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Did the Earth move ?

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#1 Thomaslee

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Posted 29 September 2022 - 10:03 AM

I wonder in light of the recent Asteroid hit - do you think the Earth moved a bit, when the 'Dinosaur' meteor hit ?



#2 csrlice12

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Posted 29 September 2022 - 10:22 AM

The Earth has always been moving, it's moving even now....it probably did, the dinos got smaller and grew feathers.  The age of dinos, the age of man, the age of bugs is probably next.


Edited by csrlice12, 29 September 2022 - 10:23 AM.

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#3 JohnBear

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Posted 29 September 2022 - 10:55 AM

Of course - it did alter our orbit - at least a very, very tiny bit. Unfortunately mammals at the time did not have the proper tools to do the measurements (our ancestors were more like gophers back then - which is good 'cuz it helped us survive as a species.

 

Here is a good discussion of how fast we really are moving thu space"

https://www.space.com/33527-how-fast-is-earth-moving.html  



#4 TOMDEY

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Posted 29 September 2022 - 10:57 AM

Everything moves it... but generally not by much.    Tom



#5 Thomaslee

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Posted 29 September 2022 - 11:42 AM

Actually - what i met was  Did our  orbit change



#6 JohnBear

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Posted 29 September 2022 - 12:01 PM

Of course it altered our orbit - possibly by a few millimeters, inches, or even meters .

 

It also very likely altered the planet tilt and rotational speed of our planet by an even 'more significant' amount.  https://www.forbes.com/sites/jamiecartereurope/2022/08/03/do-we-need-the-first-ever-drop-second-a-new-wobble-by-earth-caused-the-shortest-day-since-records-began-say-scientists/



#7 MawkHawk

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Posted 29 September 2022 - 12:11 PM

"Did the Earth move ?"

 

Yes. In fact, I feel the Earth move under my feet. I also feel the sky tumblin' down.

 

Sorry, I coudn't resist.


Edited by MawkHawk, 29 September 2022 - 12:12 PM.

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#8 MellonLake

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Posted 29 September 2022 - 12:38 PM

Even a big asteroid is only a tiny fraction of the earths mass! 

 

Ceres, a dwarf planet (a really big asteroid) is 1/6500th the mass of earth and it is a whopping 1000km wide.   The one that killed the dinos was 10 to 15km wide.   A rough calculation puts the mass of the dino killing asteroid at 1/34,000,000,000th of the mass of the earth.  One 34 billionth the mass of the earth.  It would have changed the orbit... but not much.  Kind of like a mosquito hitting you, not going to do much.   


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#9 Codbear

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Posted 30 September 2022 - 12:46 AM

I will make various assumptions to try and come up with a reasonable number for how much the Earth was displaced in the direction of impact of the asteroid.

 

I will assume the asteroid pegged the Earth perpendicular to the direction of motion (at the instant of impact it doesn't matter that the Earth's motion is elliptical...it can be treated as linear) to maximize the potential Earth displacement.

 

Assumptions about the mass of the asteroid vary, but it seems that 1x10^15 kilograms is a reasonable assumption. The speed of the asteroid has been estimated to be about 30 km/sec. or 3x10^4 meters/sec.

 

I will use 6x10^24 kg as the mass of the Earth.

 

The collision was very elastic. For reference, two pool cue balls colliding would be very inelastic, whereas the asteroid melted/disintegrated/exploded on impact. A material amount of the energy of the asteroid was converted to heat, but I will assume that all the energy was transferred to the new Earth/asteroid object in order to try and come up with a maximum amount that the asteroid could have pushed the Earth in the direction the asteroid was travelling.

 

So by assuming that no kinetic energy was converted to heat (again to maximize the calculation for Earth's displacement) we can presume conservation of momentum. 

 

This means that the momentum of the asteroid before collision, which is = Mass x Velocity, will equal the momentum of the Earth/asteroid object after collision. This will allow us to solve for the velocity of the Earth/asteroid object:

 

asteroid: Mass: 1x10^15 kg           Earth: Mass = 6x10^24 kg

               Velocity: 30 km/sec                    Velocity = 0

 

Momentum before collision = (1x10^15 kg) x (3x10^4 meters/sec) =  3 x 10^19 kg m/sec

Momentum after collision    = (1x10^15 kg + 6x10^24 kg) x New Velocity.

