For example, if we are able to gather all the available nuclear weapons in place and target towards a weak point, say an earthquake fault line, will the entire planet be blown to bits?

  • $\begingroup$ If yes, then it would've happened already. Try looking up 10 ways destroy earth on youtube. $\endgroup$
    – BCLC
    Jan 15 at 8:12
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    $\begingroup$ I’m voting to close this question because it has nothing to do with astronomy. $\endgroup$ Jan 15 at 9:54
  • 1
    $\begingroup$ Forget astronomy, this isn't even remotely Worldbuilding. So no chance of migration. Vote to close $\endgroup$ Jan 15 at 10:27

2 Answers 2


The definition of "blown to bits" is that you would need to supply sufficient energy to remove all the mass components of Earth to infinity - this is the gravitational binding energy.

The calculation has astronomical relevance if the energy were supplied by say an incoming impactor, rather than nuclear weapons.

Roughly, this is $3GM^2/5R$ for a uniform sphere of mass $M$ and radius $R$.

For the Earth, this works out as $2\times 10^{32}$ J or the equivalent of $5\times 10^{16}$ Megatons of nuclear weapon yield.

Since there are about 10,000 warheads in existence and the average yield of each is is possibly a little less than a megaton, then exploding them all falls short by 12 orders of magnitude from "blowing the Earth to bits".

A more interesting energy calculation is to work out the energy of an impacting asteroid at say 30 km/s (the speed of the Earth's orbit around the Sun). The impactor would need a mass of $4\times 10^{23}$ kg to have "blowing to bits capability", which is five times the mass of the Moon.



All the nuclear bombs together have only a minute fraction of the energy needed to overcome the gravitation attraction holding Earth together and shatter it.

I'm sure someone else will come along and calculate an answer showing how many bombs would be necessary, and that number of bombs will have many, many digits.

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    $\begingroup$ Downvoted because it's best not to answer blatantly off-topic questions. What does this question (or this answer, for that matter) have to do with astronomy? $\endgroup$ Jan 15 at 9:55

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