Question on Gravity at the equator. more or less?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by nebel, Oct 13, 2018 at 4:24 AM.

  1. nebel Valued Senior Member

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    All references maintain that there is more gravity at the poles because they are closer to the centre. and
    You weigh less at the equator because your are higher, with additional weight loss because of the spin getting you nearer to escape velocity.! But is this not incorrect, despite the majority opinion?
    There is more mass under you at the equator, less at the poles. since mass increases with the cube, or even eight fold, and distance action only by the square, should not gravity be higher at the equator? ignoring the spin lift?
     
    Last edited: Oct 13, 2018 at 4:34 AM
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  3. NotEinstein Valued Senior Member

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    There is also more mass further away from you: the "bulge" at the other side of the planet. In fact, the same amount of mass that's closer to you on this side of the planet, is also further from you at the other side of the planet. And the remaining mass of the planet ("core"? the part not counting the bulges) is also further away. So you're right that there's one part of the planet that will exert a stronger gravitational force on you, but there are two parts that'll exert a weaker gravitational force on you. Turns out, when you do the math, the weakening effect is larger.
     
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  5. nebel Valued Senior Member

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    Thank you for taking up the challenge!
    Standing as Willem Barentz would, way up north, closer to the Earth's core than in Indonesia, the equatorial "bulge", bulging there would be all around , down, sideways of him, not opposite. so:
    Let us please do a thought experiment, "Gedankenexperiment" as A.E., your namesake would say:
    Let us proceed to shrinking the polar regions downward, until this malleable Earth resembles a disk. There would be zero gravity at the center, and measured along the now very short axis, very little acceleration. certainly not ~ 10 metre/sec^2.
    This process of declining to a disk and accompanying decrease in gravity, started at the present size, but it has to be a decline from the very start of that shrinking from near equatorial size !! is that fine?
    The Shell theorem too demands , that there be less gravity,-- not more,-- below the Equatorial size , which is the case at the poles.
    The point of pinpointing gravity as acting only from the centre is pointless in this case. imho.
    All the references that maintain the opposite of the present presentation --- just might be wrong.
    The majority opinion is not always right. or?
     
    Last edited: Oct 13, 2018 at 2:33 PM
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  7. NotEinstein Valued Senior Member

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    I didn't realize there was a challenge? If so, I'm out.

    Just to clarify: when you say "zero gravity", you mean "no net gravitational force", right?

    Why do you assume it's a linear relation?

    A disc does not contain shells, so it doesn't apply in that case.

    And thus you should agree with me that the shell theorem doesn't apply.

    Or, it's just you that's wrong. Well, I'd say: do the math, and find out!

    Nope, but the math doesn't lie. Do it, and you'll find your answer.
     
  8. Janus58 Valued Senior Member

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    If you push the poles inward, then to maintain the same mass, you have to push the equator further from the center which means further from the center of mass and a lessening in gravity there.
    The shell theorem doesn't apply here as the equatorial bulges aren't spherically arranged.

    Below we have a shell theorem situation on the left. For a object sitting at the black dot, the matter spread out spherically shown by area in between the surrounding blue circle and black circle cancels itself out gravitationally adding no net addition to the gravity felt by the dot.

    On the right we have a an equatorial bulge situation. Here the gravitational pull from the matter in the bulges only cancel each other out in the horizontal direction leaving a net pull towards the center of the sphere as shown by the blue arrow.



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    The simple answer as to whether gravity is stronger at the poles vs. equator is to look at the actual numbers as physically measured. Here the values are 9.780 m/s^2 at the equator and 9.832 m/s^2 at the poles.

    Of course a part of the lessening at the poles is due to the spin of the Earth, but we can account for that and it works to to be 0.0337 m/s^2. Adding this back to the the measured equatorial value, we get 9.8137 m/s^2 which is still less than the 9.832 m/s^2 at the poles. So theoretical considerations aside, actual measurement confirms the assertion that gravity is stronger at the poles than at the equator due to the Earth's oblateness regardless of its spin.

    Another problem with your squashing the Earth to nearly disk shape argument is that it deals with extremes. Looking at the extreme ends of the spectrum won't always give you a good picture of what is happening in the region you are interested in. It is very unlikely that the ratio of polar gravity to equatorial gravity for various polar to equatorial radii ratios is a linear function. It might even be possible that if the equatorial radius becomes really large compared to the polar radius that you will start to see higher equatorial gravity. But this would only occur if the oblateness exceeds some value, a value that the Earth's shape comes nowhere near.
     
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  9. sculptor Valued Senior Member

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  10. nebel Valued Senior Member

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    that is an interesting proposition, it would imply that the maximum gravity or potential does not peak at the surface, which at least at the oceans is defined as nearly spherical. Dealing with ideal models here)
    I can not argue with that. but: Could it be that the bulging of the equator leaves the inner, more than twice as massive core unaffected, and therefore closer to the surface, giving those readings? and,
    at some point in the situation between a near perfect globe like the Sun, and thin disks , like some galaxies, there must be a change in formulation from increasing gravity with decreasing size to the zero value you would expect at a thin perfect disk's centre.
     
    Last edited: Oct 13, 2018 at 7:47 PM
  11. nebel Valued Senior Member

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    The increased mass between the extended, more distant equator points would trump the square over distance decrease every time. Example: doubling the Earth's size would note decrease the surface gravity, it would more likely double it.
     
  12. nebel Valued Senior Member

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    You are right, that assumption was wrong, I was thinking in ideal rotating spheroid terms.
    yes, and the math is being done, with the right answers, thank you.
     
  13. NotEinstein Valued Senior Member

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    Glad I could help you correct your misunderstanding of this aspect of Newtonian gravity.

