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jaydnul

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- Thread starter jaydnul
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- #1

jaydnul

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- #2

jaydnul

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what speed* (at the end)

- #3

willem2

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,

Maybe a few small pieces of the moon would fall. It's actually easier to blow up the moon than to stop it.

For the moon to crash into the earth, you need to cancel nearly all of the 1 km/sec orbital speed, or you'll get just an elliptic orbit.

The momentum of this asteroid must be equal to that of the moon but in the opposite direction.

This means that (mass of asteroid) * (speed of asteroid relative to the moon) =

(mass of of moon) * ( 1 km/sec)

The maximum speed relative to the moon (or the earth) that an asteroid in an orbit around the sun can have is about 70 km/s. (retrograde orbit. with the apogee in the kuiper belt). Such an object would have a mass of at least (1/70) of the the moon.

The diameter would be 800 km, if it had the same density of the moon.

The kinetic energy would be 2.5*10^30 J, more than a 100 times the gravitaional energy needed to blow up the moon into tiny bits. Everyone on the near side of the earth would be killed by the flash of the explosion, and everyone else after at least 0.1% of the remains of the moon would land on the earth.

A smaller asteroid would have to have even more energy to stop the moon and would produce an even bigger bang.

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- #4

BruceW

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yes. If the asteroid had equal and opposite momentum to the moon, then when they 'stick together' due to collision, then the new object would not be moving, so it would fall into the earth. (This type of collision is called inelastic, because the mechanical energy is clearly not conserved). In reality, energy is conserved, but it is simply turned into other forms, e.g. heat.

And for your other question: We assume the moon is initially in a circular orbit, and then we want to imagine an asteroid (moving in the same direction as the moon), collides with it, and we want to know the momentum of the asteroid which we require to cause the new 'asteroid+moon' object to completely fly out of orbit of the earth.

Ok, so first let's name the variables. [itex]m_m \ , \ v_m[/itex] to be the initial mass and speed of the moon and [itex]m_a \ , \ v_a[/itex] to be the initial mass and speed of the asteroid. And finally, [itex]m_e[/itex] to be the mass of the earth, and [itex]r[/itex] to be the initial distance between the moon and earth.

First, we assumed the moon was initially in circular orbit around the earth, so we have:

[tex]m_m \frac{{v_m}^2}{r} = \frac{m_e m_m G}{r^2} [/tex]

(Where [itex]G[/itex] is the gravitational constant). And rearranging:

[tex]v_m = \sqrt{\frac{m_e G}{r}} [/tex]

So that's fine, for the initial speed of the moon. Now we can also define [itex]m_c \ , \ v_c [/itex] are the mass and speed of the 'combined asteroid and moon' object, then we want the kinetic energy plus gravitational energy of the new object to equal zero, since this is the least amount of energy required for the object to escape earth's gravity. So:

[tex]-\frac{m_e m_c G}{r} +\frac{1}{2} m_c {v_c}^2 = 0 [/tex]

This last equation is the total energy of the new object, which we want to equal (at least) zero. Also, notice that the gravitational energy of the new object must be negative, this is because I am defining the gravitational energy to equal zero at infinity. Now, rearranging the last equation:

[tex]v_c = \sqrt{\frac{2 m_e G}{r}} [/tex]

Right. So, to make it clear, this is the speed of the 'combined moon and asteroid object', just after the collision has happened. Also, there is another equation we can use: conservation of momentum:

[tex]m_m v_m + m_a v_a = m_c v_c [/tex]

And now, if we use our equations for [itex]v_m[/itex] and [itex]v_c[/itex], we get:

[tex]m_m \sqrt{\frac{m_e G}{r}} + m_a v_a = m_c \sqrt{\frac{2 m_e G}{r}} [/tex]

We can use one more equation, which is conservation of mass:

[tex]m_a + m_m = m_c [/tex]

And we can use this for the value of [itex]m_c[/itex], so then we have:

[tex]m_m \sqrt{\frac{m_e G}{r}} + m_a v_a = (m_m + m_a) \sqrt{2} \sqrt{\frac{m_e G}{r}} [/tex]

And from here, doing some rearranging, we get:

[tex] v_a = (\frac{m_m}{m_a} (\sqrt{2} - 1) + \sqrt{2} ) \sqrt{\frac{m_e G}{r}} [/tex]

So this equation tells us what the minimum speed of the asteroid needs to be, in terms of the other variables. Does my working look good? Hopefully I haven't made any mistakes, It is 4am where I am :)

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- #5

K^2

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BruceW, the largest asteroid in Solar System is Vesta. Please, compute the amount of energy released in above collision with m

- #6

BruceW

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hmm. The equation for the energy released in the collision (assuming completely inelastic), is:

[tex]E_{release} = \frac{1}{2} \frac{m_m m_a}{m_m + m_a} (v_m - v_a )^2 [/tex]

And using m

I'll have to come back to try to work out the total gravitational energy of the moon :) I have run out of time right now.

