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Does constructing huge buildings affect the rotation of the Earth, similar to skater whose angular rotation increases when her arms are closed comparatively than open?

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I appreciate your question - Yes, but probably not for a few centuries before it becomes significant ;-) –  Waffle's Crazy Peanut May 25 '13 at 14:41
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Definitely. I wonder though how much more we're slowing down the Earth by burning oil. We take it from deep underground and put most of it way high up in the atmosphere. Surely that slows down the Earth more than buildings do. –  Brandon Enright May 25 '13 at 16:02
    
@BrandonEnright But in that case there is no bulk motion, since the atmosphere's density stratification is in steady state. For every amount of mass you send up, an equal amount comes down. –  Chris White May 25 '13 at 19:44
    
@ChrisWhite That thought occurred to me but I wasn't sure if that happens or not. Sounds like I need to read a bit and then possibly ask a question. –  Brandon Enright May 25 '13 at 19:49
    
Related: physics.stackexchange.com/q/56245/2451 and links therein. –  Qmechanic May 26 '13 at 14:58
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2 Answers

From conservation of angular momentum we have $(I+\Delta I)(\omega+\Delta \omega) = I\omega,$ or $$\frac{\Delta \omega}{\omega} = - \frac{\Delta I}{I+\Delta I} \simeq -\frac{\Delta I}{I}.$$ We make the following simplifying assumptions:

  • The earth is a sphere of uniform density of mass $M$ and radius $R$.

  • The building is constructed on the equator by digging out a sphere of earth of mass $m$ and radius $r$ and raising it a distance $2r$. We assume $r\ll R$.

With these assumptions we find $$\begin{eqnarray*} \frac{\Delta \omega}{\omega} &\simeq& - \frac{m(R+r)^2-m (R-r)^2}{\frac{2}{5}M R^2} \\ &\simeq& - 10\frac{r^4}{R^4}. \end{eqnarray*}$$

Assuming that $r$ is 200 m (the geometric mean of 100 m, 100 m, and 750 m) we find
$$\begin{eqnarray*} \frac{\Delta \omega}{\omega} &\simeq& -10^{-17}. \end{eqnarray*}$$ Atomic clocks are accurate to about one part in $10^{14}$, so there is no hope in measuring such an effect.

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Well, if we make a quick estimate of the mass of a huge building.

Let's say the building has a base of $100\times100 \;\text{m}^2$ and a height of $1500 \;\text{m}$, this is already substantially bigger than the current biggest building. Then we have a volume of $1.5\times 10^7\text{m}^3$. If we make the assumption, again very rough and on the high side, that the tower is solid concrete with a density of $2400\; \text{kg}/\text{m}^3$ then the total mass is $3.6\times 10^{10}\;\text{kg}$.

If we compare this to the mass of the earth which is $\approx 6\times 10^{24}\;\text{kg}$ you can already see that it is unlikely to have much influence.

Now since we're dealing with rotation we should actually look at the center of mass of the tower. The CM will be located at half the height in our example so at only $750\;\text{m}$ height. Now this is significantly lower than the average mountain so I think it is safe to say that the effect of tall buildings on the rotation of the earth is negligible.

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This building would probably lengthen a year by a few nanoseconds. With the accuracy of atomic clocks I wonder if it'd be enough to measure? –  Brandon Enright May 25 '13 at 16:01
    
It is all about scale. Building may seem big. Even Huge. But even from the airliners fly they start to get very small, and by the time you can encompass a large part of the Earth in your view (i.e. from low Earth orbit) they disappear. –  dmckee May 25 '13 at 16:19
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@BrandonEnright compared to the mass of rain in a tropical storm it is tiny so it would be lost in the random noise of earth's rotation changes –  Martin Beckett May 26 '13 at 0:35
    
Just comparing masses isn't informative. Neither is comparing to a mountain. The building represents a change in mass distribution, while mountains generally don't move on the same time scale. –  DarenW May 26 '13 at 6:14
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protected by Qmechanic May 26 '13 at 14:55

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