We walk or ride on our vehicles to our destinations daily. Does our movement have any effect on the rotation of the earth according to Newton's law? What will be the effect if we move all the peoples along with their vehicles at their maximum velocity in one line in one direction along the equator? How much effect will it make?

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    $\begingroup$ The BBC documentary "super sized earth" claims that all the river dams that we built have changed the rotation speed of the earth and changed the length of a day with "a fraction of a second" $\endgroup$ Commented Mar 8, 2013 at 6:53
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    $\begingroup$ See xkcd.com/162. $\endgroup$
    – joriki
    Commented May 29, 2016 at 23:56
  • $\begingroup$ related: physics.stackexchange.com/q/6400 $\endgroup$
    – njzk2
    Commented Jul 7, 2016 at 16:01

3 Answers 3


In principle, yes, but the effects are almost completely negligible. As objects on the surface of the earth move around, the Earth's moment of inertia changes by minute amounts, and this affects its rotation. However, performing an order of magnitude estimate on the ratio of the contribution to the moment of inertia of a person-sized object on the equator to the Earth's moment of inertia gives $$ \frac{I_p}{I_E} \sim \frac{M_pR_E^2}{M_ER_E^2} \sim \frac{10^2\,\mathrm{kg}}{10^{24} \,\mathrm{kg}}\sim 10^{-20}, $$ and even if we were to take into account the entire population of the Earth plus their vehicles (by conservatively multiplying the numerator by $10^{11}$ or so), we would still get a number on the order of $10^{-10}$. We can thus see that such small objects on the surface will have an entirely negligible effect on the Earth's rotation (at least in the sense that our motions have a negligible affect on changing the Earth's moment of inertia.)

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    $\begingroup$ not to mention that our movement on the Earth is essentially random and counteracting, your answer makes the worst case that we all organise and run in the same direction... $\endgroup$
    – Nic
    Commented Mar 8, 2013 at 12:51
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    $\begingroup$ Maby as a followup question: Could one or more of our strongest nuke's make a significant change? $\endgroup$
    – user17615
    Commented Mar 8, 2013 at 13:17
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    $\begingroup$ @MichielT Well according to hypertextbook.com/facts/2000/MuhammadKaleem.shtml the energy released in a nuclear blast is on the order of tens of thousands of terajoules. Let's be extremely generous and say that we could produce a nuclear explosion that releases $10^{20}\mathrm J$. The rotational energy of the Earth is on the order of $I_E \omega^2 \sim 10^{28}\,\mathrm J$ so the energy of the blast has around a factor of $10^{8}$ less energy. $\endgroup$ Commented Mar 8, 2013 at 17:06
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    $\begingroup$ (contd.) Even if every bit of this energy were used to slow down the Earth's rotation, the day would only lengthen by about one tenth of a millisecond (I did a very rough calculation to determine this that I can post if you want.) $\endgroup$ Commented Mar 8, 2013 at 17:06
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    $\begingroup$ @MichielT You inspired me to write this blog post joshphysics.com/2013/03/10/… $\endgroup$ Commented Mar 11, 2013 at 22:38

It does, but changes are so miniscule that they don't really count.

Even if everyone in the world started running in the same direction (East, for example), their total angular momentum would be on the order of $ 10^{18} \ \text{N m s} $, in comparison to the earth, which is on the order of $ 10^{33} \ \text{N m s} $. We would change the earth's rotation by less than a trillionth of a percent.

Some crude calculations:

$\text{Population of the earth} = \text{about 7 billion}$

$\text{Average human mass} = 62 \ \text{kg}$

$\text{Average running speed} = 5 \ \text{m/s} $ (ish)

$\text{Mass of everyone in the earth} = 4.3 * 10^{11}\ \text{kg} $

$\text{Radius of the earth} = 6.37 * 10^6\ \text{m}$

$\text{Mass of the earth} = 6 *10^{24}\ \text{kg}$

$\text{Angular velocity of the earth (its spin)} = 7.29 * 10^{-5}\ \text{s}^{-1}$

$\text{Angular velocity of the human "race"}= 7.8 * 10^{-7}\ \text{s}^{-1}$

$\text{Their moment of inertia, assuming spherical distribution}= 1.2 * 10^{25}\ \text{kg m}^2$

$\text{Their angular momentum:}\ 9.2 * 10^{18}\ \text{kg m}^2\ \text s^{-1}$

$\text{Moment of inertia of the earth} = 9.7 * 10^{37}\ \text{kg m}^2$

$\text{The earth's angular momentum:}\ 7.1 * 10^{33}\ \text{kg m}^2\ \text s^{-1}$

  • $\begingroup$ Can you also calculate how much longer or shorter the day would get? :) $\endgroup$
    – Bernhard
    Commented Mar 8, 2013 at 6:39
  • $\begingroup$ You calculated it for human's but what if we ran all the vehical's and those don't have vehical's on their foot in one direction $\endgroup$
    – Deiknymi
    Commented Mar 8, 2013 at 8:07
  • $\begingroup$ Yeah, I kinda forgot about that. I suppose we might get up to around a billionth of a percent. And @bernhard I believe that the fractional change in the length of a day would be the same as the fraction of our angular momentums, so it would be on the order of nanoseconds. $\endgroup$
    – krs013
    Commented Mar 8, 2013 at 13:10

The major external effect slowing the earth's rotation is interaction with the moon's orbital motion through tidal friction. It is possible that damming of water in major areas of tidal friction could change the rate of slowing.

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    $\begingroup$ Comment to the answer (v1): Speaking as a user (as opposed to a moderator), you really should consider adding a rough back-of-the-envelope calculation to estimate the effect, or perhaps point to a reference. $\endgroup$
    – Qmechanic
    Commented Jul 2, 2013 at 0:06

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