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Thread: Earth faster rotation

  1. #1 Earth faster rotation 
    Forum Cosmic Wizard icewendigo's Avatar
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    apparently the earth used to rotate faster at the time of dinosaurs.


    Would there be any physical or biological effects if the earth right now were to spin lets say 6 times faster(a 4h day)? Would we feel gravity in the same way?


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    YES


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    A quick calculation says we'd feel a reduction in Earth's gravity from centrifugal force of 1.1 m/s^2. That's assuming I did the calculation right.

    radius of Earth = 6000000 meters

    velocity = 2 pi r / (24 * 60 * 60 * 1/6)

    velocity ^ 2 / r = acceleration in rotating frame of reference away from the Earth.

    If nothing else, it would exaggerate the Coriolis effect on weather patterns.
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    I thing Numsgil has calculated it correctly, but the length of the day is not 6 times as long now. According to this Wikipedia article, the day was 21.9 hours long 620 million years ago.
    http://en.wikipedia.org/wiki/Tidal_acceleration
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  6. #5  
    Moderator Moderator Janus's Avatar
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    Of course, that 1.1 m/s difference would only be felt at the equator and would decrease as one went to higher latitudes.

    In reality, the difference at the equator would be larger, since this fast a spin would distort the Earth into a more pronounced oblate spheroid than it is now, increasing the equatorial radius to polar radius ratio. The Oceans would all shift to a band girdling the Equator.

    This is a moot point however as the Earth would not have been spinning that fast during the time of the dinosaurs. 230 million years ago, the Earth would have only been spinning 1.041 times faster than it is now (the day would have been 57.5 min.s shorter).

    This gives only a difference at the equator of 0.0026 m/s from what it is now.

    For comparison the present difference between polar gravity and equatorial gravity is on the order of 0.034 m/s, so this gives a much smaller change in gravity at the equator over 230 million years than exists between the poles and equator today.
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    Forum Junior c186282's Avatar
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    I realize this was not part of the question but none the less it is interesting.

    The reason the day is getting longer is because of what is called tidal lock. The tides we see in the ocean also happen with the surface of the earth and the atmosphere. These tides lag from where one would expect them to be based on a quick understanding of the interaction with the moon's gravitational field. This results in a torque on the earth that slows down the earths rotation. But don't worry the angular momentum of the earth moon system is being conserved. This results in the moon moving away from the earth at about an inch (2.54cm) a year. The energy of the earth-moon system is being dissipated in the form of heat. Once the earth is in tidal lock with the moon the loss of energy will stop and the day will be as long as a month.
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    Moderator Moderator Janus's Avatar
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    Quote Originally Posted by c186282
    I realize this was not part of the question but none the less it is interesting.

    The reason the day is getting longer is because of what is called tidal lock. The tides we see in the ocean also happen with the surface of the earth and the atmosphere. These tides lag from where one would expect them to be based on a quick understanding of the interaction with the moon's gravitational field. This results in a torque on the earth that slows down the earths rotation. But don't worry the angular momentum of the earth moon system is being conserved. This results in the moon moving away from the earth at about an inch (2.54cm) a year. The energy of the earth-moon system is being dissipated in the form of heat. Once the earth is in tidal lock with the moon the loss of energy will stop and the day will be as long as a month.
    Even then we haven't reached the end of the saga. The Sun also exerts a tidal influence on the Earth, though half of that of the moon at present.

    So after the Earth matches its rotation to the Moon's orbit, the Sun will still be trying to lock the Earth to it. So a new tug of war ensues, with the Sun trying to lock the Earth to it, and the Moon trying to maintain its lock.

    So now, the Moon pulls forward on the Earth in an attempt to prevent it slowing. as a result the Moon gives up energy to the Earth, and falls into a lower, faster orbit.

    The Moon will continue to spiral in until it reaches the Roche limit and breaks up into a ring around the Earth.

    Of course all of this assumes that in the meantime the Earth and Moon haven't been consumed by the Sun as it expands into a red giant.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


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    Quote Originally Posted by Janus
    Even then we haven't reached the end of the saga. The Sun also exerts a tidal influence on the Earth, though half of that of the moon at present.

    So after the Earth matches its rotation to the Moon's orbit, the Sun will still be trying to lock the Earth to it. So a new tug of war ensues, with the Sun trying to lock the Earth to it, and the Moon trying to maintain its lock.

    So now, the Moon pulls forward on the Earth in an attempt to prevent it slowing. as a result the Moon gives up energy to the Earth, and falls into a lower, faster orbit.

    The Moon will continue to spiral in until it reaches the Roche limit and breaks up into a ring around the Earth.

