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Asteroid Impact, Shorter Days

  • Thread starter BOAS
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  • #26
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KE is not going to help. Huge amounts of mechanical energy will be converted to heat. Angular momentum is the only way.
I don't think so... In the problem's statement: (...)'an asteroid traveling straight toward the centre of the earth were to collide
with our planet at the equator'
(...); 'straight toward the center' would mean following any of the Earth's radii... Thus, for an observer close to the impact point, the asteroid would be seen as coming down from the zenith, and -on contact- no angular moment transfer would take place at all, as with any object falling from the zenith... If you add that -always according to the original problem statement, (...) 'and bury itself just below the surface' (...), then very little dissipation of energy could be expected...
 
  • #27
haruspex
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I don't think so... In the problem's statement: (...)'an asteroid traveling straight toward the centre of the earth were to collide
with our planet at the equator'
(...); 'straight toward the center' would mean following any of the Earth's radii... Thus, for an observer close to the impact point, the asteroid would be seen as coming down from the zenith, and -on contact- no angular moment transfer would take place at all,
The impact is at the equator. To an observer there, the asteroid would have a horizontal speed of 1670 kilometers/hour relative to the surface of the Earth. The angular momentum of the Earth+asteroid system does not change, but the moment of inertia of the Earth increases by acquisition of the extra mass, so the angular velocity decreases. That is the whole point of the question.
If you add that -always according to the original problem statement, (...) 'and bury itself just below the surface' (...), then very little dissipation of energy could be expected...
It's the rotational KE of the Earth that is relevant, agreed? Forget the vertical collision, that's not relevant here - there is a massive tangential collision. You could lower the asteroid gently onto the surface and get the same result.
Try your energy-conserving approach. You will get quite a different answer from the (completely reliable) angular momentum approach, and it is therefore invalid.
 
  • #28
BiGyElLoWhAt
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Homework Statement


[/B]
Suppose that an asteroid ...
Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.
I don't think it's that simple.

For a sizable mass located a distance away from the axis of rotation...
The mass is most definitely not sizeable, however, it is massive (perhaps this is what you meant?). It's meant to be treated as a point mass. ##M_a R_{earth}^2 + I_e = I_{combined}##
 
  • #29
SteamKing
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The mass is most definitely not sizeable, however, it is massive (perhaps this is what you meant?). It's meant to be treated as a point mass. ##M_a R_{earth}^2 + I_e = I_{combined}##
An asteroid 16% of earth's mass sounds both sizable and massive. After all, the moon is only 1.2% of earth's mass, and no one neglects it.

By comparison, the biggest asteroid (dwarf planet?) known is Ceres, and it has a mass of only 0.015% of earth.

http://en.wikipedia.org/wiki/Ceres_(dwarf_planet)

If you throw Pluto into the mix, its mass is 0.22% of earth.

http://en.wikipedia.org/wiki/Pluto
 
  • #30
BiGyElLoWhAt
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That's all fine and dandy, but the problem statement says that the size of the asteroid is neglegable. This is so you can treat it as a point mass. Since when do Mechanics 1 problems need to be realistic?
 
  • #31
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That's all fine and dandy, but the problem statement says that the size of the asteroid is neglegable. This is so you can treat it as a point mass. Since when do Mechanics 1 problems need to be realistic?
Turns out whoever wrote this problem either lied or didn't know what the answer was. Having something 10 times as massive as the moon hitting earth is not 'negligible' in any sense of the word.
 
  • #32
BiGyElLoWhAt
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Ok, so then op needs to integrate r^2 dm over the density function of the asteroid (which we don't have) and sum that with the earth.
[edit] [earth]'s moment.
 
  • #33
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Ok, so then op needs to integrate r^2 dm over the density function of the asteroid (which we don't have) and sum that with the earth.
You can make a list of things we don't know about this asteroid, which is why this problem is poorly constructed.
 
  • #34
BiGyElLoWhAt
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This was just a problem that someone pulled out of their rear end. Think about it. An asteroid hits the earth aimed at towards the center of the earth, and causes an increase in our day by 28% (!!!!!). That's huge for an asteroid that exerts 0 torque. It's all about recognizing you need to use angular momentum and applying it properly.
 
  • #35
haruspex
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28.0% longer than it presently is
BOAS, are you sure it's 28%, not perhaps 0.28%?
 
  • #36
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The asteroid is a point mass: "Assume that the asteroid is very small compared to the earth" - sure it is completely unrealistic, but we can ignore that issue.

Such a massive asteroid (Mars just has 10% of the mass of earth, our "asteroid" is larger!) would change the center of rotation of earth. This makes the new moment of inertia a bit smaller than calculated before.

Also, a day 28% longer than before does not correspond to a new angular velocity 28% smaller than before.

Putting all together, I get about 12.6% of the mass of earth for the asteroid.

18.4% with the 28%-mistake, 11.2% with the wrong center of rotation and 15.56% with both at the same time.


By the way, you do not need any data about earth (apart from the constant density) - the result is true independent of its size, mass or length of a day.
 
  • #37
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BOAS, are you sure it's 28%, not perhaps 0.28%?
It's definitely 28.0%.
 
  • #38
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Also, a day 28% longer than before does not correspond to a new angular velocity 28% smaller than before.
Ah, you're right. it's 2pi radians in 1.28*86400 seconds
 

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