Does nuclear weapon can destory the moon?

[magnetar]

My friends and I are talking about this question.From the point of my view,I do not think that man-made nuclear weapons can destory the moon! But one of my friends insist that man-made nuclear weapons can do that job!what is your opinion?


Thank you in advance!

 

[Astronuc]

The largest nuclear weapon detonated (Tsar Bomba, "King of Bombs") was on the order of 50 MT TNT, and that is not big enough to destroy the moon. One could however make a very large crater.

http://en.wikipedia.org/wiki/Tsar_Bomba

http://nuclearweaponarchive.org/Russia/TsarBomba.html

http://www.atomicforum.org/russia/tsarbomba.html

 

[JesseM]

This page has some calculations of the amount of energy that would be needed to "destroy" a planet, and they offer an online calculator here where you can plug in the size and surface gravity of a planet to see the minimum energy needed to destroy it (they assume the planet is of uniform density, so the actual energy would be somewhat higher). The moon's mean radius is given as 1737.1 km (for a diameter of 3474.2 km) here, and its surface gravity is 1.62 m/s^2 which is 0.165 the surface gravity on Earth (9.81 m/s^2)...using these numbers, the calculator gives a lower limit for the energy as 1.234 * 10^29 Joules. This page says a megaton of TNT would release an energy of 4.184 * 10^15 Joules, so ignoring details like how the bombs would be distributed, you'd need around 30 trillion megatons of explosives to destroy the moon...the page above says the biggest nuclear bombs currently in the US arsenal are about 3 megatons, and the Soviet Union once developed a prototype bomb with 50-100 megatons. Anyway, the number of nuclear bombs never exceeded 100,000 at any point during the cold war, so it definitely falls short of the trillions you'd theoretically need to destroy the moon.

 

[Chronos]

Do the math, as JesseM alluded to. Fortunately, planet buster technology is still beyond our reach.

 

[jazzdude9792]

Even if we could create something that could destroy the moon, we wouldn't want to use it because of consequences of the lunar fragments hitting the earth.

 

[pervect]

My friends and I are talking about this question.From the point of my view,I do not think that man-made nuclear weapons can destory the moon! But one of my friends insist that man-made nuclear weapons can do that job!what is your opinion?


Thank you in advance!

The question reminds me of a rather bad SF show, "Space 1999".

As other posters have calculated, nuclear weapons don't have sufficeint energy to blow up (gravitationally disrupt) the moon.

 

[Esnas]

Even if we could create something that could destroy the moon, we wouldn't want to use it because of consequences of the lunar fragments hitting the earth.

Destroying the moon would also have other devastating effects. A few to start with would be:

1. Major disruption of the oceanic tides
2. Change in the Earth's "wobble"
3. Probable change in the tilt of the Earth's axis and attending axle procession
4. Changes in the Earth's rotation and orbit of revolution around the sun

 

[Chronos]

The moons mass would remain in orbit, even if it were disintigrated [which is highly unlikely], so these are fictional scenarios.

 

[JesseM]

The moons mass would remain in orbit, even if it were disintigrated [which is highly unlikely], so these are fictional scenarios.

Wouldn't a substantial fraction of it end up falling to Earth? Also, what's the escape velocity from the Earth's gravitational field for an object starting at the moon's distance from the Earth? Someone would have to do the calculations, but it might be smaller than the escape velocity needed for most of the particles that make up the moon to escape the center of gravity of the moon itself, since they'd be much closer to this center of gravity than to the Earth's (and it is assumed that they reach this escape velocity in the calculation of the energy needed to 'destroy' the moon, since otherwise the fragments would all just fall back and re-form into a moon-sized sphere).

