Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: katieHaylor on 20/08/2018 09:29:44
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Caroline asks:
If a gigantic asteroid either:
1. hit the moon just enough to knock it slightly out of its orbit but not enough to send it it totally off into space, or
2. hit the moon and the impact broke the moon into several large chunks which continued to orbit the Earth,
what would happen on Earth in both cases?
Also, would we be able to see the impact without telescopes?
What do you think?
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The moon is not accidentally located in the orbit of the earth, and no accidents, it will not remove it from the earth's orbit.
Protecting the earth from asteroids, this is the direct purpose of the moon.
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The moon is not accidentally located in the orbit of the earth, and no accidents, it will not remove it from the earth's orbit.
Protecting the earth from asteroids, this is the direct purpose of the moon.
Please elaborate enough to justify this assertion. Thanks.
To the OP - All depends on the size of the asteroid. If it was large enough to knock the moon out of orbit then it would effect earth. If it changed its orbit it would effect earth. I don't recall how.
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If a gigantic asteroid ..hit the moon.. would we be able to see the impact without telescopes?
Yes - if it were big enough, on our side of the Moon (preferably in shadow), and it was night where the person was viewing it from.
I wonder if the traditional crescent-Moon-and-star symbol on some flags represents one such event?
- Astronomers at the time would have known that a star on the shadowed part of the Moon's face is "impossible"
- So they recorded the event as something significant
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If a gigantic asteroid ...hit the moon... what would happen on Earth?
Some massive asteroids have hit the Moon. A number of the craters visible in a small telescope are over 200km across.
The impact would have blasted debris into space, some of which would have fallen to Earth over the following days and months. Most of it would be dust to gravel size, which would have burnt up in the atmosphere; but some larger chunks would have made it to ground level.
To smash the Moon into several large pieces, the impact crater would have to be a significant fraction of the Moon's diameter - and there is one at the Moon's south pole which is about 2,000km across (70% of the Moon's diameter). However, this is only visible from satellites orbiting the Moon, and is not visible from Earth's surface.
If an asteroid smashed the Moon into pieces, some very large pieces would almost certainly impact the Earth, causing an event bigger than the dinosaur extinction.
But the asteroid Vesta survived an impact which created a crater which is 90% of it's diameter. So even the South Pole Lunar impact was not nearly enough to split the Moon into pieces.
See: https://en.wikipedia.org/wiki/List_of_largest_craters_in_the_Solar_System
Protecting the earth from asteroids, this is the direct purpose of the moon
I believe you are mistaken.
- It is estimated that around 100 tons of meteoroids hit the Earth's atmosphere every day. Since most of them are dust-sized, that translates into a trillion or so impacts.
- The Moon does not deflect these dust-sized meteoroids away from Earth - in fact it will deflect about as many towards the Earth as it deflects away from the Earth.
- The point is that the Moon will do exactly the same deflection regardless of whether the incoming object is a piece of dust, or a dinosaur-killing 10km across.
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If a gigantic asteroid ..hit the moon.. would we be able to see the impact without telescopes?
Yes - if it were big enough, on our side of the Moon (preferably in shadow), and it was night where the person was viewing it from.
I wonder if the traditional crescent-Moon-and-star symbol on some flags represents one such event?
- Astronomers at the time would have known that a star on the shadowed part of the Moon's face is "impossible"
- So they recorded the event as something significant
[ Invalid Attachment ]
If a gigantic asteroid ...hit the moon... what would happen on Earth?
Some massive asteroids have hit the Moon. A number of the craters visible in a small telescope are over 200km across.
The impact would have blasted debris into space, some of which would have fallen to Earth over the following days and months. Most of it would be dust to gravel size, which would have burnt up in the atmosphere; but some larger chunks would have made it to ground level.
To smash the Moon into several large pieces, the impact crater would have to be a significant fraction of the Moon's diameter - and there is one at the Moon's south pole which is about 2,000km across (70% of the Moon's diameter). However, this is only visible from satellites orbiting the Moon, and is not visible from Earth's surface.
If an asteroid smashed the Moon into pieces, some very large pieces would almost certainly impact the Earth, causing an event bigger than the dinosaur extinction.
But the asteroid Vesta survived an impact which created a crater which is 90% of it's diameter. So even the South Pole Lunar impact was not nearly enough to split the Moon into pieces.
See: https://en.wikipedia.org/wiki/List_of_largest_craters_in_the_Solar_System
Protecting the earth from asteroids, this is the direct purpose of the moon
I believe you are mistaken.
- It is estimated that around 100 tons of meteoroids hit the Earth's atmosphere every day. Since most of them are dust-sized, that translates into a trillion or so impacts.
- The Moon does not deflect these dust-sized meteoroids away from Earth - in fact it will deflect about as many towards the Earth as it deflects away from the Earth.
- The point is that the Moon will do exactly the same deflection regardless of whether the incoming object is a piece of dust, or a dinosaur-killing 10km across.
Great response, as expected! :)
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If an asteroid smashed the Moon into pieces, some very large pieces would almost certainly impact the Earth, causing an event bigger than the dinosaur extinction.
