mass × velocity. Every kinetmatics-related property in linear motion has an equivalent in rotational motion. For example, Force ~ torque, displacement ~ angular displacement ( θ ), velocity ~ angular velocity w , mass ~ moment of inertia.
Hence angular momentum = (moment of inertia) × (angular velocity).
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Thanks Parth...I understand the logic, but I’d love to be able to intuitively “get” it!
I "second" Bob Lord: if Kinetic Rotacional Energy is 2 1 I ω 2 and I , for a sphere, is proportional to r 2 ( I = 5 2 M r 2 ), then if we halve the radius, the Rotational Momentum I gets smaller by 4 1 . If the (Kinetic Rotational) Energy is constant, then the term w 2 must get bigger by 4 , and so w must get bigger by 2 and not by 4 as stated in the solution. End result is a 12 hours day and not a 6 hours day. Please do correct me if I'm wrong, because I don't see where this reasoning is at fault.
I just used V=(4/3)•pi•r^3 and multiplied both sides by 1/8, which gives (1/8)V=(4/3)•pi•((1/2)r))^3. The radius is multiplied by 1/2 so the circumference would be C=2•pi•(1/2)r or C=pi•r. The circumference is halved and so should the rotation be. Thus a day would be 24/2=12 hours. Why is this wrong?
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this captures the reduced travel distance at the surface (circumference), but neglects the increased surface velocity
Are we conserving energy or angular momentum? The question is unclear.
If angular momentum is conserved then the overall energy associated with the rotation is also conserved..or am I wrong?
It can be done in another way. From the law of conservation of angular momentum; I1w1 = I2w2 (where I1 and I2 represent moment of inertia and w1 and w2 represent angular velocities) Now since volume becomes 1/8th of whole, 4/3 π r3 = (1/8) 4/3πR3 Or r = 1/2 R; Now Moment of inertia of sphere equals 2/5 M R2 and angular velocity equals 2π/T ; 2/5MR2 * 2π/ T1 = 2/5M*(R2/4) * 2π/T2 Or T2 = 1/4T1 (T1 = 24hrs) or T2 = 6hrs
I solved the problem through the Universal Gravity Law, resulting that T(final)^2=1/8*T(initial)^2. The result in my case was approximately 8.49 hours per day. Edit: I think the statement in this problem was wrongly read and solved because it says that the VOLUME shrinks 1/8 from the Earth’s original one, but it hasn’t ever been said that the RADIUS is 1/8 from its original one. Please take it in mind next time.
Volume goes as R 3 , so R will have to halve. Angular momentum ( I ω ) should be conserved where I is proportional to m R 2 , so ω must get four times als large as it was, changing a day to 4 2 4 = 6 hours.
and I note that the length of an earth year does not change since this is only dependent on (i) G (ii) Mass of Sun (iii) Distance between centres of earth and sun. And we finish up with 365x4 earth days in an earth year.
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Hi I got the same answer but I think I must be doing something wrong.I claimed the rotational energy would stay constant 1/2 Iw^2. I integrated the moment of inertia for each in shells with the second interval going to RE/2 and density = 8x. No I made a mistake. I did not take the sqrt.
Revised question: what would be the length of a day if (a) radius decreased by 50%, without any change in mass, ; and (b) the centripitol force on the surface reduced by 50%, only as a result of increased earth rotation (the angular velocity it produced to counter gravity)?
@Alexander Best These questions are very similar, use the same ideas!.
Since only internal forces are shrinking it, the angular momentum of Earth will remain constant.
Volume is proportional to r 3
Therefore, 8 1 t h volume = ( 8 1 ) ⅓ r = 2 1 r
Angular momentum (L) = I ω , where I is the moment of inertia and w is the angular velocity
I α m r 2
If radius is halved, I will become ¼th, thus w should become 4 times.
Hence, the earth will spin 4 times as fast, making the time of a day = 4 2 4 = 6 hours.