 

[Lockheed]

Wouldn't a substantial fraction of it end up falling to Earth? Also, what's the escape velocity from the Earth's gravitational field for an object starting at the moon's distance from the Earth? Someone would have to do the calculations, but it might be smaller than the escape velocity needed for most of the particles that make up the moon to escape the center of gravity of the moon itself, since they'd be much closer to this center of gravity than to the Earth's (and it is assumed that they reach this escape velocity in the calculation of the energy needed to 'destroy' the moon, since otherwise the fragments would all just fall back and re-form into a moon-sized sphere).

that depends on how completely you destroy the moon. If it was vaporized (which is what I think the 30 trillion megatons is based on), then none of it will fall back to Earth. If you were to blow it up into fragments, which would take up far less explosives (but that's still so much more then we can ever produce...), all of the fragments would fly everywhere in all directions (assuming they are moving fast enough so that the fragments don't fall back together under their own gravity), so the fragments that will hit Earth will be ones that are either going toward Earth or are going in a direction that is close enough to be affected by the gravitational field. The gravitational field of Earth is quite weak and its influence drops off over a short distance. And since the mass of the fragments will be much lower then that of the moon, the gravitational pull on them from Earth will be much weaker.

For reference, the acceleration due to gravity on the Earth at the Moon's distance is about 2.7x10^-3 m/(s^2)

 

[JesseM]

that depends on how completely you destroy the moon. If it was vaporized (which is what I think the 30 trillion megatons is based on)

The 30 million megatons is just based on all the particles of the moon having sufficient escape velocity so that the moon doesn't re-form under its own gravity. It doesn't specifically say whether the atoms are part of chunks or whether all the atomic bonds have been destroyed.

then none of it will fall back to Earth.

Why do you say that? All the individual atoms that made up the moon will still be present if the moon is vaporized, even if the bonds between them are destroyed, and if the explosion is radially symmetric some will be flying directly at the Earth, and many others might have initial directions that are close enough to Earth that they would be pulled towards it rather than going into orbit or going past the Earth with a radial velocity larger than or equal to the Earth's escape velocity at the point of nearest approach.

The gravitational field of Earth is quite weak and its influence drops off over a short distance. And since the mass of the fragments will be much lower then that of the moon, the gravitational pull on them from Earth will be much weaker.

For reference, the acceleration due to gravity on the Earth at the Moon's distance is about 2.7x10^-3 m/(s^2)

Even with that seemingly weak pull, the escape velocity can still be pretty high--according to this escape velocity calculator the escape velocity from Earth's gravity would be around 1440 m/s for an object at the distance of the moon's center. For comparison, the escape velocity to escape from the moon's gravity for an object leaving from the moon's surface would be around 2375 m/s. So a piece escaping from the surface of an exploding moon in exactly the opposite direction from the Earth's center would escape, but for pieces heading out from the surface at other angles, the radial velocity away from the Earth can be lower than 1440 m/s. And the "destroying the moon" calculation assumes each piece of the moon must achieve escape velocity, not from the entire moon, but only from the fraction of the moon's mass which is closer to the center than that piece (since the gravity from all parts at a greater distance from the center will cancel out in Newtonian gravity), so pieces significantly below the surface may be heading away from the moon's center with a velocity significantly less than 2375 m/s relative to the center.

Of course, of all the pieces that fail to achieve escape velocity, not all will fall to Earth--a significant fraction will go into orbit (perhaps forming a spectacular ring system!)

 

[Lockheed]

The 30 million megatons is just based on all the particles of the moon having sufficient escape velocity so that the moon doesn't re-form under its own gravity. It doesn't specifically say whether the atoms are part of chunks or whether all the atomic bonds have been destroyed.

Well then the OP is going to have to be far more specific, because particles can have different sizes. I assumed the moon would be vaporized because that is typically what they calculate for when talking about destruction of any body, whether it be celestial or an Earth-bound object such as a bunker.

Why do you say that? All the individual atoms that made up the moon will still be present if the moon is vaporized, even if the bonds between them are destroyed, and if the explosion is radially symmetric some will be flying directly at the Earth, and many others might have initial directions that are close enough to Earth that they would be pulled towards it rather than going into orbit or going past the Earth with a radial velocity larger than or equal to the Earth's escape velocity at the point of nearest approach.

I had assumed you meant pieces large enough to be actually seen by humans. Of course, that also means that I have to be more precise with my terms.

