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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    Good catch. The PO is not about an infinite entity but the Earth as it is, or as a comparable ideal globe.
    The increase of detected gravity at the poles is actually in great part due to the the presence of the solid nickel-iron core, which as a shell would have its greatest strength at its surface, ~ 1/3 up. That massive gravity will increase as the pole radius shrinks, while the whole, the mantle's contribution decreases, as Janus58's graph shows.
     
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  3. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    Nope. Even with a sphere of constant density the poles on a squashed sphere would still have higher gravity.
    No. Shells have nothing to do with this.
    As Janus58 clearly stated the graph is based on a constant density. So it has nothing to do with the mantle.

    This is typical of you. You don't understand what you read and then go half cocked off into never-never land.

    You are a very frustrating individual. Still can't figure out if you are a troll or just goofy.
     
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  5. nebel

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    Right, what I implied was, that his good graphs showed a decline. but there is a measured comparative increase. so The Earth is not of constant density. Gravity must increase at the poles as there distance shrink toward the solid, heavy core.
    I am definitely myself, goofy, good natured, trying to fill my bucket list of science queries as I enter my 89st orbit around Sol.
     
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  7. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    I hope you continue to enjoy life. I am frustrated trying to communicate with you about science so it is best if I just bid you adieu.
     
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  8. NotEinstein Valued Senior Member

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    My personal opinion: not a troll. nebel truly means what he writes.
     
  9. nebel

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    1,883
    You were of the greatest help, and underestimate the good you did. Even your opinion that I did not learn from you is clouded by the way I write, couching it in terms that sometimes require a re-take to see what I mean, like humour., which is a great teaching tool, because it will be remembered.so: A-Dieu " to God," yes, later, when we are sure what we mean by that. so, where were we?oh yeah,
    We need two tracings in a new graph, One of the gravity of the mantle, a shell. 3.8 kg per litre, and another one of the core, ~ 8kg per 1000 cc. a solid ball. The mantles's value will be O inside the core, which' s gravity will grow from maximum under our feet to the highest position on the iron surface.
    If gravity is higher at the poles, it is because of the core.
     
  10. nebel

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    1,883
     
  11. nebel

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    1,883
    Here is an answer why gravity is higher at the poles, because it is closer to the outer core.
    credit to

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    Tony Fredericks, former PC Technician (1989-2007) at Quora fair use of cr.

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  12. exchemist Valued Senior Member

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    Join the club.

    Now I know he is 88 I am a bit less harshly disposed towards him.

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    But he stays on Ignore.
     
  13. NotEinstein Valued Senior Member

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  14. nebel

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    1,883
    as you said, I got that wrong, in the be ginning, and am learning, but
    if you look at the blue line, it shows that it is only because of the surface gravity of the outer core , that Earth's gravity increases with depth, not because of a reduced radius per se, as was stated in the OP examples. In an ideal globe, the green line dives straight down , therefore gravity would decreases at the lower poles. In the constant density case, Gravity would remain high on the equator, even increase. My OP question was posed with an constant density globe in mind. The typical, ideal, simplistic nebel mindset.
    Hope I am not the only one that learned.
    exchemist said:
    "But he stays on Ignore."
    At first I quick misread that to mean : "stays in ignorance." no, I am learning.
     
    Last edited: Nov 14, 2018
  15. NotEinstein Valued Senior Member

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    No, you are still wrong. You said A, but your own source says B and C.

    But let's look a bit closer that picture:
    https://en.wikipedia.org/wiki/Gravity_of_Earth#/media/File:EarthGravityPREM.svg
    Note the usage of the word "sphere" in "other curves" #1 in the description.
    (Also note the complete absence of credits referring to Tony Fredericks. Still having trouble with copyright law?

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    )

    It also leads us to:
    https://en.wikipedia.org/wiki/Preliminary_reference_Earth_model
    And I quote: "a one-dimensional model"

    Please explain how you are mapping an oblate spheroid onto a one-dimensional line. Or put differently: your graph shows perfect spheres, not something that's flatter at the poles. So your explanation that goes with the graph is wrong.
     
  16. nebel

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    1,883
    You made me extra careful.so,

    please look at post 68. the credit even showed his picture.

    I thing some people have a deep bias to grasp that what is called a "Schwerpunkt" is actually the complete opposite. Even all the dark matter of the halo, if packed into the center of the Earth, would not be "schwerer", but light, weightless.
    Everything else follows from there. Zero gravity at the centre of a perfect disk too.
    The Frederick's blue line shows one slice of the gravity gradient of the cored Earth, picture all the slices lined up in a circle, like making a cake out of the traditional cake mold. From the high point, the surface, at the apex after the concave outside curve, it still keeps climbing, but the green constant density falls right away. linearly. The polar surface gravity has to fall to zero somewhere in the process to becoming a disk. A disk is a really flat oblate spheroid.
     