    Of course all of this assumes that in the meantime the Earth and Moon haven't been consumed by the Sun as it expands into a red giant.
    Yes, I did leave the sun out of the story. For a moment lets forget about the red giant problem. I do not see the dynamics that would bring the moon back in. We know the the influence of tides is dependent not of the force of gravity but instead the gradient. If I set the gradients of the gravity of the sun and moon equal and solve for the distant to the moon I get:

    which gives

    compared with the current distance to the moon which is

    The moon just needs to go 1.28 times its current distance and the tides due to the moon will be the same size as the tides due to the sun.

    So once the shape dipoles induced in the earth by the sun and the moon are of the same magnitude it seems to me that the torque on the orbit of the moon would average to zero over a month which would bring an end to the earth moon tidal lock process. (note I have not done the math) However if the earth is still spinning there would still be a loss of energy resulting in a loss of spin angular momentum which would still have to be picked up by orbital angular momentum. My guess would be that the earth would start to drift away form the sun.

    I did a real quick search on three body tidal lock but did not find anything. Do you have a reference for your claim?
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  10. #9  
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    Quote Originally Posted by c186282
    Quote Originally Posted by Janus
    Even then we haven't reached the end of the saga. The Sun also exerts a tidal influence on the Earth, though half of that of the moon at present.

    So after the Earth matches its rotation to the Moon's orbit, the Sun will still be trying to lock the Earth to it. So a new tug of war ensues, with the Sun trying to lock the Earth to it, and the Moon trying to maintain its lock.

    So now, the Moon pulls forward on the Earth in an attempt to prevent it slowing. as a result the Moon gives up energy to the Earth, and falls into a lower, faster orbit.

    The Moon will continue to spiral in until it reaches the Roche limit and breaks up into a ring around the Earth.

    Of course all of this assumes that in the meantime the Earth and Moon haven't been consumed by the Sun as it expands into a red giant.
    Yes, I did leave the sun out of the story. For a moment lets forget about the red giant problem. I do not see the dynamics that would bring the moon back in. We know the the influence of tides is dependent not of the force of gravity but instead the gradient. If I set the gradients of the gravity of the sun and moon equal and solve for the distant to the moon I get:

    which gives

    compared with the current distance to the moon which is

    The moon just needs to go 1.28 times its current distance and the tides due to the moon will be the same size as the tides due to the sun.

    So once the shape dipoles induced in the earth by the sun and the moon are of the same magnitude it seems to me that the torque on the orbit of the moon would average to zero over a month which would bring an end to the earth moon tidal lock process. (note I have not done the math) However if the earth is still spinning there would still be a loss of energy resulting in a loss of spin angular momentum which would still have to be picked up by orbital angular momentum. My guess would be that the earth would start to drift away form the sun.

    I did a real quick search on three body tidal lock but did not find anything. Do you have a reference for your claim?
    As the Sun's tidal effect continues to slow the Earth's rotation we start to get a situation where the Moon orbits the Earth in less time than the Earth rotates. Under these conditions(or under conditions where a satellite orbits retrograde to the rotation of the planet, The tidal interaction between the Earth-Moon system, while weaker, but still exists will cause the Moon to be drawn in towards the Earth. It is the reason that Phobos is spiraling in towards Mars and why it is easier for planets to capture bodies into retrograde orbits. This is known as tidal deceleration:

    http://en.wikipedia.org/wiki/Tidal_acceleration

    Tidal deceleration

    This comes in two varieties:

    1. Fast satellites: Some inner moons of the gas giant planets and Phobos orbit within the synchronous orbit radius so that their orbital period is shorter than their planet's rotation. In this case the tidal bulges raised by the moon on their planet lag behind the moon, and act to decelerate it in its orbit. The net effect is a decay of that moon's orbit as it gradually spirals towards the planet. The planet's rotation also speeds up slightly in the process. In the distant future these moons will impact the planet or cross within their Roche limit and be tidally disrupted into fragments. However, all such moons in the solar system are very small bodies and the tidal bulges raised by them on the planet are also small, so the effect is usually weak and the orbit decays slowly. The moons affected are:
    * Around Mars: Phobos
    * Around Jupiter: Metis and Adrastea
    * Around Saturn: none (like Jupiter, Saturn is a very rapid rotator but has no satellites close enough)
    * Around Uranus: Cordelia, Ophelia, Bianca, Cressida, Desdemona, Juliet, Portia, Rosalind, Cupid, Belinda, and Perdita
    * Around Neptune: Naiad, Thalassa, Despina, Galatea and Larissa;
    2. Retrograde satellites: All retrograde satellites experience tidal deceleration to some degree because the moon's orbital motion and the planet's rotation are in opposite directions, causing restoring forces from their tidal bulges. A difference to the previous "fast satellite" case here is that the planet's rotation is also slowed down rather than sped up (angular momentum is still conserved because in such a case the values for the planet's rotation and the moon's revolution have opposite signs). The only satellite in the Solar System for which this effect is non-negligible is Neptune's moon Triton. All the other retrograde satellites are on distant orbits and tidal forces between them and the planet are negligible.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


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  11. #10  
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    Janus, I followed the link and read the contents. I understand the dynamics of the tidal deceleration and tidal acceleration.