Even with that seemingly weak pull, the escape velocity can still be pretty high--according to this escape velocity calculator the escape velocity from Earth's gravity would be around 1440 m/s for an object at the distance of the moon's center. For comparison, the escape velocity to escape from the moon's gravity for an object leaving from the moon's surface would be around 2375 m/s.So a piece escaping from the surface of an exploding moon in exactly the opposite direction from the Earth's center would escape, but for pieces heading out from the surface at other angles, the radial velocity away from the Earth can be lower than 1440 m/s. And the "destroying the moon" calculation assumes each piece of the moon must achieve escape velocity, not from the entire moon, but only from the fraction of the moon's mass which is closer to the center than that piece (since the gravity from all parts at a greater distance from the center will cancel out in Newtonian gravity), so pieces significantly below the surface may be heading away from the moon's center with a velocity significantly less than 2375 m/s relative to the center.

Yes, but its the magnitude of the force that determines whether or not it will fall back to Earth. And the magnitude of the force is determined by its acceleration. So even if it doesn't start out at escape velocity, it can still escape Earth's influence if the magnitude of the acceleration is greater or equal to that of Earth's gravitational field. And as I pointed out, it doesn't have to be that large. In short, even if it is going at escape velocity, it can still fall back if the magnitude of the acceleration is smaller than the acceleration due to gravity.

Also, if you are assuming that the pieces don't fall back to the center (of the moon that is), the pieces have to be moving really really fast. The magnitude force from the explosion would have to be far larger than the force of gravity from the Earth, or that of the moon's. Plus you also have to take into account the gravitational binding energy, which is the energy required to keep it together or break it apart. So even if you technically had enough weapons to deliver enough force to break it apart, it may still not be destroyed unless all of its energy is directed at the moon itself. So that estimate may be off by an order of magnitude.

Of course, of all the pieces that fail to achieve escape velocity, not all will fall to Earth--a significant fraction will go into orbit (perhaps forming a spectacular ring system!)

And that is certainly true. Not all is lost if the moon is suddenly destroyed.

 

[JesseM]

Yes, but its the magnitude of the force that determines whether or not it will fall back to Earth.

No it isn't, it's the gravitational acceleration that determines its path, and the gravitational acceleration is independent of the object's mass. Two objects of totally different masses which start at the same position and initial velocity in a gravitational field will have exactly the same path.

So even if it doesn't start out at escape velocity, it can still escape Earth's influence if the magnitude of the acceleration is greater or equal to that of Earth's gravitational field.

No, that's impossible just because of the definition of escape velocity--the escape velocity is explicitly based on calculating how the object's velocity changes over time due to the gravitational pull, with "escape velocity" being the lower limit for an initial velocity large enough that the outward velocity never drops to zero due to the continued gravitational pull as the object moves away (of course if the outward velocity drops to zero, then the object will begin to fall back towards the source of gravity). No object whose initial velocity is lower than the escape velocity at that distance can ever escape, not unless some other force besides the source of gravity is acting on it.

Also, if you are assuming that the pieces don't fall back to the center (of the moon that is), the pieces have to be moving really really fast.

They have to be moving at the escape velocity for the moon's gravity, which I already found was 2375 m/s for a piece leaving from the surface (although as I pointed out it could be lower for a piece that started out below the surface, since if the explosion is radially symmetric, the gravitational influence of all the pieces that started out at a greater distance from the center will cancel out in Newtonian gravity--see gravitational force inside a spherical shell).

Plus you also have to take into account the gravitational binding energy, which is the energy required to keep it together or break it apart.

The total gravitational binding energy for a planet is defined in terms of the energy that you'd need to give every piece the escape velocity, so that the pieces would not re-form due to their mutal gravity. See the section on escape velocity here, for example:

Gravitational binding energy represents the energy required to eject a body out of the influence of a gravitational field. It is equal to the energy of the system, but opposite in sign. In the absence of friction, this energy is the mechanical energy (sum of potential and kinetic energy) in gravitational field.

Now, it is clear from the definition of binding energy itself that the initial kinetic energy of the projection should be equal to the binding energy of the body in order that it moves out of the gravitational influence. Now, the body to escape is at rest before being initiated in projection. Thus, its binding energy is equal to potential energy only.

