# Does cutting of trees affect spin angular momentum of earth?

Cutting trees reduces earth's moment of inertia. So the spinning velocity of earth should be reduced day by day. Does it really happen?

• See also: physics.stackexchange.com/q/65883 physics.stackexchange.com/q/56245 and several others linked therein. Basically this kind of question comes down to scale, and you have to take a look at some of the figures before you'll begin to comprehend just how puny our works are. Jul 7 '16 at 1:12
• Do you intend to mean the earth spins up due to the trees no longer standing up (i.e. the density of the earth increases as the trees are now lying horizontally on the earth rather than certically)? Jul 7 '16 at 3:13
• Your day-by-day wording suggests a cumulative effect, but overall, the post ignores the original growth that moved the mass from the ground upward. Jul 7 '16 at 15:05

Model the tree as a point mass $m$ located some height $h$ above the ground --- that is, forget the mass of the trunk and assume all the mass of the tree is in the branches and leaves above the ground. Then the moments of inertia of the tree before and after felling are \begin{align} I_\text{tree,up} &= m \left( (R+h)\cos\theta \right)^2 \\ I_\text{tree,down} &= m \left( R\cos\theta \right)^2 \end{align} where $R$ is the radius of Earth and $\theta$ the latitude of the tree.

The moment of inertia for the rest of the Earth is $$I_\text{Earth} \approx \frac25 MR^2$$ if Earth's mass is $M$ and we pretend Earth is a uniformly-dense sphere. (It isn't, which reduces the fraction out front from $\frac25$ to perhaps something like $\frac15$ --- I haven't done the math carefully or looked it up, and you'll see in a moment it doesn't matter. We'll use the simple assumption.)

Angular momentum is conserved when the tree falls, so the frequency $\omega$ of Earth's rotation changes: $$(I_\text{up} + I_\text{Earth})\omega_\text{up} = (I_\text{down} + I_\text{Earth})\omega_\text{down}$$ We can figure out how much it changes. Let's fell a tree on the equator, where $\cos\theta=1$, and figure out how much the ratio is: \begin{align} \frac{\omega_\text{down}}{\omega_\text{up}} &= \frac {\frac25M + m(1+\frac hR)^2} {\frac25M + m} \cdot \left(\frac RR\right)^2 \\ &\approx \frac {\frac25M + m + 2m\frac hR} {\frac25M + m} \\ &\approx 1 + 2\frac {m} {\frac25M+m} \cdot \frac hR = 1 + 5\frac{mh}{MR} \end{align} So a six-ton ($m/M = 10^{-20}$), sixty-meter ($h/R = 10^{-5}$) behemoth of a tree on the equator would change the length of a day starting in the 24th significant figure or so. Attoseconds. Felling a big tree would change the length of a day by some attoseconds.

Bigger effects include relative motion of Earth's core and mantle, tectonic shifts, evaporation of equatorial seawater and atmospheric transit of water vapor from tropical to temperate latitude zones or vice-versa. This image suggests there are daily fluctuations in day length of about a millisecond, averaging to about a half-millisecond seasonal variation. That's a seasonal change to the length of a day in the eighth or ninth significant figure.

A fun Fermi problem is to notice there are probably more deciduous trees in the northern than southern hemisphere, and assume they all drop their leaves at once in October; even that is a pretty small change in the length of the day.

