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So, consider the theoretical situation in which the earth is no longer spinning, and we want to spin it up by running in the opposite direction (Yes, this question was inspired by the Nike commercial). Suppose that you have enough people running that you could do it in a reasonable amount of time, and that they could all run at 10 mph indefinitely. Would it be (theoretically) possible for them to get the surface of the earth spinning at a rate faster than 10 mph, or would the angular velocity of the earth stop accelerating once the surface velocity reached 10 mph?

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marked as duplicate by Norbert Schuch, Emilio Pisanty, Sebastian Riese, Jon Custer, sammy gerbil Jun 6 '18 at 20:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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I think there are two possible interpretations of what OP has asked.

  • Runners run 10 mph relative to the surface.

Angular momenta in the system are $$ L_{earth} = -I_E \frac{v}{R}$$ and $$ L_{runners} = N \bar{m} R (-v + \Delta v)$$ Where $I_E = 2/5 M_E R^2$ is the moment of inertia of Earth (around its centre). $R = 6400 km$ its radius and $\Delta v = 10 mph$ the difference. $N$ is the number of people running, $\bar{m}$ their mean mass. $v$ is the unknown surface velocity. We assume all the runners run west-east across the equator. This convention is why $L_{earth}$ is negative. The conservation of angular momentum $$ L_{earth} + L_{runners} = 0$$ yields $$ v = \frac{1}{\frac{I_E}{N \bar{m} R^2} + 1} \Delta v$$ Therefore only when the moment of inertia of the runners $N\bar{m} R^2$ becomes comparable to the moment of inertia of Earth, can the Earth really speed up. Notice that no matter the combined moment of inertia of all runners $N \bar{m} R^2$, the velocity cannot exceed $\Delta v$, as OP suspected.

  • Runners run 10 mph relative to centre of Earth.

This changes the runners' angular momentum $$L_{runners} = N \bar{m} R \Delta v$$ which yields $$ v = \frac{N\bar{m} R^2}{I_E} \Delta v$$ This result is identical to the previous case in the limit $\frac{N\bar{m} R^2}{I_E} \ll 1$. This case however allows $v > \Delta v$. To put in concrete numbers - assume all the people do the run, hence $N = 7\cdot 10^9$ with mean mass (upper estimate?) of about $\bar{m} = 70 kg$. Fractional moment of inertia $I_{runners}/I_E$ is $$ \frac{5}{2} \frac{N \bar{m}}{M_E} \sim 10^{-13}$$ which would make the Earth rotate with period $$ T = \frac{2\pi R}{v}\sim 10^{13} \,\mathrm{years}$$

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  • $\begingroup$ Got a minus sign wrong that pretty much changed the whole dynamic. Edited now. $\endgroup$ – DrLRX Jun 6 '18 at 11:31

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