    But what I'm not convened of is that the moon will move out then back in. Doing the full orbital mechanics is out of my league (and attention span). But I think we could look at the two extreme cases:
    1) Earth tidal locked with sun and moon tidal locked to earth and very close
    2) Earth tidal locked to the sun and the moon tidal locked to the earth and the moon very far away.

    If we set these two cases up with conservation of angular momentum would you agree that the most probable case would be the one with the least final energy?
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  12. #11  
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    Quote Originally Posted by c186282
    Janus, I followed the link and read the contents. I understand the dynamics of the tidal deceleration and tidal acceleration.

    But what I'm not convened of is that the moon will move out then back in. Doing the full orbital mechanics is out of my league (and attention span). But I think we could look at the two extreme cases:
    1) Earth tidal locked with sun and moon tidal locked to earth and very close
    2) Earth tidal locked to the sun and the moon tidal locked to the earth and the moon very far away.

    If we set these two cases up with conservation of angular momentum would you agree that the most probable case would be the one with the least final energy?
    Number 1 is the least energy system.

    The orbital energy of the Moon is due to the total of its kinetic energy and gravitational potential energy:



    This energy remains constant for any fixed orbit and can also be stated as



    where 'a' is the semi-major axis of the orbit.

    So as 'a' decreases, the total energy decreases. (becomes more negative.)


    In terms if angular momentum for an orbiting body:



    and for a circular orbit:




    so



    and angular momentum decreases with orbital distance.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


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    Nothing you did there was wrong and I agree with your math but you have not proven that the moon goes out then back in. However, that said I'm starting to see the possibility for your claim. I have not done the math yet. But to do so we need to set up the initial configuration with the total system angular momentum and energy then compare to the final states. We need to include the three bodies and the energy in the rotation of the bodies. One thing I do not know how to estimate is if the moon comes back in there will come a point when the tidal torque on the earth is greater from the moon than the sun and the moon will start to speed the earth up again adding to the rotational energy of the earth.
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  14. #13  
    Moderator Moderator John Galt's Avatar
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    I can't do the math - without severe disruption to my psyche - but I have read of the lunar movement out and then back in again in dozens of astronomy text books over many years, so I incline to believe it.
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    Quote Originally Posted by John Galt
    I can't do the math - without severe disruption to my psyche - but I have read of the lunar movement out and then back in again in dozens of astronomy text books over many years, so I incline to believe it.
    Math has a way of doing that to one's psyche.

    Do you have a reference?

    I'm beginning to see that it is possible but I would like to understand more of the details. I'm also not convinced my energy condition presented earlier would close the question.

    Another mistake I made was the use of the word "dipole". One can have a dipole in electrostatic because of the existence of positive and negative charge but with gravity one can not have a dipole. The first non-zero term is the quadrapole term.
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  16. #15  
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    Quote Originally Posted by c186282
    One thing I do not know how to estimate is if the moon comes back in there will come a point when the tidal torque on the earth is greater from the moon than the sun and the moon will start to speed the earth up again adding to the rotational energy of the earth.
    It seems pretty obvious that such a point would be reached. Of course, this will not mean that the Earth would ever spin back up to its former speed. The tidal coupling between the Earth and Moon is pretty inefficient in transfering energy between the two, the vast majority is lost to tidal friction and radiated away as heat. Just compare the amount of energy lost by the Earth in a year, with the day lengthening by 1.5 millisecond per century, and the amount of energy gained by the Moon by climbing to a higher orbit at the rate of 3.8 cm per year.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


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  17. #16  
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    Quote Originally Posted by Janus
    It seems pretty obvious that such a point would be reached.
    I'm beginning to agree but I still what to do some math. I will have some time this Friday. My approach will be to write down the total energy for the system and force conservation of angular momentum. If I can get your end state without ever going up an energy hill, I'll buy in. I will post my analysis when I'm done.
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    In researching how to approach doing the mathematics I have come across the following book, Solar System Dynamics. I have gotten myself a copy and plan to push through the chapter on Tides, Rotation and Shape along with some others. This will not be a quick process but I have been wanting to learn more about orbital dynamics for a while so here I go.

    I will post someday with what I learn about the possibility of a moon moving out then back in. I will start a new post at that time.
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