- #7

K^2

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BruceW, use the equation for v

The question is hypothetical. Can an asteroid knock the Moon out of its orbit? A heavier asteroid is going to do the job much easier than the light one. That's the only reason I'm taking Vesta's mass as m

Question, given that it's moving fast enough, will it actually knock the moon out of the orbit, or will it blast it into pieces?

Similarly, say the asteroid is going the other way fast enough to hypothetically stop the moon completely in an inelastic collision. Should we be worried about the Moon just falling down on us in this scenario? Or will something different happen?

- #8

voko

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I do not think it is possible to stop the Moon by Vesta. Vesta is about 300 times less massive, so it would have to go at 300 km/s so that the resultant momentum be zero. Because it is 300 times less massive, its frontal projection area is about 50 times smaller than that of the Moon. I think Vesta will just pierce through the Moon and exit as ionized gas, and the Moon might break into a number of large pieces, but they will keep moving with about the same velocity.

- #9

K^2

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Well, that's the other extreme case, but it doesn't really matter. Either it punches clean through, taking most of energy with it, or it is mostly stopped, releasing most of the energy into the Moon's interior. Wherever it actually falls between these two, the net result isn't an arrested Moon. It's some fraction of the Moon still in orbit, with the rest blasted out into space.

- #10

BruceW

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Now that I think about it, an asteroid in orbit around the sun couldn't have enough energy to knock the moon out of the earth's orbit. If the asteroid was only just being kept in orbit around the sun, then we have the equation:

[tex]\frac{1}{2} {v_a}^2 = \frac{m_{sun} G}{r} [/tex]

And rearranging:

[tex]v_a = \sqrt{\frac{2 m_{sun} G}{r}} [/tex]

And when r is the distance from the sun to the moon, then the speed of the asteroid is roughly 42 km/s So is this going fast enough? Well using the equation for the minimum required speed of the asteroid to knock the moon out of the earth's orbit in a completely inelastic collision:

[tex]v_a = ( \frac{m_m}{m_a}(\sqrt{2}-1)+ \sqrt{2})v_m [/tex]

Gives us a minimum required speed of 113 km/s (For an asteroid which is the mass of vesta). So even the most massive asteroid, going at the fastest possible speed while still being kept in the sun's orbit, wouldn't be going fast enough to knock the moon out of the earth's orbit.

One thing I haven't thought of is if the asteroid was slingshot by passing very close to a planet, then continued on to crash into the moon. Maybe that would get the asteroid to move fast enough...

- #11

K^2

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None of the planets have high enough escape velocity to slingshot the asteroid fast enough either. That's why I said earlier that it'd have to be an extrasolar asteroid. The only reason I'm using mass of Vesta is because anything significantly heavier would be classified as a rogue planet.

- #12

BruceW

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Sorry to post again after such a long time on this thread, but I got round to working out the gravitational binding energy of the moon, which is roughly 10

- #13

K^2

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You are still comparing it to kinetic energy of Vesta in its current orbit, which has nothing to do with the question. You need to compare it to the energy released in impact that transfers enough momentum to stop the Moon or to eject it from orbit. Again, the mass of Vesta is only used as an upper limit of what we are willing to call an asteroid. We are not interested in its orbital characteristics, as an asteroid required for this collision is obviously extrasolar.^{29}Joules. This is more than the energy produced by collision, but not by much. It is interesting that the two numbers are so close. I guess that since the energy produced by collision is almost as much as the binding energy, then the moon would get pretty much obliterated. Poor guy.

- #14

BruceW

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Oh yeah, that's true. Duh. Okay, for a collision between the moon and an object with the mass of Vesta, but a speed of 113km/s (enough to eject the moon from the earth's orbit), the energy released would be roughly 2x10

So this means that any object that can be classified as 'asteroid' (as far as mass is concerned), is more likely to cause the moon to be smashed apart than to be pushed out of its orbit of the earth. Not a very surprising result, but it is good to see the physics.

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