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Destroying a Planet - How Much Energy?

Sooner or later maybe somebody will press the start button on an experiment, and we are all destroyed...
Sure, but most of the things that might do that propagate at or near c, so the first inkling we will have is the complete cessation of our existence.

Given that we will all likely die, and that for most, death will be rather drawn out and unpleasant, I am not convinced that it would be the end of the world if we accidentally destroyed the universe.

I mean, obviously it would literally be the end of the world; But it could be a great deal worse.
 
Life on earth will endure in some form until it gets absorbed into the red giant that our sun will become in a few billion years.
At least that’s my guess. Sue me if I’m wrong.

Wrong. Even the "best" case projections leave the Earth completely sterilized by the red giant phase.

While the energy of a single event tends to radiate into space rather than cook the planet deep enough the energy from the red giant phase is continued--the heat will melt the surface and everything deeper will inherently be warmer. (I won't say it will melt the whole planet because there's the issue of things staying solid because of pressure even when they're above their normal "melting" point.) Nothing lives in lava.

Okay… maybe life is gone a little before it gets consumed. What’s a few hundred million years between friends?
 
So the take away is eat,drink, get high, and fornicate....enjoy it while it lasts?

One of those pesky killer atseroids an sneak up on us any time.
 
The temperature of the core is higher than the surface so heat flows from the core to the surfcae ultimately seeking equilibrium with the CMBR as we are in a vaccuun.

Two objects close to each other in a vacuum with different temperatures. Like two flat plates parallel to each other. Ignoring the CMBR the hotter plate will radiate more energy to the colder plathan the colder plate to the hotter plate. Whrn they are at equilibrium. At that they radiate and absorb the same amount of energy from rach oyjer while loosing energy to the background..The amount of energy radiated by the Earth is set by the albedo.

In the plate example the rate of energy radiated depends on temperature difference and the emissivity of plates. Emissivity of 1 is a perfect blackbody radiator. Emissivity is in watts/m^2.

An estimate of the net of emiissivity of molten rock is .6-.75.





I think as the sun expands it becomes conduction as well as radiation. Hot gasses in contact with the Earth.

Temperature rise of an object
q = mcdT where q is heat in Joules T in kelvins c specific heat m mass
 
The temperature of a planetary core isn't determined solely by losses of heat to space, because conservation of energy doesn't apply - there's significant heating due to radioactive decay, and additional radiative energy from the nearby Sun, so energy is constantly being added to the system.

Conservation laws don't apply to open systems.

Conversely, if the Earth were in free space far from any star, and at some distant future time when radioactive decay had finished and thermal equilibrium achieved with the CMBR, there would be no loss of energy from that system to "the background", because the CMBR is the background. The universe IS a closed system, so conservation of energy DOES apply.

It's quite impressive to make two exactly opposite errors in the same post; But then, that's to be expected when engineers try to do physics.
 
(antiprotons for destroying a planet...)
You'll get a bit of extra energy out of it--when that antiproton beam hits normal matter it will be moving things up the periodic table and some of the resulting isotopes will decay. I think an antineutron beam would be far more effective as a weapon, though.
This is antiprotons, not ordinary protons. They will annihilate with a proton or a neutron in the nucleus and the resulting energy will go into boiling off nucleons and small nuclei.

Antineutrons will do the same thing when they annihilate, but they will have greater penetrating power, since they are electrically neutral, and they won't repel each other.
When an antiproton hits a nucleus it will move it one spot up the periodic table while reducing it's atomic weight by one. The new atom might be radioactive--a bit more energy.
 
When an antiproton hits a nucleus it will move it one spot up the periodic table while reducing it's atomic weight by one.
Not for long it won't. An instant later, the energy released from the annihilation will blast the nucleus into fragments.

Or, as @lpetrich correctly (but less dramatically) observed,
boiling off nucleons and small nuclei.
 
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