What is the minimum amount of energy required to
irreversibly
blow up our lovely planet into smithereens?
Details and assumptions
Input your answer in Joules
Model the Earth as sphere of uniform volume density composed of tiny neutral pieces of mass.
The radius of the Earth is R e = 6 3 7 1 k m .
The mass of the Earth is M e = 5 . 9 7 2 × 1 0 2 4 k g
Newton's gravitational constant is G = 6 . 6 7 3 × 1 0 − 1 1 m 3 k g − 1 s − 2
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I must say that the gif is great !!! Would you mind adding it to the note Brilliant GIFs by Julian Poon ?
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Thank you. I found it after a few google searches. Sure, feel free to add it. The GIF on this problem was also popular a year back.
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Thanks :)
Yeah , about the other question , I won't comment abt the GIF , but it sure is a damn good question .
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@A Former Brilliant Member – Wow. I completely forgot about that note. :D. I've added it already btw
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@Julian Poon – Oops , I had forgotten to add the GIF ! Thanks for adding it .
Nice explanation and thanks for appreciation. Picture is really KABOOM.
Would dimensional analysis give us the constant 5 3 ? Is it possible to solve it just using dimensional analysis?
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nope, no way
Dimensional analysis never gives you the value of any constant.
Lovely lsolution I am finally able to solve thisafter so long.
Wasn't your location Addis Ababa some time ago, now you're in the USA ?
"The energy required to irreversibly blow up our lovely planet into smithereens" should be equal to self-gravitation potential energy of sphere because it is written that earth is made of neutral particles. From energy conservation Self gravitational potential enrgy+Energy supplied=0 At infinity GPE of a body is treated as zero hence it will be irreversible. Self GPE = − 5 R 3 G M 2 (for proof see this )
Hence 5 R 3 G M 2 = e n e r g y s u p p l i e d substitute values and we get answer 2 . 2 4 1 × 1 0 3 2
Hi , how about you use this . 2.241\cdot 10^{32} to get 2 . 2 4 1 ⋅ 1 0 3 2 ?
It'll look better :)
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Thanks!!Done.
This avoids confusion between decimal and multiplication:
2 . 2 4 × 1 0 3 2
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Yeah, I thińk that. Your idea is even better .
It's just that I am more in the habit of using cdot instead of times , so you see :)
seems good to me .
But, Shouldn't we also take Earth's translational and rotational kinetic energies into consideration????
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but who knows earths rotational and translational kinetic energy,they are dependent on frame of reference.Then we can say universe is expanding and hence it will be irrelevant to consider kinetic energy here and secondly as frame of reference is not given and as well as not the angular and translational speeds we shall not consider it.
Pls ask if any doubts . your question was good .i didn't give it a thought in that way( may be because of jee pattern preparation ->"if you are not given a certain thing you shall not consider it")
Note : self GPE is independent of reference frame.
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No..i do get about the GPE!! But, see...if you even consider the expansion of space, and that too at the speeds at 90% the speed of light. Will the blasted remains ever have zero potential energy???
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@A Former Brilliant Member – Yeah! that is the point that's why i will not consider KE. there is uncertainity. what KE would you take?
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@Gautam Sharma – See, since the initial state is the Earth's revolution and Earth's rotation...So it's total ke is ...It's translational KE around SUN + It's rotational KE around SUN + It's rotational KE around it's own axis which will be equal to zero as they have asked for minimum, and all the particles will come to rest, in the final state...
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@A Former Brilliant Member – What about revolution of milky way? why sun only and what about gpe due to other planets? maybe particles will collide on their way to infinity(pun intended.)
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@Gautam Sharma – If you take all these measures in mind, then i say, that the given answer is also not correct! Because, we know that our universe is infinite. That means the potential energy will be zero only at that point...Agreed?