 

The first clue that this is going to be an extremely small number is that the combined Earth/asteroid mass

is 6.000000001 x 10^24 kg. that is how insignificant the mass of the asteroid is compared to the mass of the Earth.

 

We can now set up the equation to solve for the Earth/asteroid velocity:

 

Momentum before equals momentum after: 3x10^19 kg m/sec = 6.000000001x10^24 kg x Velocity (V)

 

V = (3x10^19 kg m/sec) / 6.000000001x10^24 kg = 5x10^-6 meters which = .000005 meters ------>

 

which = 1/200th of a millimeter

 

Taking into account that in reality much of the momentum was converted to heat (think BBQ'd dinos), the distance was probably more like 1/500th to 1/1000th millimeters.

 

Compare that to the diameter of the Earth, which is about 13 billion millimeters, and you have Mellonlake's mosquito!


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#10 dave253

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Posted 30 September 2022 - 12:56 AM

Brilliant work, Mr Bear! 


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#11 KBHornblower

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Posted 30 September 2022 - 09:11 AM

Codbear, re post 9

 

That final number should be millimeters per second, not just millimeters.    In a frame of reference in which the Earth is stationary just before the impact, it will be drifting away about 1/200 mm/sec afterward.

 

V = (3x10^19 kg m/sec) / 6.000000001x10^24 kg = 5x10^-6 meters/sec

 

Transforming back to heliocentric frame of reference, the Earth's orbit will be changed a tiny amount.

 

You had the concepts of elastic and inelastic impacts backward.  An impact in which the projectile bounces off the target with no generation of heat is by definition 100% elastic.  In this case both momentum and kinetic energy are conserved.  When a bullet is embedded in the target with no debris flying off, it is by definition a totally inelastic impact, and it generates the maximum amount of heat.  Here the total momentum is conserved but the kinetic energy is not.  Since total energy is conserved, we get heat.  Billiard balls are highly elastic but probably a bit less than 100%, so they will be heated a little bit by the impact.  Just because a target can be deformed does not mean it is elastic.  It has to spring back to its original state with no heat to be considered totally elastic.

 

Momentum = mv or the sum of the mv components for multiple objects.

Kinetic energy = 0.5mv2 or the sum of the components for multiple objects.

With the velocities being observed, we get simultaneous equations from which we can calculate the amount of heat generated by an impact.


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#12 MeteorBoy

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Posted 16 October 2022 - 03:52 PM

Going a step further to Codbear's brilliant work..., how much has the Earth been displaced in space since that fateful impact 65m years ago?

 

Converting 65m years into seconds and then multiplying it by codbear's Earth impact movement of 1/750mm per second yields ~1,500,000km of Earth's displacement in space to date.  If the asteroid's impact was near 0 or 180 degrees to the Earth's movement around the sun it results in a still circular orbit but moves Earth either slightly closer or slightly further away from the Sun.  Perpendicular impact angles would push the Earth into a slightly elliptical orbit.

 

Seeing as the Earth's circumference around the Sun is ~460m km, moving it 1.5m km in space over all that time doesn't make much difference to us or to life on Earth.

 

Ken



#13 garret

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Posted 17 October 2022 - 10:32 AM

Remember that the earth rotates like a gyroscope, what changes with a large impact is the angle of the axis of rotation, the other effects are negligible.



#14 KBHornblower

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Posted 18 October 2022 - 12:42 PM

Remember that the earth rotates like a gyroscope, what changes with a large impact is the angle of the axis of rotation, the other effects are negligible.

I am as sure as physics 101 can make me that you are mistaken in general.  Suppose the Earth is spinning but otherwise stationary in our chosen frame of reference.  It will have zero net momentum, as each increment of it is balanced by one on the other side.  The incoming projectile will have a non-zero vector quantity of momentum which will be conserved in the impact.  Whether the projectile embeds itself, bounces off, or splatters in an intermediate state, its share of the total momentum is changed.  Thus the Earth has move slowly to conserve the momentum.  Depending on the trajectory of the projectile, the spin axis angle may or may not change.  This is a matter of angular momentum, which is closely related to what we simply call momentum but is distinctly different in detail and has different units of measure.



