How much TNT do you need to blow up the moon if you place it exactly in the middle? Just think of the explosion -- that there's no air in space doesn't matter!
What do you think is it possible and how much TNT do you need for this?
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Sign up to join this communityHow much TNT do you need to blow up the moon if you place it exactly in the middle? Just think of the explosion -- that there's no air in space doesn't matter!
What do you think is it possible and how much TNT do you need for this?
To destroy the moon, you would need to provide at least $1.24\times10^{29}$J of energy to exceed the Moon's gravitational binding energy. (This provides a lower bound on the energy to "blow up" the moon.) A megaton of TNT releases 4.184 PJ of energy.
Put this together, and you would need at least: $2.96\times10^{13}$ megatons of TNT.
Said another way: you would need some 30 trillion million tons of TNT.
If you would like to perform this calculation yourself, see the Planetary Parameter Calculator. Based on a couple inputs, it will calculate the gravitational binding energy of a body.
Nick2253 gave a good start to the answer but that's only part of it. When you look further you find it's impossible.
Note cpast's comment--it's 1/3 of the moon's mass in TNT. But that's just the energy to blow up the moon, not the energy to blow up the moon and the TNT you used to blow up the moon.
We need to increase the TNT by 1/3 to account for the TNT itself--but we need to add another 1/9 to blow that up and so on--this sequence sums to 1/2. Thus we need half the moon's mass in TNT to blow it up.
Oops--now we have increased the binding energy by 50%. We need still more TNT to overcome that. This sequence sums to 100%--now we are up to the entire moon's mass in TNT. We also need to add enough TNT to blow up the TNT we just added.
Edit: Given Nick's comments I tried to work it out by brute force. My feeling the sequence didn't converge was right. The closest it comes is when the mass of TNT exactly matches the mass of the moon, this provides 62% of the energy needed. (Note, however, that the curve is quite flat--within the margin of error of the data--for quite a range around 1.0.)