By looking at the rotating car washing brushes, it can be seen that its volume increase with rotation and decrease when it's stopped.

So, just for curiosity, would the earth volume decrease if it would stop rotating? I suppose so, and not only for solid sections. But I'm not sure. Perhaps it is even measurable!


2 Answers 2


By looking at the rotating car washing brushes, it can be seen that its volume increase with rotation and decrease when it's stopped.

If you measured the volume of the individual bristles then no the volume of the brush doesn't increase. If you had a cylinder of steel the same size, or a water heater, and spun it at high speed the volume would increase because the length is greater than the diameter.

The difference between a sphere (equal width and height) and the shape of the Earth is ~0.0035452% the spinning causes the equator to increase it's diameter but the sphere flattens at the poles - so the volume doesn't change on that basis, but you would slightly decrease the density of the mass thus causing a small increase in the volume; less than a cylinder of the same mass.

On another note, if the Earth weighed two million times as much it's volume would be smaller, because it would be a black hole.


Brushes are not a good model of the Earth, since they are not very stiff nor under compressive force. Still, the question is a good one. Let's divide it into (1) how much compression is there due to gravity, (2) how much does rotation matter, and (3) does the shape matter? Quick answer: less than 2.6% volume change.

The interior of a non-rotating Earth is going to be under pressure since every piece will be carrying the weight of all the other stuff above it. This means that it is going to be compressed. The pressure makes the interior material volume shrink based on its bulk modulus, $$\Delta V = V_0 \frac{P}{B}.$$ An estimate of the bulk modulus of the Earth is maybe 40 GPa.

The pressure behaves as $$\frac{dP}{dr}=-\frac{GM(r)\rho(r)}{r^2}$$ where $M(r)=\int_0^r 4\pi r^2 \rho(r) dr$; if we assume density is constant $M(r)=4\pi\rho r^3/3$ and we can integrate the differential equation to $P(r)=C-G\rho^2 (2\pi/3) r^2$. The integration constant $C$ must obviously be set so $P(R)=0$, $C=(2\pi/3)\rho^2 R^2$. This also gives the central pressure, in this case 172 GPa. The actual central pressure is higher, about 360 GPa since Earth is not constant density. At 360 GPa Earth material would have a relative volume shift $\Delta V/V=9$: were the core as compressible as the outer layers it would be seriously compressed (the density is 13 compared to the average 5.51). Were we to turn off gravity Earth would certainly expand a fair bit.

This was all due to gravity. How much does rotation change things? The acceleration due to rotation is $a=v^2/r=4\pi^2 r /T^2$ where $T$ is 24 hours and $r$ here means the distance from the axis. At the equator $a= 0.0337$ m/s$^2$, 0.3% of the gravitational acceleration. Were gravity everywhere reduced by 0.3% we should expect a volume change near the core by $0.003\times 9=0.027$, about 2.7%. But obviously this overestimates things since (1) not all parts of Earth are that compressed, (2) rotation acceleration is largest further away from the rotation axis. So the overall answer is: no, changing rotation of earth will not change the volume much.

One could try integrating pressure with rotation ($dP/dr = -GM(r)\rho(r)/r^2 - 4\pi^2 r/T^2$) but this would only be valid along the equator. The actual behaviour of Earth is to form an ellipsoid with a constant gravitational+rotation potential surface. If $R_1$ is the equator radius and $R_2$ is the polar radius the ratio is $R_1/R_2=1.0034$ - not much of a difference.

One can solve this shape problem too for constant density, and it gives us a central pressure $$p_0 = \frac{\rho}{2} \left (\frac{3}{2}GM \alpha_1 - \omega^2\right )R_1^2,$$ where $$\alpha_1 = \int_0^\infty \frac{du}{(R_1^2+u)\sqrt{(R_1^2+u)^2(R_2^2+u)}},$$ and $\omega=2\pi/T$.

I numerically get $\alpha_1=2.8350\times 10^{-27}$ which seems about a million times less than it should be, but adjusting for this discrepancy (likely I have some cgs to mks conversion error somewhere) the relative effect of the rotation on central pressure in this model is to reduce it from 190.3959 to 189.3748 GPa, a difference of 1.02 GPa. This takes into account that the shape would change (I used $R=(R_1^2R_2)^{1/3}$ for the new radius). That would give us 2.6% volume change in the core if rotation stopped, about the same as the above estimation (and again, much less in total since this is the maximum pressure; how much changes can be estimated by integrating equation 2.110 throughout the ellipsoid, something I leave to ambitious readers).

Obviously, if you are a mad genius bent on stopping Earth's rotation and want to calculate property inflation you may want to do these calculations far more carefully (inhomogeneous ellipsoidal earth with different modulus at different depth, and so on) but it is worth noting that suddenly stopping the rotation is likely to expand the Earth for another reason: there is a lot of energy in the angular momentum, and were that released the planet would melt and expand by sheer thermal expansion. But that is another question.


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