#15 spereira

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Posted 18 October 2022 - 02:15 PM

If this is becoming a discussion about Physics, perhaps you should move it over to The Science, Astronomy and Space Exploration forum.

 

smp



#16 csrlice12

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Posted 19 October 2022 - 12:03 AM

I just wonder if NASA is continuing to monitor the moonlet as well as the asteroid it is orbiting....what long term effect the moonlet has on its mother asteroid.

 

If our moon's orbit changed from 28 days to 24 days, what would be the effect on Earth's orbit...obvious changes in tides, but what possible effects might it have on Earth's molten core?  Could this effect change Earth's orbit.



#17 spereira

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Posted 19 October 2022 - 07:12 AM

Given the drift, I'm moving this over to the Science, Astronomy & Space Exploration topic.

 

smp


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#18 rhetfield

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Posted 21 October 2022 - 10:19 AM

Reminds me of a problem in my college dynamics class. 

 

It involved two frictionless wedges sitting on top of each other on a frictionless table.  We were to figure out how the wedges would slide apart.  Can't remember if they were looking for speed, distance or what.  Do remember that everybody could get to a certain point in the calcs, then were stuck - a key bit of data was missing - likely only determinable through experimentation.  None of us had a set of frictionless wedges. 

 

The class was taught by an adjunct professor (an aerospace contractor who had lost his job to defense cuts) who did not have office hours.  Since the prof was unavailable, I went to the department head (an MIT physicist - our school was small and the engineering program was under the auspices of the physics department).  He was not able to come up with an immediate answer.  Next session of class, the professor tried to work the problem and got stuck the same place as the rest of us.  He looked in the answer book from the textbook authors.  It detailed the same calcs that we had all done, then a ""poof" - here's the answer and the end of the calcs is left as an exercise for the students". 

 

The department head later told me that he was fascinated by the problem and had devoted a great deal of time to trying to figure out the answer.  The only solution he could come up with was to factor in the mass and movement of the earth.  Since the wedges were much smaller than the dinosaur asteroid, the planetary motion would be too small to give any sort of accurate answer.  Probably too small to be displayed by the calculators of the era.


Edited by rhetfield, 21 October 2022 - 10:19 AM.


#19 KBHornblower

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Posted 21 October 2022 - 01:53 PM

Reminds me of a problem in my college dynamics class. 

 

It involved two frictionless wedges sitting on top of each other on a frictionless table.  We were to figure out how the wedges would slide apart.  Can't remember if they were looking for speed, distance or what.  Do remember that everybody could get to a certain point in the calcs, then were stuck - a key bit of data was missing - likely only determinable through experimentation.  None of us had a set of frictionless wedges. 

 

The class was taught by an adjunct professor (an aerospace contractor who had lost his job to defense cuts) who did not have office hours.  Since the prof was unavailable, I went to the department head (an MIT physicist - our school was small and the engineering program was under the auspices of the physics department).  He was not able to come up with an immediate answer.  Next session of class, the professor tried to work the problem and got stuck the same place as the rest of us.  He looked in the answer book from the textbook authors.  It detailed the same calcs that we had all done, then a ""poof" - here's the answer and the end of the calcs is left as an exercise for the students". 

 

The department head later told me that he was fascinated by the problem and had devoted a great deal of time to trying to figure out the answer.  The only solution he could come up with was to factor in the mass and movement of the earth.  Since the wedges were much smaller than the dinosaur asteroid, the planetary motion would be too small to give any sort of accurate answer.  Probably too small to be displayed by the calculators of the era.

To me this looks like an exercise in conservation of momentum and total energy.  I cannot imagine what sort of data might be missing without seeing the exact setup.  The upper wedge would slide down the face of the lower one and in the process push it sideways in the opposite direction.  In principle the Earth would rise slightly as the upper wedge settles to its final elevation, but I would expect it to be negligible, given the vastness of Earth's mass compared with any laboratory apparatus.  The change in gravitational potential energy as the upper wedge comes down would be reflected in the total kinetic energy acquired by the wedges, and the final velocity of each wedge should be inversely proportional to its mass.




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