Change of the Earth's angular velocity made by humanity accelerating around the Equator Consider this scenario:
Every person on Earth (~8 billion people each averaging 70 kg) get together in a ring around the Equator and start accelerating 10 m/$s^2$ Westward, how would the Earth's angular velocity change? Is their mass relevant to know exactly how much will it change?
I think that, given the lineal acceleration was of 10 m/$s^2$, we only need to obtain the angular acceleration $\alpha = -1.54\times10^{-6} rad/s$ to know how much will the Earth's angular velocity can change.
Is my assumption correct?
Thanks a lot.
 A: Let's assume the Earth has a moment of inertia of $I=Cmr^{2}$ where $m$ is the mass and $r$ is the radius and $C$ is a constant (2/5 for a sphere, 0.33 if you know the center of the Earth is denser)
And the force is $F = Ma$ where $M$ is the mass of the people and $a$ is the acceleration.
So the torque is given by $Fr = T = I\dot{\omega} = Cmr^{2}\dot{\omega}$
So $\dot{\omega} = \frac{Ma}{Cmr}$
As you can see, the mass of the people was important
Which is $4.4 \times 10^{-19} s^{-2}$ for the given values.
The other important takeaway is that the units of acceleration are $s^{-2}$.
The Earth is $10^{13}$ times as massive as all the people. And $6\times10^{6}$ meters in radius. These factors result in a very small change.
We can ask how much a day will change.
$1 + \frac{dP}{P} = \frac{P+dP}{P} = \frac{\omega_{0}}{\omega_{0} + d\omega} = \frac{1}{1+dw/\omega_{0}} \sim 1-d\omega/\omega{0}$
So
$dP \sim -P^{2}\dot{\omega}t$
Using $P=$ day, after an entire year of accelerating, the length of a day would change by 0.1 seconds!
A: I think an important point to be made here first, is that the total angular momentum of any isolated system remains constant, and this system involving the earth plus people can be considered isolated for this exercise.
This conservation law for isolated systems holds for energy and momentum as well as angular momentum.
We can also see here that once the people start running, while they all push on the ground and exert a torque,  when they stop running, they will all exert a push in the opposite direction, which will cancel the starting push, so that in the end, the net result will be no increase in angular velocity - unless they can continue to run with an acceleration forever!
But even still, it's probably unlikely there would be any detectable effect on the Earth’s rotation even if all 8 billion people were running with an acceleration of $10ms^{-2}$, or even if they ran with a higher acceleration. This is because the mass of all the people is still much less than the mass of Earth.
The mass of the Earth is  $\approx 6\times 10^{24}kg$ and given 8 billion people each of mass $70kg$ comes to about $5.6\times 10^{11}kg$, meaning earth is still $10^{13}$ times more massive than all these people.
All these people running together with an acceleration of $10ms^{-2}$ produces a torque $$\tau = r\times F= r_{\text{earth}}\times (5.6\times 10^{11})\times 10 \approx 3.4\times 10^{19}Nm$$
We can compute the torque needed to get the earth moving from rest, to its current angular velocity $4.2\times 10^{-4}$ degrees per sec. We can get$^1$ this by using $$\tau=\frac{\Delta L}{dt}\approx 7.050 \times 10^{33} Nm$$ So if we use both these two results, we get a net torque given by $$\Delta \tau=7.049\times 10^{33}Nm$$ which is actually very small change indeed (note that I have rounded off and on the calculator there were more significant figures). Needless to say, the change in angular velocity will be laughably small.
$^1$ We can get the moment of inertia of a solid sphere, and the change in angular momentum over one second by a quick google search.
A: The earth/people system wouldn’t be spinning any faster. But to us it would.
The earth (excluding the people) would indeed accelerate.
Until the people decelerate and undo all their hard work.
