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Gravity on surface of sphere

  1. Nov 17, 2011 #1
    Howdy. I'm not a physics expert, but I am checking for solutions to this problem.

    For an individual standing on the surface of a sphere (assume a perfectly spherical earth),
    gravity of all atoms "pull" as per their fixed relation to that individual and their distance.
    While the only perceivable force is "down", there is a pull nearly sideways and and all angles downward. The question is, will this force (negated for its even-on-all-sides pull) continue
    at a gradient from weak to strong until it reaches its maximum gradient at straight down, or
    will it have "spikes" of stronger pull at definite angles downward?
     
  2. jcsd
  3. Nov 17, 2011 #2

    berkeman

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    What are your thoughts? Why would there be spikes?
     
  4. Nov 17, 2011 #3
    One thread mentioned that gravity is null in the center of the sphere (earth) but 109 percent
    at 55 percent of the distance towards the surface (from the center). At a certain angle
    from a fixed point on the surface there might be a greater concentration of mass (like a giant doughnut) where distance and the essential pull of gravity is actually larger than at a further point on the sphere (closer to straight down). I imagine drawing a series of straight lines (chords) through the sphere to make a cone. While each line individually cuts through fewer points on the line (atoms) than a straight line to the bottom, the wider circumference allows for
    a greater number of these lines, and so more points exerting gravitational pull. A longer cone (to the perpendicular surface) has longer lines, but fewer of them (assuming a point distance between each line origination). There must be a solution where the number of lines and their combined distance allows for the greatest number of points (assuming some fixed length for each point). Assuming the above is correct, what is the angle of the cone? I'm curious about what I observe in nature, and am looking for reasons why things appear as they do.

    It might be also good to think of each cone in terms of volume.
     
    Last edited: Nov 17, 2011
  5. Nov 17, 2011 #4

    DaveC426913

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    It is important o point out that you cannot experience "gross" pull of gravity. You only experience net gravity.

    It'll be a vector, having a magnitude and a direction. One direction, not many. It'll be straight down.
     
  6. Nov 17, 2011 #5
    I appreciate your answer, and agree with the comment on net pull. I'd like to make this a hypothetical question. Assuming that I was standing on the north pole, and the earth suddenly became a stable (and perfect) hemisphere from north to south, at what angle would the pull from gravity be the strongest?
     
  7. Nov 17, 2011 #6

    DaveC426913

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    What do you think?
     
  8. Nov 17, 2011 #7
    Guess answer is 45 degrees to the equator. I don't have the math to discover whether
    a 1/2 conical section at 45 degrees will have more area/volume than one at 55 or 65.

    OK, here is a way to solve for volume: http://www.karlscalculus.org/pr_conesp.html [Broken]
    I'm not a calculus expert but I might be able to get through it. Still looking for surface area.
     
    Last edited by a moderator: May 5, 2017
  9. Nov 17, 2011 #8

    DaveC426913

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    Wait. Do you mean if the Earth were sliced in half? Like the entire Western Hemisphere vanished in a puff of smoke? Ah. I see. That makes more sense.
     
  10. Nov 17, 2011 #9

    DaveC426913

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    You should look up Newton's Shell Theorem. The answer you seek is much simpler than this.

    The net force of gravity at any depth on or in a sphere is
    1] always straight toward the centre of mass
    2] directly a result of the mass below the point of observation*

    i.e if you are at a depth of 1000 miles standing on a planet that is 4000 miles in radius, It is exactly as if you were standing on a planet 3000 miles in radius.
     
  11. Nov 17, 2011 #10
    Assuming that they were peering off onto the hypothetical flat section, they would be standing
    at an angle that was not directly north-south. Were they to step over on the flat section, they would be at some angle (not necessarily 45 degrees) initially. As they walked toward the center of the flat section some 4000 miles away, "down" would become more and more perpendicular to the ground. It would be like walking down a steep slope to a flat level plain.

    Actually this thought picture does reinforce 45 degrees as a guess.
     
    Last edited: Nov 17, 2011
  12. Nov 17, 2011 #11

    DaveC426913

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    You are right about what they would experience. Walking on the flat surface would be like an extremely steep mountainside.

    But the angle will be much more vertical than 45 degrees. Attached is a diagram that spitballs it.

    A quick Google reveals this:
    A solid hemisphere of radius r has its centre of mass 3r/8 from the centre of the circular face, on the axis of symmetry.
    That will tell you exactly.


    What I don't understand is what this has to do with your initial question. We're kind of going off-topic.
     

    Attached Files:

  13. Nov 17, 2011 #12
    I'm creating a hypothetical to get around the net effect of gravity as might be found in Newton's shell theorem. While the net effect is straight down, a section of the sphere would reveal the angle at which that (eventually) cumulative effect might "pull" the strongest.

    3r/8. So with a radius of ~4000 miles, the center of mass would be 1500 miles from the center of the circular face. I'm not sure where this would show up on the spherical surface, or the angle of the chord from the north pole to that spot.


    Appreciate this forum's help.
     
  14. Nov 17, 2011 #13

    DaveC426913

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    I am still not sure it makes any sense. But whatever.

    Well that would depend on where you were standing. But since you now have measurements, it's pretty simple to work out the geometry. Take the diagram I mocked up and apply a few right triangles to it.

    Easy. It forms a right triangle 4000 mi by 1500 mi. Angle is then trivial to work out.
     
  15. Nov 17, 2011 #14
    Thanks for your help. This is a great forum. Signing out.
     
  16. Nov 17, 2011 #15

    DaveC426913

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    Don't be a stranger. :smile: That was an intriguing question.
     
  17. Nov 18, 2011 #16
    An interesting, semi-related thing, I was trying to figure up some stuff about gravity, using two dimenstions: If you had a sector of a circle, whole angle was a constant α, and that circle had uniform density d(kg/m2, the area of the sector would be:

    A = (α/360)∏r2

    The center of gravity is located at r/sqrt(2) (*)

    So given that the mass is determined by A*d and A is determined by r, the mass increases and center of gravity increase in such a fashion that no matter the r chosen, the newtonian gravitation pull will always be the same at the point that would be the center of the circle

    Acceleration, a = mG/r12

    m = dA = d∏r2α/360

    a = (Gd∏r2α/360)/(r/sqrt(2))2

    which simplifies to

    a = 2Gd∏α/360



    (*) the area of an annulus between r1 and r2 is: ∏r22(α/360) - r12(α/360)

    the shape resulting from cutting an equal area annulus off the original sector should be another small sector using r1 as it's radius, the center of gravity should be r1 in this equation for a given r and α

    ∏r12(α/360) = ∏r2(α/360) - r12(α/360)

    Multiplying everything by 360/∏α gives

    r12 = r2 - r12

    add r12 to each side, and then divide and take the square root, you get

    r1 = r/sqrt(2)
     
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