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Earth angle - Gravity or acceleration?

  1. May 7, 2014 #1
    I hope this is interesting....
    I am standing on the Earth's surface, I want to know the angle created with an Isosceles triangle 10' across by 3,960 miles on each side. I know the answer is just under 90 degrees, but need a more precise number.

    The problem came up when discussing whether a person in a room could tell if they were under the influence of Earth's gravity or 1 g of acceleration.

    Note: The machine that is 10' across is supposed to be a 'gravit-o-meter' - It does not measure the force of gravity, only its apparent source - hence the 3,960 miles - the radius of the Earth.

    Thank you for your help.

    GBSmith
     
  2. jcsd
  3. May 8, 2014 #2

    jbunniii

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    I assume you mean the angle ##\theta## depicted in the attached picture? Assuming so, divide the triangle into two right triangles. Consider one of the right triangles. Which trig function relates the 5' side to the 3960 mile side?
     

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  4. May 10, 2014 #3
    After converting the miles into feet - about 21 million feet, I used the calculator TI-BA2Plus - using tan -1, came up with 89.9999852 degrees. Thus proving the person is sitting in a sealed room on Earth, and not in a space ship being constantly accelerated at 1g.
     
  5. May 10, 2014 #4

    Mentallic

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    In reality, the centre of mass of the Earth is not exactly at the centre of the core. Also, assuming that this device is perfect and can measure such a minute difference in angle, and the Earth is perfect in its spherical symmetry and density throughout, then if you measured from each of the two ends of the room you'll find that the focal point of gravity will be exactly perpendicular from both positions. This is because the room will ever so slightly curve with the curvature of the Earth.
     
  6. May 17, 2014 #5
    But if I took the room and placed it in space, the difference between point source gravity and uni-directional force acceleration should register two different results. With gravity measuring less than 90 degrees, and constant acceleration (1 g) measuring a true 90 degrees.

    Actually think of a huge grav-o-meter, stretching between NY and Paris, it will definitely get an angular measurement of <90 degrees, while one in space will have an 'absolute' 90 degree reading.
     
    Last edited: May 17, 2014
  7. May 17, 2014 #6

    haruspex

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    But the apparatus is unspecified. It need not depend on the room geometry. You could in principle construct a quadrilateral with A 1m directly above B, D 1m above C, B and C level, A and D level. Then measure the difference between lengths AD and BC.
    (The 211 m towers of the 1.3 km long Verrazano-Narrows bridge are 41 mm further apart at the top than at the bottom.)
     
  8. May 17, 2014 #7

    mfb

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