Would any amount of activity (such as an explosion) cause the moon to crash into Earth?
If we were to set off a bomb on the non-Earth-facing side (aka “the (not really) dark side”) of the moon, how big would such an explosion have to be for the moon to fall to Earth?
First of all, you’re hitting it on the wrong side. Think of a plane in flight. If you want it to hit the ground, you don’t smash it on top. You smash into it in the front, to stop it, and gravity will do the rest. So you’d want to hit the moon off to the side, in the direction it’s orbiting.
As for how big the explosion would have to be, 18,000,000,000,000 megatons. Because people like to see numbers like this, that’s about 3.6 billion (3,600,000,000) times the combined global nuclear arsenal.
But shouldn’t you have to factor into account that both the moon and the earth are subject to the Sun’s gravitational field?
I agree with Achernar that the “bomb” gives the most bang per buck if detonated at a point on the leading face of the moon since you must reduce it’s orbital velocity before it can fall to earth. As to the amount of energy, Achernar, is this figure calculated to bring the moon to a dead stop, or just enough to reduce the orbital radius to approximately the sum of the radia of the earth and the moon?
Achenar, you’re mistaken. You need to give the moon a net velocity towards the Earth. Deceleration orthogonal to the orbit may do this so long as gravitation acceleration overcomes the orbital velocity, but acceleration perpendicular to the orbital axis will do a better job as it directly assists the gravitational acceleration.
As for the amount, well an instantaneous impact would have to be very large: the moon’s mass is 7.35x10^22 kg i.e. 7.35x10^13 megatonnes, so Achenar’s explosion would smash it to smithereens.
Paging The Bad Astronomer for chapter and verse.
I’d think that amount of force would be enough to shatter the moon completely.
First off, I made a mistake. I forgot to divide by two. So it’s only 9,000,000,000,000 megatons.
Second, moes lotion brings up a good point. My figure (GM[sub]Earth[/sub]M[sub]Moon[/sub]/2d[sub]moon[/sub]) is just to bring the moon to a stop, and let it fall in. The reason this would work, qts, is that the orbital speed is now 0, and so of course gravity would overcome it.
What you could do instead is take away just enough kinetic energy that it enters a highly eccentric orbit and scrapes the surface of the Earth at the next perigee. This would be the least amount of energy you would have to expend to get the moon to touch the Earth. I’ll work out what it is…
Wow. That would be a bigger catastrophe even than someone unleashing a battleship-sized 1920’s-Style Death Ray on the world from space.
What we want is an orbit with the current distance of the moon (d[sub]moon[/sub] = 384,467 km) as the apogee (A) and the sum of the Earth and moon radii (8116.24 km) as the perigee §. Such an orbit would have eccentricity e = (A-P)/(A+P) = 0.959. The orbital speed at apogee is v[sub]A[/sub][sup]2[/sup] = GM/A × (1-e), and the orbital kinetic energy is GMm(1-e)/2A.
The current orbital kinetic energy of the moon is GMm/2A (this is the 9 trillion megatons). The difference, then, is GMme/2A, which is only 8.7 trillion megatons.
Tusculan, I think you might need to account for the sun, but not to first order. (I mean, technically, a projectile on the Earth is in orbit around the sun, in a different orbit as the Earth, but we never bother with that.)
qts, I’m not sure how you converted the moon’s mass into megatons, but these values are of course comparable (or equal) to the orbital kinetic energy of the moon. They have to be. You could do it with a bunch of smaller explosions, I guess, but I don’t think the sum of their energies could be less than the above value.
If we wanted to crash the moon into the Earth using the method in the OP, pushing it from the far side, it would take a bit more energy. This can be described by an orbit with the same angular momentum as the present one, but with a perigee at P = 8116km.
I get that this would take (to first order) GMm/2d × (d/P)[sup]2[/sup] = 20,000 trillion megatons.
Achernar, you’re assuming that the energy needed in the explosion would be the same as the orbital energy of the Moon (or the difference in orbital enegy of the Moon, before and after the explosion), but this is not the case. That’d only work if you were hitting the Moon with another Moon-sized mass. The most efficient way to use your available energy is actually to eject material from the surface of the Moon at the Moon’s escape speed. In a post from a few years ago, I calculated how much energy it would take to lower the Moon’s orbit to the Earth’s Roche limit (this is the point at which the Moon would break apart from the tidal stresses). It would take a bit more energy to bring it all the way down to the Earth’s surface, but not much. I don’t recall the binding energy of the Moon offhand (the amount of energy needed to blow the Moon to smithereens where it is), but it’s a lot higher.
(apologies for the lack of numbers… I meant to include them and a link to the previous thread, but the search engine doesn’t seem to like me today)
What would be the effects on the earth as the moon approaches? In his book, our own Bad Astronomer describes the effects of a close encounter (~ 1000km) with a planet like venus - the whole sky blocked out for days or weeks, kilometer-high tides, uniminaginable earthquakes…
I’ll be happy to do the math, but I’m not sure what you’re saying. Can you descibe it a little more?
An Atlanta Falcons Superbowl victory. Rest easy, my child, it ain’t gonna happen.
I don’t think we need to start all this again…
Well, thanks for all the info… I was just curious to know if any amount of activity done on the moon in the process of building a lunar base would affect the moon’s orbit at all. But that doesn’t seem to be the case.
The process of constructing a lunar base would certainly not affect the Moon’s orbital velocity to any significant degree.
Look at it another way; the process of building all kinds of crap on Earth hasn’t affected its orbit around the sun.
DarrenS, I suspect the results of a close shave by the Moon would be along the same lines. The radical shifts caused by the huge tidal force would certainly cause absolute catastrophe - monster shifting of the Earth’s crust and constinental plates.
Finally found those old calculations, and I got that it would take 6.6710[sup]28[/sup]J, or nearly 1.6710[sup]13[/sup]megatons, to bring the Moon down to the Roche limit (ignore the figure of 9.56*10[sup]6[/sup] in the linked thread; I just now noticed that that was a mistake).
Achernar, your mistake was in just calculating the change in energy of the Moon, without considering how you’d bring about that change in energy. Remember, you also have to conserve momentum. So to change the momentum of the Moon, you also need to give momentum to some reaction mass, in the opposite direction, and you lose energy doing that, too. The lower the velocity of your reaction mass, the less energy you waste on moving it. But if you use any velocity less than lunar escape speed, the reaction mass will just end up crashing back onto the Moon, undoing all of your change in momentum. Once you’ve gotten all that figured out, it’s just a fairly straightforward collision problem with conservation of momentum to figure out the mass you need to eject, and once you have the mass and the speed, you can find the energy you need.
You give me a couple mil, and I’ll make it happen, my friend.
Don’t forget to correct for wind resistance.