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I know when cars drive, there is action and reaction pair on the wheel, then is it possible to speed up or slow down the rotation of Earth if all cars and other machines travel towards one direction together?

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    $\begingroup$ It's called a reaction wheel and they are very popular on spacecraft: en.wikipedia.org/wiki/Reaction_wheel $\endgroup$ – CuriousOne Sep 17 '15 at 2:09
  • $\begingroup$ When they all hit the ocean? $\endgroup$ – paparazzo Sep 17 '15 at 2:37
  • $\begingroup$ yes u can.... imagine the mass of all the cars, ships etc to make up a massive sphere which is rolling on earth's surface, then u realize u can. $\endgroup$ – Shubham Sep 19 '15 at 17:48
  • $\begingroup$ I had thought about this kind of ounce if cars traction act as motors, propelling the earth relative to the car. But I would first assume in total, all cars net direction is random and likely mostly cancels out. Additionally if the cars typically return to same spot the force should cancel to the amount of f small differwnces in position where they are parked. However, if cars ultimately drive to one location from another, like junk yard, then there could be more significant effect. However cars are usually transported not by their own power from their initial and final positions. $\endgroup$ – marshal craft Jun 22 '18 at 10:55
  • $\begingroup$ But I suppose their would still be a momentum difference, so I presume approximately the net forces would only come from where cars are manufactured and where they come to rest finally when they are scrapped. If say most cars come from japan, travel east, get on boat and unloaded in LA then transported eastward in united States where they average out to get scrapped in Midwest, then you would have a significant net force. This is likely very small effect but probably unknown it's effects in general. NASA prob studies these things. $\endgroup$ – marshal craft Jun 22 '18 at 11:02
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Earth + machines is an isolated system so angular momentum is conserved \begin{align} I_{\text{Earth}}\omega_{\text{Earth}}^{\text{before}} &= I_{\text{Machine}}\omega_{\text{Machine}} + I_{\text{Earth}}\omega_{\text{Earth}}^{\text{after}}\\ \Rightarrow &\frac{\omega_{\text{Earth}}^{\text{after}}}{\omega_{\text{Earth}}^{\text{before}}} =1 - \frac{I_{\text{Machine}}\omega_{\text{Machine}}}{I_{\text{Earth}}\omega_{\text{Earth}}^{\text{before}}} \end{align} Therefore the percent change in earths angular rotation speed is \begin{align} \frac{\omega^{\text{after}}_{\text{earth}}}{\omega^{\text{before}}_{\text{earth}}}=\frac{I_{\text{Machine}}\omega_{\text{Machine}}}{I_{\text{Earth}}\omega_{\text{Earth}}^{\text{before}}} &=\frac{5}{2}\frac{M_{\text{Machine}}}{M_{\text{Earth}}} \frac{v_{\text{Machine}}}{R\omega_{\text{Earth}}^{\text{after}}} \approx \left(10^{-27} \text{sec}/ \text{kg}\cdot\text{m}\right) \times p_{\text{Machine}} \end{align} In other words assuming all machines are moving at the equator, their momentum must be of the order of $10^{27}\text{kg}\cdot\text{m}/\text{sec}$ to get a $1\%$ change in earths angular speed, i.e. to increase/decrease the day by around a quarter of an hour.

In comparison notice that if all the cars on earth where to move at a very illegal speed, their combined momentum wont exceed $10^{14}\text{kg}\cdot\text{m}/\text{sec}$ which could only change the length of the day by less than $10$ nano seconds...

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