## The End of the Aztec Calendar…

September 15

So as I’m sitting here trying to figure out if I have to go to school tomorrow, or if I can put in a full eight at work, my roommate tells me an awesome story. He knows someone who believes the world will end in 2012. Now, at first I pass this off as the regular old spiritual BS, but this friend has details!

She knows, because aliens speak to her through the Internet. In 2012, a “Planet X”, which is currently orbiting on the opposite side of the sun at the exact same radial velocity as the Earth, will accelerate past the Earth, and Planet X’s gravitational pull will cause the Earth to start rotating the opposite direction. This will happen so fast, that the atmosphere will continue on at its same speed, causing 600 mph winds and general mayhem across the world.

Now, I knew this was absurd, but I couldn’t prove it. So I did.

The Earth’s equatorial diameter is about $40.0 \times 10^{6} m$, in Flagstaff, Arizona (about where she lives) which is at 35° of Latitude, that diameter is $D_{equator} \cdot cos(35^{\circ}) = 32.8 \times 10^{6} m$

The Earth rotates once a day, giving us an angular velocity (at $35^{\circ}$) of:

$72.7 \times 10^{-6} \frac{rad}{s} = 379 \frac{m}{s} = 878 \frac{mi}{hr}$

So stopping the Earth instantly would produce 878mph winds. I’m not sure of the fluid dynamics of the atmosphere, and so can’t estimate the slowdown time that would give a $\Delta V$ of 600mph, but it doesn’t matter because:

The kinetic energy of a rotating sphere is:

$E_k = \frac{1}{2} I \omega^2$

where I is the moment of inertia of a solid sphere, and ω is the angular velocity in radians per second. The moment of inertia is calculated as:

$I_{sphere} = \frac{2}{5} m R^2$

which leads to the expanded equation of:

$E_k = \frac{1}{5} m R^2 \omega^2$

The Earth’s mass is approximately $5.97 \times 10^{24} kg$ and its radius about $6.38 \times 10^{6} m$, so, the energy required to stop the Earth’s spin is:

$E_k = \frac{1}{5}(5.97 \times 10^{24} kg)(6.38 \times 10^{6} m)^2(72.7 \times 10^{-6} \frac{rad}{s})^2 = 257 \times 10^{27} J$

That’s 257 octillion or 257 million-billion-billion Joules. Putting that in perspective, we receive about $1.37 \times 10^{3} \frac{W}{m^2}$ from the sun, or $701 \times 10^{15} \frac{J}{s}$, so the energy required to stop the Earth’s rotation in 1 day is enough to heat and light the Earth for eleven thousand years. Assuming that the energy transfer from “Planet X” or the aliens is 90% efficient, that leaves $25.7 \times 10^{27} J$ of waste energy, probably being converted to heat. That’s enough energy to raise the temperature of all the Earth’s oceans 4,527°C or to 8,211°F, which is just colder than the surface of the Sun.

And if you followed that, ladies and gentlemen, you are either a physics major, or an engineer.

This is a historical cross-post of a short-lived feature I posted on Facebook called “You Might Be an Engineer If…” with a slight edit. The original post can be found (for now) here.

Posted by on September 15, 2009 in You Might be an Engineer If...

1 Comment

### One response to “The End of the Aztec Calendar…”

1. September 16, 2009 at 9:20 am

LaTeX module looks pretty nice.