• Luckily, we can measure down to attosecond resolution, or at least a few tens of attoseconds, and zeptosecond science is being thought about seriously. What else do you need? Jul 7 '16 at 15:42
• on the other hand, the tree grew before it was fell. Therefore it added the same amount of momentum first. Jul 7 '16 at 15:56
• @EmilioPisanty That paper is a part-per-thousand measurement of a femtosecond process. There is no measured quantity in the sciences which is known to twenty-four significant figures. Even LIGO, with all its interferometric trickery, detects strains larger than $10^{-21}$.
– rob
Jul 7 '16 at 16:25
• @njzk2 Exactly correct. Thus the most you can get is a fluctuation, as in the figure.
– rob
Jul 7 '16 at 16:27
• @rob Yes, of course; perhaps I didn't make the tone tongue-in-cheek enough. In fact, it's not even metrologically meaningful to talk about temporal variations of ~1 as/day, since the caesium SI second is only stable to about a tenth of a picosecond per second. Moving to an optical new-SI second would get you to a standard that was stable to about an attosecond per second, which is still ~100,000 times too small. Jul 7 '16 at 16:32

Cutting the trees and leaving them flat instead of vertical will diminish the moment of inertia of earth. The angular momentum of course will not change, but the speed of rotation will increase. However, I do not believe that the change is measurable with current instruments.

• Supposedly the effect of filling a large dam reservoir is measurable. It's not inconceivable that the change due to a whole year's lumber production might be within a few orders of magnitude. The question is if it's above the noise floor due to all other things that might affect it. The answer is probably not. Jul 7 '16 at 23:39

I know the example given sounds crazy but the physics behind it might be useful for someone learning rotational dynamics.

The angular momentum of a system does not change if there are no external torques, since $$\frac{d\vec L}{dt}=\vec \tau,$$ where $\vec L$ is the total angular momentum and $\vec \tau$ is the total external torque. So if you cut the trees the angular momentum of the system (Earth+trees) is constant. Leaving the cut trees on Earth does not change the Earth's angular momentum, but since the inertia momentum $I$ has decreased the relation $$L=I\omega,$$ shows that the angular velocity has to increase. The day would be shorter.

On the other hand if you cut and throw them in orbit (just as a didactic example) the trees by themselves might have an arbitrarily given angular momentum. So the Earth's angular momentum would change in order to keep the angular momentum of the system constant.

You might hear the story about figure skating. When a rotating person expands his/her arm, he/she can slow down rotation. Same thing can happen with earth. Assuming the tree is trillion tons and you cut it and lift it up, you can slow down the earth.

To formalize @knzhou's comment:

The answer, in a nutshell, is no.

Your basic assumption that cutting trees reduces the Earth's mass is wrong, because trees don't leave the Earth when they are cut!

Even if all trees left the Earth when cut,

1. a lot of tree cutters plant trees to replace what they cut, and
2. each tree is such a tiny amount of mass compared to Earth's mass that it doesn't really affect anything.

Hope this helps!

@Rob's calculations are the best place to look for a better explanation.

• He's not assuming that the Earth's mass changes, but that its moment of inertia, or rotational inertia, changes. And it does, since the moment of inertia is determined by the distance of the mass from the axis of rotation. But he's wrong that the rotation would slow down, it would speed up, as rob's calculation shows. Jul 7 '16 at 4:59
• @Previous The original version of the question talked about the earth's mass decreasing. Jul 7 '16 at 14:51
• Oops, my mistake! Apologies to heather, I didn't notice that. Jul 7 '16 at 14:59
• Okay, so the question has been updated to ask about inertia. Thanks. Jul 7 '16 at 15:04

Displacing some mass closer to the axis of rotation reduces the moment of inertia $I$. Considering Earth as an isolated system (which is not), its angular momentum $L$ must be conserved:

$$L=I\omega = \text{const}$$

Therefore if $I$ goes down, the rotation frequency $\omega$ must increase. But if we should also consider the reduction of $I$ while the tree was growing, displacing mass away from the axis of rotation, so by cutting a tree to the ground we are simply bringing back $\omega$ to the value it had before the tree was born.

If the average number of trees is going down, then this contributes to speed up the Earth's rotation. By how much? A tiny amount hidden into a bunch of stronger perturbations which still are really weak.

I would think that the whole atmosphere surrounding the earth is far heavier than the trees that were cut. The atmosphere turns with the earth and the changed position of trees would not even be noticed.