# End of the Earth IV - Shocking Destruction

Earth day is upon us once more. So many other namby-pamby bloggers out
there (don't hurt me!) are writing about how wonderful the earth is and
how great earth day is. We here at The Virtuosi take a more hardline
approach. Today I'm going to tell you how to destroy the earth.
Completely and totally. Unlike
last
year's
methods,
this one should work. In fact, this method is so simple that I can tell
you what to do right now. Just tweak the charge on the electron so it is
a bit out of balance with the charge on the proton. Just a little bit.
How little a bit, you might ask? A very little bit. Really, this doesn't
sound hard. I mean, sure, you have to do it for all of the electrons in
the earth, but we're talking about a very very small percentage change.
Not convinced? Let me show you just how small a change we're talking. If
there is a charge imbalance in the electron and the proton, this will
give the earth a net charge, throughout it's volume. I've got to make a
few assumptions about the earth here, so hold on. I'm going to assume
that the earth is a uniform density everywhere, and I'm going to assume
that the earth is made entirely of iron.* Now, the net charge of any
iron atom will be $$ (q_e-q_p)Z=(q_e-q_p)26$$ where Z is the atomic
number of iron, the number of protons (and electrons) the atom has. The
net charge of the earth, Q, is the number of iron atoms, N, times this
charge, $$Q=(q_e-q_p)ZN$$ I've previously
estimated
that N is about 3*10^50 atoms. Now, the electric potential energy of a
sphere of radius r with charge q uniformly distributed throughout it's
volume is $$U_e=\frac{3kq^2}{5r}$$ where k is the coulomb constant.
Dissolution of the earth will occur when the electrostatic energy of the
earth equals the gravitational potential energy of the earth. The
gravitational bound energy of the earth is given by
$$U_g=\frac{3GM^2}{5R}$$ Where M is the mass of the earth, G is
Newton's gravitational constant, and R is the radius of the earth.
Setting this equal to the electrostatic energy of the earth,
$$\frac{3GM^2}{5R}=\frac{3kQ^2}{5R}$$ $$Q^2=\frac{G}{k}M^2$$ so
$$(q_e-q_p)ZN=\left(\frac{G}{k}\right)^{1/2}M$$ Now, N is given,
in our approximations, by $$N=\frac{M_{earth}}{m_{iron}}$$ so
$$q_e-q_p=\left(\frac{G}{k}\right)^{1/2}\frac{m_{iron}}{Z_{iron}}$$
Now we can plug in some numbers. G=6.7*10^-11 m^3*kg^-1*s^-2, k=
9*10^9 m^3*kg*s^-2*C^-2, m_iron=9*10^-26 kg, Z=26. Thus,
$$q_e-q_p=3*10^{-37} C$$ To put this in perspective, the charge on
the electron is 1.6*10^-19 C, so this is roughly 10^18 times less
than that charge. Put another way, if the charge on the electron was
imbalanced from that of the proton by roughly 1 part in 10^18, the
earth would cease to exist due to electrostatic repulsion.** As I told
you at the beginning, you only have to change the charge by a very small
amount! So get working. There are only about
1000000000000000000000000000000000000000000000000000 electrons you need
to modify! *According to the internet, the density of the earth, on
average, is roughly 5.5 g/cm^3. The density of iron is 7.9 g/cm^3 at
room temperature, and the density of water is 1 g/cm^3 at room
temperature. So, while the earth is not entirely iron (of course), it is
a better approximation to assume the earth is iron than the earth is
water. And those, of course, were really our only two choices. ** It
turns out that this is a good argument for the charge balance of the
electron and the proton.

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