Re: Rotational KE of Earth


<BretCahill@xxxxxxxxxxxx> wrote in message news:7c315b1b-c9c4-482e-87fa-f1483ee15fba@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Before we start extracting kinetic energy from the earth's rotation
and upset anyone's 24 hr. sleep cycle, we need to double check some
calculations.

First we'll need the density, mass, volume and centroid of the
earth.

If the radius is 4000 miles or 6 million meters, then the vol. is
216,000,000,000,000,000,000 million m^3.

(Check the order of magnitude.)

According to Zarathustra the heart of the earth is of gold.

The mass is therefore, 3,000,000,000,000,000,000,000,000 kg.

(Check the order of magnitude.)

Omega = 0.00007/sec

KE = 1/2 m (R omega)^2 =

100,000,000,000,000,000,000,000,000,000 (some units maybe joules).

OK let's do it!


Not even the right formula. For rotating rigid bodies, you must use the moment of inertia for the object. The formula you're using will tell you the KE of a concentrated mass rotating not around it's own diameter, but around a point in space R distance away from the mass.

The moment of inertia for a perfect sphere of uniform density (which the earth is neither perfectly spherical, nor uniform density) would be....

I=2/5 mR^2

Google says the mass is 5.97e24 kg and the radius is about 6.370e6 m. so...

I = 2/5 * 5.97e24 * (6.37e6)^2 = 9.7e37 kg-m^2

KE = 1/2 I omega^2 = 1/2 *9.08e55* (7.3e-5)^2 = 2.56e29 Joules or 2.56e11 ExaJoules.

Now, go 'do it' if you can!

daestrom

.

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