Earth's rotation slows down slightly over time -- How much energy is lost?

In summary: We want the energy to change by ##\delta T##, not ##m\delta T##. So we need to use the chain rule: ##E=mT+c##, which is the same as saying ##E=mT-c##.
  • #1
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Homework Statement
The rotation slows down slightly, about 2.5ms increase for a full rotation after 100 years. Calculate Hoe much rotation energy is lost during 100 years?
Relevant Equations
inertia, I=2/5 mr².
Inertia I = 2/5 mr² m=5.98* 10^24 kg.
r=6.38* 10^6 kg.

I= 9,736 * 10^37 kg.

Earth rotation is V= (2 * pi* R)/T = (2*pi*6.38*10^6m)/(24*3600s)=463 m/s
angular velocity w= V/r =463m/s/(6.38*10^6m)=7.27*10^-5 rad/s
Enerrgy, E= 0.5* I*w²=2.57*10^29 j
I get the same value when period is 24h+2.5ms. How to I calculate energy loss due to Earth rotation slows down.I don't understand how do I get the value of Energy loss? The change of
 
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  • #2
You have made several mistakes with units and use of standard form (scientific notation). Can you spot them?

The difficulty is that you are dealing with a small difference between two very similar values.

E.g.
x = 1234567890123456789
y = 1234567890123456788
x – y = 1

But if you are trying to work out x – y and you only use x and y to 3 significant figures:
x = 1.23x10¹⁸
y = 1.23x10¹⁸
then x – y will give you zero because of the lack of precision.

The answer is to use some simple calculus.

##E = \frac 1 2 I \omega^2## (which you used in your method)
and
##\omega = \frac {2\pi}{T}##

Can you find ##\frac {dE}{dT}## and take it from there?

EDITED. Example changed. Errors correct.
 
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  • #3
I get the value -5.959 *10^24 j. Is it lost energy? From my point of view that it is lost of energy. what does it mean that energy is negative?
 
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  • #4
robax25 said:
I get the value 5.959 *10^24 j. Is it lost enegry?
EDIT, Sorry, my mistake. It might be correct - how did you calculate it?
 
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  • #5
yes I get it. The energy that I get after rotation have changed and the lost energy would be 2.56*10^29J. But it is almost same if rotation is 24 hour.If period is 24 h, than the energy is 2.57*10^29j.Lost Energy = (2.57*10^29J)- (5.959*10^24J) =2.5699*10^29j
 
  • #6
robax25 said:
yes I get it. The energy that I get after rotation have changed and the lost energy would be 2.56*10^29J. But it is almost same if rotation is 24 hour.
Yes, the energy with T = 24 hours and with T = 24hours+2.5ms are very similar. But the difference (energy lost) can be calculated.

Have you read and understood my Post #2?
 
  • #7
I read it but I have already calculated the value. However, I get the same value. If I miss something, please indicate me.
 
  • #8
robax25 said:
I read it but I have already calculated the value. However, I get the same value. If I miss something, please indicate me.
EDIT. Sorry I misunderstood and have changed this Post. I will post again when I have looked at your figures more carefully.
 
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  • #9
Sorry about errors in previous posts.

As far as I can see, you have calculated the energy-loss to be 5.959*10^24J.

You have not shown your working so I can't check. I get a different value - of the order 10^22J.
 
  • #10
The calculation procedure: If I do derivative of the Energy dE/dt= .5 * I * 4pi^2*(-2)(1/T^3)

I= 9.736* 10^37 kg-m²

dE/dt=.5 * I* 4*pi²*(-2)*(24*3600s)^-3 =-5.959*10^24 J
 
  • #11
robax25 said:
The calculation procedure: If I do derivative of the Energy dE/dt= .5 * I * 4pi^2*(-2)(1/T^3)

I= 9.736* 10^37 kg-m²

dE/dt=.5 * I* 4*pi²*(-2)*(24*3600s)^-3 =-5.959*10^24 J
Upper case for period. Not dE/dt, but dE/dT.

What units does dE/dT have? (it's not J!)

What does dE/dT mean? Answer: it is the rate of change in energy with respect to the period.

If we consider a very small finite change of period δT, we can write
δE/δT = 5.959*10^24

We can now find the energy-change (δE) resulting from a small period change (δT).

Can you complete the problem? I've got to go out soon so may not be able to reply for a while.
 
  • #12
should I multiply with 2.5 ms.? Then, I get the result.
 
  • #13
I guess the question is, do you understand why multiplying by 2.5ms gets you the answer?
 
  • #14
because the rate of energy per second.
 
  • #15
Ibix said:
I guess the question is, do you understand why multiplying by 2.5ms gets you the answer?
I was about to post a full explanation, but in view of this question, I reconsidered and provide a hint instead: Chain rule.
 
  • #16
robax25 said:
because the rate of energy per second.
Kind of.

The point is that if you imagine drawing a graph of ##E## as a function of ##T## it'll be ##E\propto 1/T^2##, right? We can look on the graph and see what energy the Earth has if the period is ##T_0##, and we want to know how much that energy changes by if the period increases a small amount ##\delta T##. If the graph were a straight line, ##E=mT+c##, it'd be easy - the change would be ##\left(m(T_0+\delta T)+c\right)-\left(mT_0+c\right)##, which is ##m\delta T##. Now our function isn't a straight line, but if ##\delta T## is very small, it's very, very well approximated by one. So for any line, the change would be ##\approx m\delta T##. But what's ##m##? It's the gradient of the straight line approximating the curve - so it should be the gradient of the curve. So the answer you are looking for is ##\frac{dE}{dT}\delta T##, plus some very small correction because the line isn't actually straight.

That's why you multiply by ##\delta T##. Keep the trick in mind - for any function ##f(x)## a small change ##\delta x## gives a change ##\delta f=\frac{df}{dx}\delta x##, which is further generalised as the chain rule @kuruman mentions.
 

1. How does Earth's rotation slow down over time?

Earth's rotation slows down due to the tidal forces exerted by the Moon and the Sun. These forces cause a slight bulge on the Earth's surface, which creates friction and slows down the rotation.

2. How much does Earth's rotation slow down each year?

The Earth's rotation slows down by about 1.4 milliseconds per day every 100 years. This may seem like a small amount, but it can add up over time.

3. What is the impact of Earth's slowing rotation?

The slowing of Earth's rotation has a minimal impact on our daily lives. However, it can affect navigation systems and the length of our days, as well as the Earth's climate and weather patterns.

4. Is Earth's rotation expected to slow down indefinitely?

No, Earth's rotation is not expected to slow down indefinitely. As the Earth's rotation slows, the Moon moves further away from the Earth, causing the tidal forces to decrease and eventually reach an equilibrium.

5. How much energy is lost due to Earth's slowing rotation?

The amount of energy lost due to Earth's slowing rotation is relatively small. It is estimated that the Earth loses about 3.3 x 10^12 joules of energy per year, which is equivalent to the energy produced by a small power plant in one hour.

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