Question on Gravity at the equator. more or less?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by nebel, Oct 13, 2018.

  1. nebel

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    If I, and other readers understood right, Janus58's position, and the Frederick graph show, that in a constant density globe/spheroid, a reduction of radius at the poles, will lead to a reduction in surface gravity there. so,
    If there is the opposite, an increase in measured gravity, the body must not have constant density, and not surprisingly, the Earth does.
    Do not ~ constant density spherical bodies form because lower areas have less gravity? would increased gravity with depth there not create ever deepening holes? once triggered, runaway oblateness without any spin?
     
    Last edited: Nov 17, 2018
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  3. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    I am not sure who these 'others' are but I can say that you have not understood right. The 'Frederick graph" has nothing to do with compressing a body at the poles. The graph is showing the acceleration due to gravity for different cases as you move outward from the center of a body.
    If any body has a reduction of the radius at the poles with an increased radius at the equator the the measured acceleration due to gravity will decrease.
    If there is enough mass for the body to form into a roughly spherical body then there will always be a higher density as you move to the center of the body.
    Holes? What are you talking about now.
    Runaway oblateness? Do you mean, "do spherical bodies sometime spontaneously turn into disks"? Uh... NO.
     
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  5. nebel

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    I am not saying that any reader agrees with my questioning, but I have raised a frequently asked question. Readers will like to see the developments in both sides of the arguments. Your previous graph, thank you, and Frederick's indicate that for a reduction in radius in a constant density gradient, there is a reduction in gravity/acceleration. Janus58 agreed with that in his graph, comment in post #41. If one considers the depressed poles as extended excavations, then going deeper, sinking, will reduce the volume directly below you, by a ratio of halving the radius is reducing the volume/mass there to 1/8 st. In a spheroid, that volume and mass is displayed toward the perimeter, further away from the poles, reducing that mass 's pull by the inverse square of the distance. . or?
    will have to sleep on the other good objections. thank you.
     
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  7. nebel

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    that is exactly what I am saying, it seems Janus58 seconds that, yet many maintain the surface gravity at the poles increase, so : the body in question must not have constant density and indeed the Earth has a core with a higher internal surface gravity than the surface. It is the core that causes the increase in surface gravity as the poles are shrunk closer to it. show me a spheroid with increased polar gravity, and i show you one with a massive core. or?
    You are right, provided that the cloud out of which the body contracts has material of different specific weights, Iron, silicates, for example., and the material is compressible. Heavier objects, even if they fall at the same speed, as the hammer and the feather demonstration by Scot with Apollo 15 on the Moon showed, the hammer fell deeper into the dust. The Titanic sank deeper that the floatsam.
    Here might be the crux of the misunderstanding, I was referring to reduction of the polar radius as such, all other factors , like density being equal, constant, --whereas the "increased gravity" advocates assume a more or less stratified body, as large astronomical bodies tend to be.
    Let us assume in a constant density, partially deformable body; -- if increased gravity with depth were true---. the material at the bottom of any depression would weigh more than the: overburden, making the pit sink even further. It would be like having a self- carving Grand Canyon without flowing water. By contrast, in an even - density body, weight would decrease with radius. The material near the center of the Earth is not pulled down by greater gravity there, (it is after all, nearly weightless!) , but compressed down only by outside pressure.
    As it is, there is no acceleration in a hole at the centre of the earth.

    I alluded to the impossibility. Obviously they do not, but they would within an "increasing gravity with depth" world, given 2 symmetrical trigger events at the poles. They can not, because gravity decreases, does not increase with decreasing radius. thankfully.
    An afterthought:
    Assuming you had a perfect, constant density globe, and then sliced a cut under the poles to form the spheroid,
    measured the gravity in the cut, and then lifted the sliced material away; theoretically, would the downward reading of your acceleration meter change? would it change the before and after reading of an equatorial instrument if you deposited the material there?
     
    Last edited: Nov 20, 2018
  8. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    Never mind.
     
    Last edited: Nov 20, 2018
  9. Janus58 Valued Senior Member

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  10. nebel

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    Kris Walker
    , BSc Adv. Physics & Astrophysics, Monash University (2021)
    Updated Feb 4, 2018 · Author has 258 answers and 152.9k answer views
    Outer core? Yes. Inner core? No.

    If the Earth had a uniform density, the gravitational acceleration would decrease linearly with depth, and so the core would have a weaker gravity. However, according to the preliminary reference Earth model (PREM),[1] the Earth’s outer core has a higher gravitational acceleration than the surfacen of the Earth.
    As you can see, gravity gets stronger between the surface and the outer core before dropping off in an approximately linear fashion towards the center. so:

    let it be resolved that in a spheroid like the Earth, gravity increases at the poles only because of the core, not the reduction in radius.
     
  11. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    So it is written, so it shall be (in your mind at least).
     
  12. NotEinstein Valued Senior Member

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    Isn't it time this thread was moved to a more appropriate section of this forum? nebel has now multiple times argued in direct conflict with his own provided authority, keeps doding me when I point that out, and has ignored my post #79 completely. There's no science going on here (anymore); just someone endlessly repeating the same self-debunked material over and over again...
     
  13. nebel

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    1,893
    In the discussion since the OP, several concepts emerged, and were pitted against each other: The situation in an ideal globe vs one with a stratified interior. the changing acceleration values at the poles and the equator as the oblateness increases. how that is affected by a solid core ----
    shutting down auch a discussion, or relegating it to the pits will serve who's interest?
    of course we were treated to such profound questions too, -- whether an image of a cited authorities is really authentic. so:
     
  14. nebel

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    1,893
    "Such a rotation means that when it runs out of fuel and begins to explode as a supernova, it will collapse at the poles before the equator, producing a gamma-ray burst."
    A snippet from<www.sciencedaily.com/releases/2018/11/181120073641.htm>.
    If it iss relevant: If one considers that a spheroid becomes more oblate because of spin, as the star mentioned above, of course the Gravity at the poles becomes more effective than at the perimeter, where centrifugal forces dominate.
    When I considered gravity, I understood just that, mass -generated pull, not the resultant force from which centrifugal/centripetal forces have been added, subtracted.
     
  15. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    That's strange considering it was mentioned on the first page of this thread.
     
  16. nebel

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    1,893
    thank you for pointing that out. I understood though, that we talked about the gravitational field, the strength of acceleration due to mass itself. If we consider the creating of a more oblate spheroid through greater spin, not "squashing", gravity itself would be hard to separate out in the equatorial measurements, and more matter driven to the extreme perimeter by the centrifugal forces, or not, if angular momentum is to be preserved. It is helpful to work through all these angles. Given enough spin, all the gravity would be in the perimeter, the central pole region void of any matter. Any eccentricity there, would propel it outward. so, even in this scenario. is not the polar region bound to lose gravity, not gain it? Leaving a central mass, Core for a planet, Black Hole/bulge for a galaxy, as only possible source of increasing gravity with reduced radius?
     
    Last edited: Nov 21, 2018

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