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# The Truth of the Day (and a Bevy of Rotational Physics to Boot!)

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First, here is some Swiss ski flying: (please note: for the purpose of this post, the mentioned video is not used for educational purposes, but instead serves as a really cool sight that is hopefully inspirational to readers)

And now for your weekly dose of physics...

Sometimes one may feel like he or she cannot count on anything, like life has no constants and is unbearably unpredictable. But humans have always taken one constant for granted: a new day is always coming. The phenomena called the "day" is the unit of time measuring how long it takes for our earth to rotate one complete revolution. When earth rotates, the sun's light hits a new section of the earth, giving us sunrise and sunset. Everything from the workday, agriculture, visibility, human instinct, and Swiss ski flyers rely on the presence of the sun to function properly.

Okay, so we get that the day is important. But what if this thing that we always take for granted was not actually inevitably reliable? That's right, the day as we know it is coming to an end. More on this later.

First of all, taking the earth out of its orbital context, the earth's rotational kinetic energy is defined by $K_{R}=\frac{1}{2}I\omega^{2}$

$I$ is the moment of inertia (used similar to mass in translational motion). The general definition of moment of inertia is $I = \int r^{2}dm$.

Assuming the earth is a solid sphere (and in this case, of uniform mass), it's moment of inertia would be $\frac{2}{5}MR^{2}$. With a mass of ~$5.9742 \cdot 10^{24}kg$ (thank you google.com) and a radius of ~$6.3781\cdot 10^{6}m$ (google.com, what don't you know?), earth must have a moment of inertia of $I_{Earth}=\frac{2}{5}(5.9742 \cdot 10^{24}kg)(6.3781\cdot 10^{6}m)^{2}\approx 9.7213\cdot 10^{37} kg\cdot m^{2}$

$\omega$ (deceivingly pronounced "oh-may-guh" and not "double-you") is the angular velocity. $\omega$ is similar to translational velocity, but instead it tracks the speed of particles as a radians/second measure around a circle. According to http://hypertextbook.com/facts/2002/JasonAtkins.shtml, the angular speed of the earth is ~$7.27\cdot 10^{-5} rad/s$

With this information, we can calculate earth's kinetic rotational energy.

$K_{REarth}=\frac{1}{2}(9.7213\cdot 10^{37})(7.27\cdot 10^{-5})^{2}\approx 2.57\cdot 10^{27}J$

That's a lot of Joules!

In a perfect world, this energy will not change and our days will remain intact. Unfortunately, perfect this world is not. What could change our lovely earth's rotational speed? One may first answer force, but in the rotational world of physics, we convert forces that cause rotation into torque (or $\tau$).

$\tau =\overset{\rightharpoonup }{F}\times \overset{\rightharpoonup }{R}$

This means that with a net force of 0, no net torque would be exerted on our planet and the day would remain constant. In reality's case, the friction of the tides being pulled by the moon is causing the day to shorten by 2 milliseconds each year. The question is: what torque are those pesky tides exerting on our beloved home? Let's find out!

For this we will need a couple more rotational formulas. Note these formulas can be derived from kinematics equations using m = I, X = $\theta$, v = $\omega$, and a = $\alpha$.

Finding the change in $\omega$:

$\Delta \omega =\omega _{f}-\omega _{o}$

$\omega _{o}=\frac{2\pi }{86400}$

$\omega _{f}=\frac{2\pi }{86399.998}$

$\therefore \Delta \omega = 1.68338\cdot 10^{-12}rad/s$

Finding $\alpha$:

$\alpha =\frac{\omega _{f}-\omega _{o}}{t}$

$\alpha =\frac{1.68338\cdot 10^{-12}rad/s}{3.1536\cdot 10^{7}}$

$\therefore \alpha =5.3380\cdot 10^{-20}rad/s^{2}$

Finding torque!

$\tau =I\alpha$

$\tau =(9.7213\cdot 10^{37} kg\cdot m^{2})(5.3380\cdot 10^{-20}rad/s^{2})$

$\therefore \tau _{Tides}\approx 5.189\cdot 10^{18}N\cdot m$

In retrospect, that is a lot of torque, even when considering that our earth is ~75% water on the surface.

Awesome analysis and a creative way of applying what we're learning to a VERY big scenario. Well done! :wave)

Very Impressive!! Would love to know how you got fancy equations into your blog

@will Indeed, that "fx" button works some wonderful magic

Thanks so much

My mind is twice blown

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