    Actually, this math was already done hundreds of years ago, but I'm glad you're finally catching up!
     
  14. nebel Valued Senior Member

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    well of course, an dI am not the only one. because the published arguments in favour of the higher gravity at the poles was based on the simple distance from the centre of mass supposition, when in fact it has to do with the fact that the massive central iron Nickel core is not bulged by the small rotational speed there, but remains closer to the Poles.
    The argument is solved by the mass solution, not the geometry. and yes. I was wrong in assuming the decline would be linear.
     
  15. RainbowSingularity Valued Senior Member

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    im wondering if you have remembered to add rotational spin.
    the faster the speed you get from going around the equator, the more energy you gain and so the more gravity you have and the more of gravity that effects you.
    soo that serves to balance its self off at the poles. you have less kinetic energy.

    the mass stays the same regardles of where you are, but the vectors of energy are vastly different wehen you are travelling at different speeds.
    as you gain rotation at the poles subject to less propulsion it would stand to reason that you would have less gravity pull as you gain more inertial energy with less velocity against the pushing force.

    this basics science E=Mc2 process is tested by children in play grounds with carousel like equipment where they spin it and hold on.
    if you are standing in the centre it is much easier to hold on.
    if you are on the outside it is very difficult to hold on.
    applying this as a physics principal at an early age is great for teaching kids.
     
  16. nebel Valued Senior Member

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    Sculptor thank you for the links. The general proposition is that higher elevation of matter, Water, rock, means greater gravity, --lowered areas, like the poles, less gravity. In extreme cases to no gravity at all. When you get closer to massive heavy bodies like the core, of course gravity rises, that how they scout for mineral deposits. The poles are closer to the core. The exception that confirms the rule.
    Rainbowsingularity, given how much energy it takes to get so little gravity, it would not be tipping the scales., My example was of the Earth, but I really meant an ideal bulging sphere. But the effects would clearly be greater in the fast spinning giants with 10 hour days, Jupiter & co. Or the Galaxies, flattened to a disk with nearly zero gravity in the very centre and most at their "Equators".
     
  17. nebel Valued Senior Member

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    Please Register or Log in to view the hidden image!

    Janus, thank you again for those images. Does it not show that the laterally displaced matter gone into the bulges is now further away from the observer? and the Blue arrow has actually become shorter too?. can I reproduce and use your work elsewhere? thank you: N.
     
  18. sculptor Valued Senior Member

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  19. Janus58 Valued Senior Member

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    One thing to note here is that the scale of the both these graphs is in milliGals, and a Gal is 1 cm/m^2
    By this measurement, the gravitational strength difference between pole and equator is 1830 mGal.
    So even the mantle anomaly variation is in total less than 1/2 of that and the variance from the mean is less than 1/4.
    So it would seem that both of these images show variation from the expected pole to equator variation over the reference geoid.
     
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  20. nebel Valued Senior Member

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    Thank you! Again, mostly we have altitude shown with higher gravity and trenches with low values. But, and it was my failure, a pity to not consider the Earth's greatest anomaly, its small core with 33% of its mass, but little volume, as the defining feature of it's gravity. I treated the question as a strictly geometric, a radius problem; Most answers on Wikipedia or Google do likewise.
    If the above illustration were a Pumpkin, with double the equatorial diameter then the polar, gravity there would be only half, with eight times less mass there than between the horizontal direction. As it is, the Earth resembles not a pumpkin or potato, but a peach,or an avocado with a real stone pit. so: Let it be resolved that
    The flattening of a homogenous globe would reduce gravity on the center of the flats; but in an entity with a heavy core
    Such reduction occurs only once that core is flattened too.
     
  21. nebel Valued Senior Member

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    would the expected Geoid not include, have figured in the existence of that dominant massive core?
     
  22. Janus58 Valued Senior Member

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    The force of gravity felt at any point of the Earth is due to the sum of the all the mass of the Earth. no matter where you are on the surface, you are pulled on by all the matter of the Earth. A person standing at the equator is just as far from the matter at the pole as a person standing at the pole is from the Equator, so the matter at the pole pulls on him just as hard as the matter at the equator pulls on the person at the pole. However the matter at a point on the equator ninety degrees away from our person on the equator is further away from him than any point of the equator is from the person on the pole. And gravity pull falls off by the square of the distance. The bulge on the other side of the Earth is further away also.
    In order to work out the total gravity pull on you, you have to take the combined effect of all the individual parts and their distances from you. And when you do that for a body with the oblateness on the order of the Earth's you find that the increase in distance from the center of mass ends up being the deciding factor.

    To get the exact value takes a complex mathematical analysis; one that is not subject to the simple analysis you are trying to use.
     
  23. nebel Valued Senior Member

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    Thank you, Interesting!, my question is: since the distance is accumulated mostly in the lighter mantle material, and the central, doubly dense core is relatively unaffected by the oblating forces, could this be a cause for the perplexing polar rise in Gravity, when overall there has to be a decline? It really would help the uneducated like me, if a graph could be produced as ORIGIN did with the earlier gravity question. That new work would be showing
    At what point the reversal occur from the rise in Gravity as the polar radius shrinks, to the inevitable decline to zero gravity at the center of any but convincingly to a truly flattened obloid. Not an abstract question, because
    Nature has produces really breathtaking thinness in the Rings of Saturn for example way more oblate than the Earth, and surely having zero gravity at their center, also with an "equator" that is only 10 meters wide!
    That hoped- for rising and falling curve of gravity as the pole to pole axis shrinks, will be the picture that is worth a thousand words.
    As an aside: Surely dark matter could not be possibly acting on that small a scale?
     
    Last edited: Oct 15, 2018 at 6:29 AM

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