# How much energy to stop the earth from turning in 1 second?¶

Key formula:

L = I x omega = moment of inertia x angular velocity

I = 0.4 x r2 x M #(for a sphere)

In [17]:
r = 6378000 # meters
M = 5.97E24 # kg


omega = 2 x pi radians per day

But what is that in radians per second?

In [18]:
omega = (2 * pi) / (60*60*24)
print omega

7.27220521664e-05


In [19]:
I = 0.4 * r**2 * M
print I # unit is m**2 * kg

9.7141174992e+37


In [20]:
L = I * omega
print L # unit is m**2 * kg * rad/s

7.06430559528e+33



Note that 1 joule = 1 kg x (m2/s2)

Notice the extra s-1 unit... that means that it should require 7E33 joules of energy applied on the earth's surface, at the equator, in the due west direction to bring the earth's angular momentum to zero in a time period of 1 second.

Interesting side note: The answer above means you'd apply 7 x 1033 joules of energy to the earth over the course of 1 second... this implies that the power you'd need is 7 x 1033 watts, or about 7 trillion trillion trillion megawatts. If you wanted to know whether your car engine could do it, you'd need to call the manufacturer and ask if your V8 can be souped up to have 936 thousand billion billion billion brake horsepower.

If you had all day to end the world as we know it instead of just 1 second, you'd only need about a hundred and ten billion billion billion brake hp as we can see by changing a few variables: (smallpaul ran out of energy here...but you can imagine it)

In [21]:
# change some variables and redo it.

In []: