Earth Exploding

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This cannot happen. Here's why:

Conservation of electric charge would be violated: The Earth is made of protons, neutrons, and electrons. Electrons are charge; protons and neutrons contain particles with charge. If you turned the Earth into a bunch of chargeless photons, all this would be gone. The only way out would be to say that the Earth previously contained an even number of protons and electrons, but this is highly unlikely.
Conservation of baryon and lepton number: Electrons are leptons, which are a certain class of elementary particle. Destroying the Earth would mean destroying electrons, which would violate conservation of lepton number. Protons and neutrons are baryons, a class of composite particles made of quarks. Destroying the Earth would mean destroying protons and neutrons, violating conservation of baryon number.

Would it vaporize our solar system?

No. Destroying the solar system would violate the above laws.

Would it destroy space-time?

No. Absolutely not. Why would it "destroy" space-time? There's no possible mechanism.

Would it destroy the sun or even nearby stars?

I highly doubt it. The photons would spread out in a spherical shell, so the vast majority of them would not come near the Sun. And by the time they reached the nearest stars (4 light-years away - the Alpha Centauri system), they would not be very dense.

How much energy would be released?

Use $E^2=p^2c^2+m^2c^4$, or $E=\sqrt{p^2c^2+m^2c^4}$. Note, though, that the Earth also has kinetic energy, which would have to be accounted for, as well as gravitational potential energy.

If all planets had life on them how close would the closest planets be that still had multi-cellular life?

Ummm. . . Not sure I understand this one.

Okay, let's do some calculations.

Mass of Earth ($M_E$): $5.97219 \times 10^{24} \text{ kg}$

Speed of Earth: The Earth completes one orbit ($2 \pi r = 2 \pi (150,000,000 \text {km}) = 300,000,000 \pi \text { km}$) in $31557600 \text { seconds}$ (using 1 years as 365.25 days. This means it has an angular velocity $\omega$ of $1.991021278 \times 10^{-7} \text { radians/second}$. Using the formula for tangential velocity, $v=r \omega$, we find $v=29865.31916 \text { m/s}$, if the Sun if the point of reference. This leads us to the kinetic energy by $KE=\frac{1}{2}mv^2=2.663409478 \times 10^{33} \text{ Joules}$.

The Earth is moving, so we have to use $$E=\sqrt{p^2c^2+m^2c^4}$$ Defining $p$ as $mv=1.783613604 \times 10^{29} \text{ kg} \text{ m/s}$, We find $$E=5.374971 \times 10^{41} \text{ Joules}$$ Add this energy to the kinetic energy to find that the total energy released is $$5.374971027 \times 10^{41} \text{ Joules}$$ Now we use

posted over 9 years ago