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Surface area of the earth

  1. Jun 1, 2008 #1
    I have a very faint idea of General Relativity.. hence this question. I think that according to General Relativity, the shortest distance between two points on this earth is along a curved path, which is the curvature of the earth [or sorta.. parallel to it]. Hence, i assumed that when on earth, we measure an area of the land, and we treat it as a rectangle, we are not making an approximation, but that, in fact is the actual area we calculate on this earth.

    so, is it true that if we are to measure the surface area of the earth while being on the earth, we will measure it to be [itex]4\pi^2 R_e^2[/itex] rather than [itex]4\pi R_e^2[/itex]??
     
    Last edited: Jun 1, 2008
  2. jcsd
  3. Jun 1, 2008 #2
    lol i realy did not know dolphins lived in iglooes.. cartman is soo smart..

    sorry but becouse of the geological imperfection of our planet neither of your formulas will work.. For one the earth is not a sphere, its an elliptoid. this means that the circumference around the equator and the circumference around the poles has a difference of thousands of miles. Also if you try to measure it manually using "rectangles" by the time u finish the surface area will be vary different and you have to start over.. someone else can help you with those formulas cause i have no clue what they mean.. ha ha they wernt on my GED test
     
  4. Jun 2, 2008 #3
    i know about the geological imperfections and how the earth is not a sphere but an ellipsoid. The question I asked was a theoretical one, so practical considerations, as you mentioned aren't necessary.

    in that case, I suggest that you should stop being a troll...
     
  5. Jun 2, 2008 #4

    tiny-tim

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    Hi rohanprabhu! :smile:

    No, General Relativity doesn't do earths.

    Loosely speaking, the "shortest distance" in General Relativity is the space-time path followed by a body under zero force (so free-falling or orbiting, or just drifting). I think. :rolleyes:
    There's no such thing as a rectangle on a sphere … the sum of the angles of any quadrilateral is always more than 360º.
    I'm not following you … where does the π² come from? :confused:
     
  6. Jun 2, 2008 #5
    first of all, thanks for the quick reply.

    From assuming that the sphere is a folded rectangle, where each side is of length [itex]2\pi R_e[/itex] (i.e. the equatorial line and one of the latitudes perpendicular to it can be spread out to form a large rectangle, which is cut into four rectangles by these axis and the intersection of these axis is the center of that rectangle). The surface area of such a rectangle would be [itex]2\pi^2 R_e^2[/itex].
     
  7. Jun 2, 2008 #6

    tiny-tim

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    Now I understand. In that case …
    No, definitely not, no way.

    General Relativity doesn't work like that, not even remotely like that.

    Area is measured by using infinitesimally small squares and pretending they're flat, not by measuring large squares and pretending they're flat. :smile:
     
  8. Jun 2, 2008 #7
    ahan.. i get it now. Originally, i wasn't thinking that we measure large squares and pretend that they are flat. What i was thinking was that due to the curvature of the spacetime deformed by the gravity, we have a different geometry local to a mass than a geometry far away from a mass.

    Something like what the density parameter defines. For ex:

    This is something I thought would happen locally. I am not saying that the reason would be the same. I am just saying that as the geometry of the Universe could be hyperbolic, I assumed that due to a mass, it changes the local system such as to make the geometry locally spherical and hence all straight lines are basically curves when viewed from an external frame and hence the 'flat' for an internal observer would appear to be curved to an external observer. In short I assumed that the geometry of the universe was relative.

    Thanks for clearing that up :D
     
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