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@A Former Brilliant Member – Yeah that's why i have taken total energy =0 at infinity. Agreed it should be mentioned in the question that earth is stationary. But while solving we have that in our mind as nothing data about that is given.
question -'What is the minimum amount of energy required to irreversibly blow up our lovely planet into smithereens?" HAHA lol i know it will arise that it is not given that we cannot take it because we have sir google
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@Gautam Sharma – Hahaha...I get it! it's alright...but i'm debating about the GPE!!You say that you have taken the GPE of every particle to be 0 at infinity. But, my question, is that , theoretically, will any particle be able to reach this infinity??
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@A Former Brilliant Member – Theoretically yes (every thing is possible in theory like air resistance goodbye), practically not at all(but remember nothing is impossible).
Sir newton took this convention and infinity is the limiting case by infinity we mean that when the gravitation pull is that small that it is relevant against when the particles are at some finite distance with appreciable gravitational pull or gpe.
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@Gautam Sharma – Yea!! But, the gravitational pull will never be that irrelevant!! See, everything is expanding further and further away...So, when the particles flew off, they all move into a new section of space, the next moment. So, according to this theory, all the planets in that section also move with the same speed. Techincally, the gravity on the particles, will never be zero.
And this has been proved by Einstein! Newton, was after all wrong!!
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@A Former Brilliant Member – my point was when earth is considered as a stationary sphere. But lemme think its particles mutual attraction these processes can be simultaneous A "rough" energy conservation
selfgpe + energy given +expansionenergy=expansionenergy particles have extra energy
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@Gautam Sharma – Well, i guess, in the end, the energy receivd in the answer will be every less than the actual answer. Anyways, it was nice debating with u man. It's nice to make new friends on here!!
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@A Former Brilliant Member – true and bye .my pleasure . True
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@Gautam Sharma – Btw, how much do you know about quantum mechanics?? Coz if you do...i would really like u to check my note on Quantum tunnelling...Please...thanks!
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@A Former Brilliant Member – A lil bit ok i will check.
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@Gautam Sharma – Thanks @GAUTAM SHARMA
@Gautam Sharma – And i want to know how will KE contribute to disintegration(just asking)?
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@Gautam Sharma – Suppose we consider the kinetic energy of translation, as an exmple. That means all the particles are movin in one direction with one linear velocity. But, after the disintegration, this might not be true. so, some energy would surely be lost to change this parameter...
"The energy required to irreversibly blow up our lovely planet into smithereens" should be equal to self-gravitation potential energy of sphere because it is written that earth is made of neutral particles. We can assemble the planet by bringing in thin shells of mass to our partially assembled mass of radius . This partial mass will cause a potential
The energy is just under the entire output of the Sun in a week. Or the total energy from the Sun arriving at the Earth during about 4 Million Years.
A megaton of TNT yields about 4.184E15 Joules of energy. A thermonuclear warhead is equivalent to about 6 times amount of energy. So, to blow up Earth , we would need about 8.93E15 warheads of that size.
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This hopefully fills out the details in Gautam's great solution above. First from the problem statement let us convince ourself that it is asking for the potential energy of the system and that this is equivalent to the gravitational binding energy. If we bash it out and write the equation for bringing each infinitesimal mass to infinity, we will discover that U is independent of the order in which the masses were disassembled. Since it is independent also of the route by which the masses were brought, U must be a unique property of the final configuration of the masses. Thus U is just the gravitational potential energy of this system. This can be found by just finding the energy of assembly and taking the negative.
Now for the actual problem, consider the diagram above. We can assemble the planet by bringing in thin shells of mass d m to our partially assembled mass m of radius r . This partial mass will cause a potential ϕ ( r ) = − ∫ ∞ r r 2 G m r ^ d r = r − G m
The energy required to bring the thin shell from infinity d u is
d U = ϕ ( r ) d m We also know that ρ is constant, thus m = ρ 3 4 π r 3 and d m = ρ d V = ρ 4 π r 2 d r
Substituting these two expressions in our first , we get
d U = 3 1 6 G ρ 2 π 2 r 4 d r
Integrating this from 0 to R and simplifing , we get
U = 5 R 3 G m 2
As an extra fun exercise, think about how we to attack this problem using just dimensional analysis.