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If an Asteroid was to strike the Earth, would it affect noticeably the Earth's rotation, and if so, how large would this Asteroid have to be?

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    $\begingroup$ It'd be the least of our worries. $\endgroup$ – gerrit Sep 16 '15 at 14:10
  • $\begingroup$ haha XD @gerrit still an interesting thought though because if it's possible, then it is also possible to have a different rotation than that existed during the dinosaur era- before an asteroid wiped them out. $\endgroup$ – JonathanScialpi Sep 16 '15 at 14:19
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    $\begingroup$ There certainly is a different rotation now that 65 million years ago, as the Earth rotation is slowing down. I don't know the answer to your question, though. $\endgroup$ – gerrit Sep 16 '15 at 14:34
  • $\begingroup$ Right, well I meant in terms of an outer space projectile directly affecting its rotation. +1 for the cool fact though $\endgroup$ – JonathanScialpi Sep 16 '15 at 14:38
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    $\begingroup$ Everything has some effect, but it's pretty tiny. Even the Chicxulub impact, about 5 miles across, the largest in over 100 million years had a tiny amount of energy compared to the Earth's rotation. I can try to run the math later, but you'd need a really big impact to have a significant effect. A 100 mile in diameter space rock would obliterate the Earth's surface and boil the oceans but it's mass would be just a bit over a millionth that of the Earth. Even at a perfect impact angle at a few tens of thousand MPH, it would only add or remove a tiny bit of rotational velocity. $\endgroup$ – userLTK Sep 16 '15 at 18:15
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To have a noticeable effect the impactor needs to be BIG.

Most questions about "what would happen if ... hits" can be answered by the "Earth impact effects program" (http://impact.ese.ic.ac.uk/)

Here are calculations for a 100km stony asteroid...

A brute like this would have a good chance of wiping out most complex life on the planet. There has been nothing like this in the last 4 billion years (or so)... It could cause the length of the day to change by "up to 2.42 seconds"

As gerrit said, it would be the last of our worries.

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An asteroid less than a kilometer wide impacting the earth could noticeably affect the rotation rate.

When an asteroid strikes the earth in an inelastic collision, momentum is conserved between the asteroid and the earth (as long as no material is ejected). That means that all the linear momentum from the asteroid is transferred into the momentum of the earth's revolution around the sun and the earth's rotation around its own axis.

The transfer of asteroid momentum into the earth's rotational angular momentum will be maximized if the asteroid's trajectory lies within the equatorial plane and strikes along the equator at a shallow angle (similar to spinning the cue ball in the game of billiards). For the below calculations, I choose asteroid trajectory, speed, and density to minimize the size of the asteroid necessary to alter the earth's rotation rate.

The rotation rate of the earth is known to within $\omega =1e-13$ radians per second https://en.wikipedia.org/wiki/Earth%27s_rotation . So we would notice via GPS measurements if the rate changed by more than that.

The moment of inertia of a solid sphere is $I=2/5MR^2$ where $M$ and $R$ are the mass and radius of the earth. Using wikipedia values, $I$ is about $9.3e37kgm^2$. So the angular momentum change to noticeable slow the earth is $I\omega = 9.3e24kgm^2/s$. To translate this to linear momentum, we can divide this by the equatorial radius of the earth to get $l=1.53e18kgm/s$. This is the linear momentum an asteroid would need to impart to the earth, along the equator, tangent to the earth's surface to noticeably alter the rotation rate.

If we take a fast meteor like Oumoamoa that was going about 50km/s at 1 AU, and suppose it has an impact trajectory opposite the earth's orbit, we can add the earth's orbital speed of 30km/s to get a whopping speed of $s = 8e5m/s$ impact. Since linear momentum $l=mv$ the mass of the asteroid is $m = l/s = 1.53e18/8e5 = 1.925e12kg$.

The above calculations assume that the asteroid strikes the earth surface at zero degrees, but such a trajectory would cause the asteroid to bounce off atmosphere back into space. A steeper angle like 45 degrees would mean only half of the asteroid's linear momentum would be transferred into the earth's angular momentum, so we would need an asteroid twice as massive or $m=3.85e12kg$.

If the asteroid is a very dense $d=9000kg/m^3$ then the volume of the asteroid is $v = m/d = 4.3e8m^3$. Since volume of a sphere is $v = 4/3\pi r^3$, if we solve for the radius, we get $r=468m$. So the diameter of the asteroid is $d=2r=936m$ wide or slightly less than a kilometer!

Note: We have never had an asteroid this large strike in human history (10,000 years ago), but at least 6 have struck the earth since humans evolved (300,000 years ago). https://en.wikipedia.org/wiki/List_of_impact_craters_on_Earth

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  • $\begingroup$ If I did the math right, this would alter the day length by about 1ms. I'm not sure this would be enough to throw off GPS satellites - the day length normally fluctuates by more than that over the course of weeks/months. See en.wikipedia.org/wiki/Day_length_fluctuations#Observations. Apparently there are tools to measure such small changes in day length, but I'm not sure how that's done. $\endgroup$ – Nuclear Hoagie Oct 30 at 19:34
  • $\begingroup$ I actually got just a little over a nanosecond per day. I don't think this would throw GPS off either, in my answer above I just meant we could use GPS to help find the change www2.unb.ca/gge/Pubs/TR171.pdf . All the day to day fluctuations average out over a long period of time since the entire earth system (land, atmosphere, and magma currents) conserves angular momentum. A change to the overall rotation rate, however, will show up as a bias instead of noise. Regardless of how it's calculated, I think the known error bound for rotation rate is the smallest noticeable change. $\endgroup$ – Connor Garcia Oct 30 at 20:30

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