    Last edited: Nov 14, 2018
  17. nebel

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    1,883
    Assuming we start out with the constant density sphere, very similar to the graph origin supplied for the "more gravity, in or out" thread, Now, let us press in the top and bottom ends. Thereby we reduce the distance between the poles, leaving less mass there. Simultaneously, the squished mass is displaced side ways, toward the Equator, away from the central pole position. Both these movements will decrease the gravity at the polar surface by way of : 1) less mass, 2) increased distance.
    as Janus58 said:
    Any reference that asserts that reducing the radius at the poles increases surface gravity there, should be amended so as to specify the reason it does, poles when flattened, are approaching the surface of the core. imho
     
    Last edited: Nov 15, 2018
  18. NotEinstein Valued Senior Member

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    Sure, except it appears it's not his picture; it's from Wikipedia.

    (Irrelevant.)

    You are still missing the point: the models used are spherically symmetrical, so there is no oblateness. In these models, Earth is a perfect, and thus surface gravity is equal all over.

    And that action invalidates your graph: you are no longer in a spherically symmetric case, and thus the graph no longer applies.

    And you are still wrong, according to your own provided authority; see post #70.

    So you think your own provided authority is wrong on this matter? Then why did you bring him up as an authority in the first place?
     
  19. nebel

    Messages:
    1,883
    Moving the radius line on Frederick's graph post #68 to the right to increase radius will roughly simulate the equator conditions as a sphere stretches into a disk. . Given that the core is now seen as solid, not subject to the factors causing oblateness, the blue line remains valid . Going toward the center to the left will give the radius closer to the core, as at the poles, and show the increasing gravity up to the outer core surface. Reducing the radius, moving toward the center, like poles, will decrease gravity, true in a sphere, it is also true also for the disk; Graphs approximately hold true.

    All your dismissible nit picking aside, increased gravity at the poles is due to the presence of the core, not the reduction of the radius alone. Even then, reducing the pole radius into the hard core, will further reduce gravity there, but surface gravity at the rim, with mass driven there, although weakened at any distinct point, be present through the vast distance of the equatorial periphery. There is only one constant point at the pole, increasing points on the equator as oblateness increases.
     
  20. NotEinstein Valued Senior Member

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    1,986
    Alright, so the graph is not enough; you have to modify it.

    Please give proof of this assertion.

    Evidence please.

    Evidence please.

    Me calling you out when you are clearly wrong is not "dismissible nit picking", because it's not "nit picking", and as you continue to demonstrate, it isn't "dismissible" either.

    Wrong according to your own authority; please re-read post #70.

    Evidence please.
     
  21. nebel

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    1,883
    nebel said: Given that the core is now seen as solid, not subject to the factors causing oblateness, the blue line remains valid .
    Australian National University. "Earth’s inner core is solid, 'J waves' suggest." ScienceDaily. ScienceDaily, 19 October 2018. <www.sciencedaily.com/releases/2018/10/181019135124.htm>
    well, here is an example, where you keep at it, even quivveling whether the copy right picture is the right on, really relevant?. NotEinstein at his finest.
    nebel:"increased gravity at the poles is due to the presence of the core, not the reduction of the radius alone."
    NotEinstein reply: "Wrong according to your own authority; please re-read post #70."

    The authority I quoted was Frederick's graph, ( not the partially cited text), very clearly showing rise in surface gravity from the pole surface as it sinks through oblateness toward the core. We assume that interested learned readers can mentally rotate the axes to have the radius vertical for the poles, or move the radius in and out. so:
    as evidence I can do no better than to invite the readers to
    manipulate Frederick's graphs to see how a reduced polar radius and an increased equatorial radius (the result of increased oblateness ) affect the surface gravity there. For an constant density spheroid, or the blue line cored one.
    Keeping in mind that the 2 polar points at the end will have nearly no unbalanced matter under them, zero gravity at the centre too, , but every and all equatorial points at the rim all the mass between them that was in the original globe.
    perhaps a polished pebble on the beach, not ein stein
     
    Last edited: Nov 16, 2018
  22. NotEinstein Valued Senior Member

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    1,986
    Right, that's about the first part of your statement. Please also show evidence of the second part of the statement.

    I don't think calling somebody out for being wrong is "quivvveling", especially seeing as you are unable to provide any evidence for your claims. I mean, look at the post you just quoted. I directly asked for evidence 4 times. You only were able to respond to one of those, and only by addressing the first, least relevant part. If I'm only "quivveling", then what is it that you are doing?

    Yeah, I know that upholding the law is rather low of your priority list; we've established that before.

    Not really; this is all so obvious, I'm not really even trying.

    Incorrect; as I explained, the picture does not contain the distinction between poles and equator.

    False; read the description on Wikipedia more carefully to understand what that graph actually shows.

    Oh sure, I have no problem with that (well, except for the obviously wrong "poles"-part). That's also not what I am complaining about.

    And as I pointed out, you can't do that, because the models used in the graph explicitly exclude that.

    (Irrelevant.)

    Care to also address all the parts of my post you skipped?
     
  23. nebel

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    I told you, here it is, the poles are closer to the hot care. the heat is even leaking out at the bottom. credit BB science.
     

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