Therefore, kinetic energy of the projection should be equal to the magnitude of potential energy on the surface of Earth,

\frac{1}{2} m v_e^2 = \frac{GMm}{R}

where "v_e" is the escape velocity.

 

[Lockheed]

Ok, I got the escape velocity wrong, but you also missed:

The escape velocity can be thought of in terms of the speed an artillery shell or bullet fired from the surface would have to travel (ignoring the effects of drag) to leave orbit, but it is not the speed a rocket or other powered object would have to travel. An object under power could leave the Earth's gravity at any speed, assuming enough fuel.

Obviously blowing the moon is going to deliver a large enough force for it to accelerate, which means that many of the fragments are in principle powered objects. Escape velocity does not necessarily apply. If an object is already going at escape velocity then it doesn't matter, but even an object not going at escape velocity can still leave Earth provided its acceleration is large enough, and as I pointed out it doesn't have to be that large. In fact, if it is accelerating, it can quite easily reach the escape velocity before it stops accelerating. In particular, KE = W = (1/2)mv^2 = F*d*cos (theta), so force is still a pretty big factor in determining whether it escapes or not.

Not only that, but if you were to detonate inside the core, there has to be enough energy to actually blow out the outer portions of the moon, otherwise it will just simply be absorbed. Which requires that the velocity (and depending on the circumstances, its acceleration too) of the core fragments to be much larger than the fragments from the outer portions, since there is less energy by the time it even reaches the surface. So it doesn't really matter if the gravity near the center or core of the moon is lower than if you were to be on its surface. It doesn't change the energy requirements, which is dependent on their speed.

Another thing that I forgot to mention is Torque, centripetal acceleration and rotational velocity, because so far all of the calculations are assuming that the moon is stationary (The moon is orbiting and so has a centripetal acceleration, a rotational velocity and a rotational acceleration, and a velocity TANGENT to its orbit). For the fragments to actually fall to the Earth the torques have to cancel out, there has to be no rotational velocity, and its centripetal acceleration has to be lower than that of Earth's gravitational acceleration. A nuclear explosion releases energy in all directions, so for it to produce any torque whatsoever, the blast from the bombs would have to travel outward primarily in a single direction tangent to its orbit (I challenge you to do that!). So even if you do end of blowing up the moon most or even none of the fragments may actually fall to the Earth because most of it would still be orbiting. This is far more complicated than just relying on escape velocity, there are lots of conditions it has to meet.

So far, my points still stand. In fact, your calculations actually support my points now that I think about it.

No it isn't, it's the gravitational acceleration that determines its path, and the gravitational acceleration is independent of the object's mass. Two objects of totally different masses which start at the same position and initial velocity in a gravitational field will have exactly the same path.

And you missed my point completely. I said that acceleration determines the magnitude of the force, not whether or not it affected mass or its path. The same applies for gravity. And I had assumed that you knew what I meant, either its path is directly away from the Earth or at an angle to it, which would then lead to messy calculations involving force components.

Actually, if you really want a precise definition, F=dp/dt. Which is the change in momentum with respect to time. If the objects have a large enough change in momentum then they can leave Earth regardless of its direction...

 

[JesseM]

Ok, I got the escape velocity wrong, but you also missed:

The escape velocity can be thought of in terms of the speed an artillery shell or bullet fired from the surface would have to travel (ignoring the effects of drag) to leave orbit, but it is not the speed a rocket or other powered object would have to travel. An object under power could leave the Earth's gravity at any speed, assuming enough fuel.

No, I didn't miss that. The only non-gravitational force applied to each piece is the initial explosion, after that the only force on it is gravity alone, so if its initial velocity immediately after the explosion is lower than the escape velocity for that distance, it will eventually fall back.

Obviously blowing the moon is going to deliver a large enough force for it to accelerate, which means that many of the fragments are in principle powered objects.

The initial blast provides a non-gravitional force, but that's over in an instant, it's not applying a continued non-gravitational force over an extended period of time.

If an object is already going at escape velocity then it doesn't matter, but even an object not going at escape velocity can still leave Earth provided its acceleration is large enough, and as I pointed out it doesn't have to be that large.

