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B Correct formula for deflection of a falling object

  1. Jun 8, 2017 #1
    Hello,

    I found a derivation on NASA and several others that say something different. Can someone tell me which is correct? They differ by a factor of 3. This is confusing because some places say an object falling from 100m should deflect by 3cm but NASA says .16mm. Those are more different than a factor of 3, so I am not understanding what the correct formula is.

    https://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/falling_eastward.htm

    http://hepweb.ucsd.edu/ph110b/110b_notes/node14.html
     
  2. jcsd
  3. Jun 8, 2017 #2

    anorlunda

    Staff: Mentor

    This is not an answer to your question, but the two linked problems are not the same. One is for 32 degrees latitude and the other for 42 degrees.
     
  4. Jun 8, 2017 #3

    jbriggs444

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    From the Nasa Article: "With w.gif = 7.27 x 10-7/sec"

    What is the angular rotation rate of the earth in radians/sec?
     
  5. Jun 8, 2017 #4
    I get 2*pi/(24*3600) = 7.3E-5. Is NASA wrong on such a basic thing?
     
  6. Jun 8, 2017 #5

    jbriggs444

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    Anyone can slip a couple of digits. Especially if they fail to sanity-check their work.

    Sanity-check: The difference in velocity of a ball at 100 m altitude and the ground at 0 m altitude is the same as that of a ball moving in a 100 m radius circle once every 24 hours (*). That's 2 pi times 100 m every 86400 seconds. Multiply that by a 4.5 second fall time and you get 3 cm. [Rather than trying to do a double integral of Coriolis acceleration in the rotating frame, I'm just using the inertial frame. We have a moving object falling onto a more slowly moving surface]

    You sanity-check the 4.5 second fall time by reasoning that a 5 meter fall time is 1 second and that 100 meters is a factor of 20 farther so the fall time should be ##\sqrt{20}## times longer. 20 is about halfway between 16 and 25, so its square root should be about halfway between 4 and 5.

    (*) We're using the back of an envelope and need not worry about sidereal versus solar days. And I'm doing my back of the envelope at the equator.
     
  7. Jun 8, 2017 #6
    Thanks! Do you know where the factor of 3 came from in the non-NASA formulas? I have found several derivations and they all include that factor.
     
  8. Jun 8, 2017 #7

    jbriggs444

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    They were integrating ##gt^2##. The integral is ##\frac{1}{3}gt^3##.

    Edit: or perhaps you were referring to this factor of 3...

    Multiply height by a factor of 10 and you've multiplied fall time by a factor of approximately 3
     
    Last edited: Jun 8, 2017
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