Not necessarily. The gravitational binding energy of the Moon is ~1.2e29J This is the minimum energy needed to "shatter" the Moon such that the resultant pieces aren't drawn back together by their mutual gravitational attraction.
That energy would be used up in just doing that. In order for any of the pieces to reach Earth, they would have to be put into an Earth surface intersecting orbit. This would require additional energy; more than what would be required to kick the same sized piece out of orbit around the Earth. (The recent Parker Solar probe illustrates this idea. It had to be given ~11km/sec just to escape the Earth, but that is all used up fighting Earth's gravity. To get to the Sun, additional delta v had to be supplied, in an amount that exceeded that which would have been needed to escape the Solar system.)
So there is a range of collision energies which could leave us with a shattered Moon, but without enough energy left over to put any large hunks on an Earth intersecting trajectory. Instead, we are left with a debris ring.
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The moon is not accidentally located in the orbit of the earth, and no accidents, it will not remove it from the earth's orbit.
Protecting the earth from asteroids, this is the direct purpose of the moon.
Do you even science?
The moon was created by the impact of a large asteroid hitting the Earth.
At best that's a strange sort of "protecting".
It's an absurd bit of wishful thinking to believe that the moon has a "purpose".
I presume that the OP (and other readers of the thread) will recognise that you are spouting utter nonsense.
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The gravitational binding energy of the Moon is ~1.2e29J This is the minimum energy needed to "shatter" the Moon such that the resultant pieces aren't drawn back together by their mutual gravitational attraction
Thanks, Janus - I agree that this is a theoretical possibility, if the energy were applied such that it is evenly divided amongst the pieces of the Moon.
However, I suspect that most real collisions between astronomical bodies will be much more "messy", with a lot of vaporized rock created at the point of impact, blasting out material sideways, greater than escape velocity (some of which probably would impact the Earth).
The behaviour will then be quite different once the impactor slows below the speed of sound in rock (typically around 3km/sec), as energy can be carried away from the impact point by earthquake (Moonquake?) waves.
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The moon was created by the impact of a large asteroid hitting the Earth.
Randomly flying asteroids can not create random satellites, in nature there are no accidents.
Randomness is chaos, and chaos does not exist in nature, chaos exists only in the minds of people. Which are based on erroneous theories.
Planets and satellites are formed on the basis of the laws of celestial mechanics, the three-body problem (and this is the mystery of the axial and orbital motion of planets and satellites, and the forces behind the movement of satellites).
Satellites of planets are formed from rings.
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The gravitational binding energy of the Moon is ~1.2e29J This is the minimum energy needed to "shatter" the Moon such that the resultant pieces aren't drawn back together by their mutual gravitational attraction
Thanks, Janus - I agree that this is a theoretical possibility, if the energy were applied such that it is evenly divided amongst the pieces of the Moon.
However, I suspect that most real collisions between astronomical bodies will be much more "messy", with a lot of vaporized rock created at the point of impact, blasting out material sideways, greater than escape velocity (some of which probably would impact the Earth).
The behaviour will then be quite different once the impactor slows below the speed of sound in rock (typically around 3km/sec), as energy can be carried away from the impact point by earthquake (Moonquake?) waves.
It's not enough for the material to reach escape velocity from the Moon, it would also have to have enough excess velocity beyond that to reach Earth. An object launched from the Moon's surface at Moon escape velocity will end up in orbit around the Earth at Moon orbit distance. In order to reach the Earth, it would have to shed a good part of of the Moon's orbital velocity of 1018 m/s. The easiest way to do this is to fire it in the opposite direction of the Moon's orbital motion, so that it has ~836 m/s relative to the Moon left over after escaping the Moon's gravity. (basically this means that it would have to leave the Moon's surface at 2518 m/s rather than the 2375 m/s moon escape velocity. In addition, there is also a maximum velocity an object launched in this direction can have relative to the Moon and still hit the Earth and that is ~1200 m/s. After that, the object is put into a retrograde orbit with a perigee that is above the Earth's surface)
Now what would happen if you fired that object directly at the Earth rather than opposite the Moon's orbital motion? It would enter an orbit with a perigee some 200,000 km short of the Earth's surface.
To actually hit the Earth aiming in this direction takes a much larger velocity with respect to the Moon than aiming "backwards" does.
So that 834 m/s left over after gravitational escape is the minimum needed for an Earth intersecting trajectory, and that's along a particular trajectory.
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The moon was created by the impact of a large asteroid hitting the Earth.
Randomly flying asteroids can not create random satellites, in nature there are no accidents.
Randomness is chaos, and chaos does not exist in nature, chaos exists only in the minds of people. Which are based on erroneous theories.
Planets and satellites are formed on the basis of the laws of celestial mechanics, the three-body problem (and this is the mystery of the axial and orbital motion of planets and satellites, and the forces behind the movement of satellites).
Satellites of planets are formed from rings.
Chaos definitely occurs in nature... you even appear to refer to it in this post as "the three body problem." You should look up the double pendulum (the third body involved is the earth).