Acceleration due to what force? Again, only a non-gravitational force like the electromagnetic or pressure waves from an explosion are providing a non-gravitational force on any given piece of the moon, and those are over very quickly, immediately after that you can look at the velocity of the piece and check if it's lower than the escape velocity, if not it will eventually fall back towards the center. Also, the total kinetic energy the explosion imparts to all the different atoms affected by it cannot exceed the energy of the explosion itself, and of course the velocity of an atom can be derived directly from its kinetic energy. So you don't really have to worry about the specific details of how long the explosion applies force to each particle, you can just figure out the total energy released by the explosion that creates the reaction, then look at the total energy needed to give every piece of the moon an escape velocity for its initial position in the moon's gravitational field, and that gives you a minimum for the total amount of energy that all the explosions would need to release to destroy the moon (although this assumes no energy is wasted). It's absolutely impossible that a set of explosions could release less than this energy and still cause the moon to blow apart without any part of it falling back towards the center of the explosion (assuming no other forces are at play besides the ones caused by the explosions and the gravity of all the atoms that make up the moon).

Another thing that I forgot to mention is Torque, centripedal acceleration and rotational velocity, because so far all of the calculations are assuming that the moon is stationary. For the fragments to actually fall to the Earth the torques have to cancel out, and there has to be no rotational velocity.

That's not true, an object can hit the Earth's surface even if it has some nonzero angular velocity around the Earth's center. If an object's angular velocity had to be zero in order for it to hit the Earth's surface, then throwing a ball at an angle instead of straight up would be sufficient to prevent it from ever falling back to the ground! Likewise, satellites in unstable orbits eventually fall back to the surface even though their angular velocity is obviously rather large, and Earth's gravity applies no torque to them.

So even if you do end of blowing up the moon none of the fragments may actually fall to the Earth because most of it would still be orbiting. This is far more complicated than just relying on escape velocity.

I agree that even if a fragment's initial velocity is lower than the escape velocity for the Earth at its initial distance, it may end up in orbit rather than falling to the surface. It would be a very complicated calculation to figure out approximately what percentage would fall, what percentage would orbit, and what percentage would escape...but I see no reason to think the percentage that fell wouldn't be large enough to be quite catastrophic for those of us on Earth.

 

[Lockheed]

No, I didn't miss that. The only non-gravitational force applied to each piece is the initial explosion, after that the only force on it is gravity alone, so if its initial velocity immediately after the explosion is lower than the escape velocity for that distance, it will eventually fall back...

The initial blast provides a non-gravitional force, but that's over in an instant, it's not applying a continued non-gravitational force over an extended period of time...

Acceleration due to what force? Again, only a non-gravitational force like the electromagnetic or pressure waves from an explosion are providing a non-gravitational force on any given piece of the moon, and those are over very quickly, immediately after that you can look at the velocity of the piece and check if it's lower than the escape velocity, if not it will eventually fall back towards the center...

Well, the force from the explosion was what I meant actually. But still, after the initial blast, the velocity will not change (we are in space after all). Since we are assuming the pieces will NOT fall back, you have said it yourself earlier:

Even with that seemingly weak pull, the escape velocity can still be pretty high--according to this escape velocity calculator the escape velocity from Earth's gravity would be around 1440 m/s for an object at the distance of the moon's center. For comparison, the escape velocity to escape from the moon's gravity for an object leaving from the moon's surface would be around 2375 m/s.

Since they already have an escape velocity higher then that of Earth (and thus a large momentum), it won't necessarily fall back, even if it is going toward it. And actually, since it is moving, like I said earlier it depends on where it moves, because the escape velocity is different the further away you are from the Earth.

==============================================
And actually, the explosion doesn't have to be over in an instant. The moon is is a very large object, and it is quite possible for the force to be acted over a considerable amount of time even if the fragments are very far apart from each other. You seem to be assuming a Star Wars like explosion here.

That's not true, an object can hit the Earth's surface even if it has some nonzero angular velocity around the Earth's center. If an object's angular velocity had to be zero in order for it to hit the Earth's surface, then throwing a ball at an angle instead of straight up would be sufficient to prevent it from ever falling back to the ground! Likewise, satellites in unstable orbits eventually fall back to the surface even though their angular velocity is obviously rather large, and Earth's gravity applies no torque to them.

