### Fermi Problem Friday: Leaf It to Me Autumn is well underway, and the leaves are beginning to turn in much of the Northern hemisphere. That means the ground will soon be carpeted in layers of brown, yellow and gold leaves.

But when a leaf falls from a tree, and its mass moves from on high to down low, what sort of effect does that have on the motion of the Earth?

More specifically, what happens when all the leaves have fallen from the trees?

Does it matter if all the leaves fall at once rather than gradually over the course of several weeks?

I haven't done any calculations yet, but I'll post my answer on Monday.

-Buzz

***

Solution:

My estimate is that the leaves falling from the trees in the autumn speeds the Earth's rotation by about 1 part in 2,000,000,000,000,000. (About 5*10^-14 percent). That's just in the range of detection by the most accurate atomic clocks.

Of course, it's a VERY rough estimate, and I haven't checked any of my calculations. If you care to run through them, and you're the first to find any given error, I'll send you a prize (probably a sticker or some other cool but worthless thing).

**************************

This is a Fermi problem, so I’m going start with some rough numbers.

A reasonable estimate is that trees’ leaves are typically in a range between 5 and 20 meters off the ground. So simplify things by saying their all 10 meters off the ground before they fall from the trees.

When I gather the leaves in my tiny, 10 square meter yard, they typically fill a bag that weighs about 20 kilograms. But they’re dry when I collect them, as opposed to being full of moisture when on the trees, so the leaves in my yard account for about 50 kilograms of leaves on the trees.

Assume, for simplicity, that the entire world is covered with trees just like mine (don’t worry, we’ll scale it back soon enough), so the Earth is covered in a perfect shell of leaves that weigh

ml=A*50 kg/10 meter^2= (4*pi*R^2)( 50 kg/10 meter^2)

Where the asterisk is a multiplication sign, the '^' symbol means 'to the power of', A is the area of the Earth, R is the radius of the Earth, and ml is the total mass of all the leaves on the Earth.

In the summer, the leaves are h=10 m above the ground, so the moment of inertia of the leaves in the summer is about

I(summer)= (2/3)*ml*(R+h)^2

And in the winter is

I(winter)= (2/3)*ml*(R)^2

So the total inertia of the Earth is

I(total summer)=I(Earth)+I(summer)

In the summer.

And

I(total winter)=I(Earth)+I(winter)

In the winter.

But only 10 percent of the Earth is covered in land, and only the leaves in the northern hemisphere fall off in our autumn. As far as I can tell from traveling around, about a fifth of the northern hemisphere is covered in trees, so the revised calculations are more like

I(total summer)=I(Earth)+I(summer)/(10*2*5) )=I(Earth)+I(summer)/(100)

In the summer.

And

I(total winter)=I(Earth)+I(winter)/(10*2*5) )=I(Earth)+I(winter)/(100)

In the winter.

The change in the total angular momentum from summer to winter is

I(total summer)- I(total winter)

And the percentage change is

100*[I(total summer)- I(total winter)]/ I(total summer)

=100*[ I(Earth)+I(summer)/(100)- I(Earth)-I(winter)/(100)]/ I(Earth)+I(summer)/(100)

=[ I(Earth)+I(summer)- I(Earth)-I(winter)]/ I(Earth)+I(summer)

=[ I(summer)-I(winter)]/[ I(Earth)+I(summer)]

Which is approximately

=[ I(summer)-I(winter)]/I(Earth)

=[ (2/3)*ml*(R+h)^2-(2/3)*ml*(R)^2]/I(Earth)

I(Earth)=(2/5)M*R^2

Where M is the mass of the Earth. I could calculate it, but I’m just gonna Google it instead.

M= 5.9742 × 10^24 kilograms which is roughly = 6 × 10^24 kilograms

Percentage change in the inertia is =(2/3)*ml*[(R+h)^2-(R)^2]/[ (2/5)M*R^2]

=(5/3)*ml*[(R+h)^2-(R)^2]/[ M*R^2]

But the radius of the Earth, R, is about 6x10^6 meters so

ml=A*50 kg/10 meters^2= (4*pi*R^2)( 50 kg/10 meters^2), which is roughly 2.4x10^15 kg

=(5/3)*ml*[(R+h)^2-(R)^2]/[ M*R^2]= (5/3)* 2.4x10^15 kg *[(6x10^6 +10)^2-(6x10^6)^2]/[ 6 × 10^24 kg*6x10^6]

=(2*ml*2*R*h)/(M*R^2)=4*ml*h/(M*R)=4*2.4x10^15 kg*10/(6× 10^24*6x10^6)

=10*15*10*10^15/(36*10^30)=1.5*10^18/(3.6*10^31)=.5*10^-13=5*10^-14

So the rotational inertia of the earth should drop by 5*10^-14 %in the fall.

Angular momentum is conserved, so if the angular inertia falls, the angular velocity must rise by the same amount. That means the day shortens by 5*10^-14 %. Which is to say the day is shorter by 1 part in 2,000,000,000,000,000.

Atomic clocks at NIST are half as accurate as this, and should be nearly able to detect an increase in the Earth’s rotation as a result of the leaves falling from the trees in the Northern Hemisphere in autumn. The prototype Cesium fountain clock at NIST is accurate enough to see a change of the order that I calculated, but I imagine I made pretty big errors and am probably off by several orders of magnitude one way or the other.

-Buzz

1. Don't forget that falling leaves in the northern hemisphere's autumn are balanced by growing leaves in the southern hemisphere's spring. (But then you have to take into account the relative N/S land masses.)

2. Interestingly, not many plants in the Southern hemisphere are winter-deciduous to drop leaves. For example there is only one native winter-deciduous plant in Australia, Fagus, which grows only on few mountain ranges within Australia's tiny island state of Tasmania.

3. Most of the land mass (nearly 70%) is in the northern hemisphere. Also, most of the large forests in the southern hemisphere are evergreen (meaning they don't shed their leaves on an anual cycle)

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