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Non inertial frame of reference(rotating)

  1. Jul 31, 2016 #1
    1. The problem statement, all variables and given/known data
    $$g_o=\frac{GM}{R^2}e_R$$
    where g0 is the gravitational acceleration, G = 6.67 * 10-11Nm2 /kg2 is the universal gravitational constant, M= 5.98 *1024 kg is the mass of the Earth, and R = 6.38 * 106 m its radius

    $$g = g_0-ω*[ω*(r+R)]$$.................(8.29)
    where gravitational acceleration g that is felt near the surface of the Earth (i.e., on the surface of the Earth we cannot discerned between gravity g0 and the centrifugal acceleration -ω*[ω*(r+R)], we can only feel the resulting acceleration g )

    The Earth’s angular velocity vector is given by ω= 7.3 * 10-5ez rad/s (i.e., it is directed along the z'-axis ), and we assume that it is a constant.

    It is to be noted that because of the presence of the centrifugal acceleration
    -ω*[ω*(r+R)] in this equation for the effective gravity, g and g0 will in general
    not point exactly in the same direction. This effect is rather small, but measurable as
    ωR^2/g_0= 0.0035 . It should also be clear from the equation (8.29) that the magnitude of the effect is a function of latitude.
    upload_2016-7-31_15-45-16.png
    Source:http://www.astro.uwo.ca/~houde/courses/PDF files/physics350/Noninertial_frames.pdf

    where R is the radius of Earth, r is the position of the object measure from the rotating frame of reference which is moving at a constant velocity and constant angular velocity, r' is the position of the object measure from the center of the Earth

    2. Relevant equations

    ωR^2/g0= 0.0035 where R is the radius of Earth.
    I have no idea how they get this equation. Besides I don't quite get what they mean about "that the magnitude of the effect is a function of latitude.".


    3. The attempt at a solution
    I have no idea how to get to this equation ωR^2/g0= 0.0035. I try to calculate the above equation but I get is 303234462.4 instead of 0.0035.
    $$ωR^2/g_0=ωR^2*\frac{R^2}{GM}=(7.32*10^{-5})(6.38*10^6)^2*\frac{(6.38*10^6)^2}{6.67*10^{-11}*5.98*10^{24}}=303234462.4$$
     

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    Last edited: Jul 31, 2016
  2. jcsd
  3. Jul 31, 2016 #2

    Simon Bridge

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    They are working out a subjective feeling of the gravitational acceleration "g" in terms of the different effects that come into play.
    One of these effects is centrifugal ... why would the centrifugal force vary with latitude and how?

    As for the equation ...
    Recall the work you have done on circular morion in an inertial frame: what is the equation for centripetal acceleration?
    What is the relationship between centripetal acceleration and centrifugal force?
    Now divide the centrifugal "acceleration" by the bare gravity to get the correct expression: compare with what the author did above.
     
  4. Jul 31, 2016 #3
    Ok thanks. I know the equatorial bulge at Earth can be explained by non-inertial frame of reference using centrifugal force. But can i explain it from the inertial frame of reference using centripetal force? Cause as we move from the equator to the poles, the centripetal force require to hold the mass that make up Earth decreases, therefore the gravitational force(remaining one not being use to provide the centripetal force) will increase from the equator to the pole, causing the bulge?
     
  5. Jul 31, 2016 #4

    TSny

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    This expression has a typographical error. The author of the notes misplaced the square.
     
  6. Aug 3, 2016 #5

    Simon Bridge

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    ... which was, of c
    In the inertial frame, gravity is the same across the surface of the Earth. You need to think about why objects inside a car going around a corner tend to slide to one side.
     
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