Ok, that's true, a little conceptual mistake on my part .

I agree that even if a fragment's initial velocity is lower than the escape velocity for the Earth at its initial distance, it may end up in orbit rather than falling to the surface. It would be a very complicated calculation to figure out approximately what percentage would fall, what percentage would orbit, and what percentage would escape...but I see no reason to think the percentage that fell wouldn't be large enough to be quite catastrophic for those of us on Earth.

Well, you never know. I suppose we actually have to do the calculations to determine that, because it depends on the situation. If the fragments are very small then we have nothing to worry about. More details are needed before we can determine that.

 

[JesseM]

Well, the force from the explosion was what I meant actually. But still, after the initial blast, the velocity will not change (we are in space after all).

But the velocity does change, because there is still a gravitational force on a given fragment from all the parts of the moon which are closer to the center. If its velocity is higher than the escape velocity, that means that if we ignore all gravitational force except for those from parts of the moon, then although the velocity will continually decrease as the fragment flies out, it will never reach zero so the fragment will never fall back towards the center.

Since we are assuming the pieces will NOT fall back, you have said it yourself earlier:

Even with that seemingly weak pull, the escape velocity can still be pretty high--according to this escape velocity calculator the escape velocity from Earth's gravity would be around 1440 m/s for an object at the distance of the moon's center. For comparison, the escape velocity to escape from the moon's gravity for an object leaving from the moon's surface would be around 2375 m/s.

Since they already have an escape velocity higher then that of Earth (and thus a large momentum), it won't necessarily fall back, even if it is going toward it.

But as I said later in that post, it's the component of the fragment's velocity that's in the radial direction from the Earth's center that determines whether it can escape the Earth's gravity or not. A fragment that is flung from the moon's surface at 2375 m/s in exactly the radial direction away from the Earth's center will escape, but if it's flung at an angle of 60 degrees from this radial axis, its velocity in the radial direction will only be 2375 * cos(60) = 2375*0.5 = 1187.5 m/s, less than the Earth's escape velocity of 1440 m/s (it works out that the velocity in the radial direction will be less than 1440 m/s if the fragment's velocity is more than 52.7 degrees off from being perfectly radial). And of course the half of the fragments from the surface of the moon on the side that's facing towards the Earth will have a radial velocity towards the Earth's center, not away from it. Finally, the 2375 m/s is only for fragments from the moon's surface, the escape velocity from fragments that come from closer to the moon's center is less because these fragments only feel a gravitational pull from the part of the moon's mass that is closer to the center than themselves (again, see gravitational force inside a spherical shell for why the gravity from the shell consisting of all parts of the moon farther from the center will cancel out).

And actually, the explosion doesn't have to be over in an instant. The moon is is a very large object, and it is quite possible for the force to be acted over a considerable amount of time even if the fragments are very far apart from each other. You seem to be assuming a Star Wars like explosion here.

But as I said, if you just think in terms of energy you can avoid having to worry about the details of the duration of the explosion--if the total energy released by the explosions is less than the gravitational binding energy (which again is based on the assumption that every bit of the moon must achieve escape velocity so that no part falls back towards the center), then it is impossible that at least part of the moon wouldn't fall back and re-form a central mass, no matter how the explosions are timed.

Well, you never know. I suppose we actually have to do the calculations to determine that, because it depends on the situation. If the fragments are very small then we have nothing to worry about. More details are needed before we can determine that.

I don't get why you think that small fragments mean we have nothing to worry about. If you dump a cloud of dust on to the Earth from space with a total mass anywhere close to the mass of the moon, that's going to cause pretty catastrophic effects on Earth--the amount of dust kicked into the atmosphere needed for a nuclear winter would be a tiny tiny fraction of the moon's mass, and I don't know how much of the dust would actually hit the surface rather than staying in the atmosphere, but the total kinetic energy of all the falling dust would be just as large as the total kinetic energy of a bunch of large chunks with the same total mass falling from the same height, that energy has to go somewhere so I imagine the Earth's surface would be heated considerably, perhaps converted to a molten state.