Claiming that naturally occurring phenomena (such as the moon's orbit) are "purposeful" is well outside of the realm of science. Please refrain from posting such comments in the threads in the Physics section. Continuing to do so will result in a temporary ban.
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[The gravitational binding energy of the Moon is ~1.2e29J This is the minimum energy needed to "shatter" the Moon such that the resultant pieces aren't drawn back together by their mutual gravitational attraction.
Help me out here. By definition, the magnitude of gravitational binding energy is the energy required to assemble a gravitational body from point particles at infinity. So by shatter you must mean either powderize it or you really meant it as an approximation. Its been years since I've followed these derivations so have mercy on me when I ask - If we break the moon into 4 pieces of equal mass and either go into orbit around the Sun or fly off to infinity, does each case require the grav-binding energy? Thanks.
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The subject not mentioned here yet is life without a moon. There's a good article on this at
https://www.scientificamerican.com/article/moon-life-tides/
When a large enough asteroid smashes into Earth which is large enough to kill all life on earth then it might require having a moon to jump start the process. See link above.
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I did some back-of-the-envelope calculations (which are probably wrong, but here goes... corrections welcome! :)
It's not enough for the material to reach escape velocity from the Moon, it would also have to have enough excess velocity beyond that to reach Earth.
Incoming velocity of an asteroid = (Lunar Escape velocity from the surface) + (Earth Escape velocity from Lunar Orbit) + (Difference in orbital velocities)
= 2.4km/s + 1.4km/s + (anything from 0 to 18km/s)
= Let's call it 4km/s collision velocity, assuming an asteroid of negligible mass compared to the Moon
To make the geometry simpler, I assumed a spherical Moon colliding head-on with a spherical asteroid of equal diameter, with a relative velocity of 4km/s.
Now, 1 second into the collision, the contact area is a circle of around 240km diameter.
In that 1 second, an attempt has been made to crush 170,000 cubic kilometers of rock into a volume about half that - a rather energetic exercise, which I expect would vaporise a lot of rock.
After 1 second, the contact area will be expanding outwards at around 15km/s, pushing a large volume of superheated, vaporised rock ahead of it. I am sure this will scoop up a lot of non-vaporised rocks and give them a significant boost.
So I am guessing that there will be a lot of rock "squeezed out" from between the colliding bodies, at greater than Lunar escape velocity.
So that 834 m/s left over after gravitational escape is the minimum needed for an Earth intersecting trajectory, and that's along a particular trajectory.
I estimate that there is an equal balance point between Earth and Moon (Lagrangian Point L1) that is about 43,000km from the Moon, so something flung off the Moon in this direction would only need to start with about 1.9km/s velocity, which is less than Lunar escape velocity. But that is also a very specific trajectory.
Comments welcome!
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I estimate that there is an equal balance point between Earth and Moon (Lagrangian Point L1) that is about 43,000km from the Moon, so something flung off the Moon in this direction would only need to start with about 1.9km/s velocity, which is less than Lunar escape velocity. But that is also a very specific trajectory.
Comments welcome!
L1, some 326054 km from the Earth, is a point where, with the help of the Moon's gravity, an object could be placed and it would stay (if not further disturbed) even though it is traveling at a speed slower than what would normally be needed to maintain a circular orbit. Reaching L1 does not mean you now have a free trip to Earth.
Moon orbital velocity is ~1019 m/s. In order to maintain relative position with the Moon, an L1 object would be moving at ~864 m/s, which is -241m/sec slower than the 1106 m/s it would have to be moving in order to maintain the same path around the Earth in a free orbit without the aide of the Moon. (V_o = sqrt(u/r), where u is the gravitational parameter for the Earth of 3.987e14 M^3/s^2)
If we were to remove the Moon, an object at L1 distance from the Earth and moving at L1 point velocity would be at the apogee of an elliptical orbit with a perigee of ~143250 km from the Earth.
(orbital energy/kg = v^2/2-u/r = u/2a, where a is the semi-major axis of the orbit.
solving solve for and then use a2-r to find the perigee distance)
Even without the Moon's aide, an object placed at the L1 point would not "fall" all the way to the Earth.
To get such an object to reach Earth surface you would have to reduce its L1 point velocity to ~217 m/s or by an additional 647 m/s.
V_r = sqrt(u(2/r-1/a) where a= (r+r_e)/2 and r__e is the radius of the Earth.
Again, this velocity change would need to be applied in the opposite direction of the orbital velocity.
From the actual L1 point, its going to take a bit more. That same gravitational "help" the Moon gives an L1 point object in allowing it to hold position, is now going to be a hindrance in trying to reach Earth.
Think of it this way, without the Moon, the object would drop to 143250 km from the Earth, with the Moon, it maintains a constant 326054 km distance. That means with the Moon, you would have to give the L1 object a push in order to force it into that elliptical orbit that it would naturally assume without the Moon. The same is true for putting it on an Earth intersecting trajectory, its going to take more effort to put it on that path with the Moon than it would have without the Moon for the Same object starting at the same point and speed.
This is orbital mechanics that we are dealing with. Intuition often leads you astray.