# Could enough people spin Earth backwards faster than they could run? [duplicate]

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?

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

• Possible duplicates: physics.stackexchange.com/q/56245/2451 , physics.stackexchange.com/q/266666/2451 and links therein. – Qmechanic Jun 6 '18 at 3:29
• Thank you, those answers are very closely related, however I could not find anything in the answers that resolved the specific question I had. – SpiralStudios Jun 6 '18 at 3:42
• From the answer by krs013 in the 1st link, you'd need to replace each person on the planet by about 110,000 times the whole Earth's population (that is multiply the current population by 770 trillion) to get the required angular momentum to spin the planet back up. Obviously, there isn't enough room to do that. ;) – PM 2Ring Jun 6 '18 at 9:24
• – Emilio Pisanty Jun 6 '18 at 11:11

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.
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 $$T = \frac{2\pi R}{v}\sim 10^{13} \,\mathrm{years}$$