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Coriolis and different refererence frames

  1. Sep 20, 2009 #1

    I am stuck on the coriolis force. If an object is dropped how do I get the eastward deflection when looking from an inertial frame? I know how from the rotating frame but for some reason I am stuck, i.e. what terms to use.
    Also, if I throw a ball straight up from earth what is the westward deflection, do I solve this twice? Once for the ball moving up and then again once it is at the max height falling down?

    Thank you
  2. jcsd
  3. Sep 20, 2009 #2


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    What you see there initially is the ball that has the same angular velocity but a higher tangential velocity than the ground. As it descends it outruns the ground due to that higher tangential velocity. The angular velocity of the ball increases to preserve it's angular momentum.
    Yes the west acceleration during rise is canceled by the east acceleration during fall, so the ball lands with zero velocity in the east-west direction, but in a point west to the throw position. The defection is the same during both phases, so you just calculate it once and multiply by two.
    Last edited: Sep 20, 2009
  4. Sep 21, 2009 #3


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    I won't give you the derivations, but you can check your results against the following Java applet that is available on my website. http://www.cleonis.nl/physics/ejs/ballistics_and_orbits_simulation.php" [Broken]
    The Coriolis effect is the main subject of my website. (http://www.cleonis.nl)

    The applet http://www.cleonis.nl/physics/ejs/ballistics_and_orbits_simulation.php" [Broken] shows the motion of a launched object. The panel on the left shows the motion with respect to the inertial coordinate system, the panel on the right shows the motion with respect to the coordinate system that is co-rotating with the Earth.

    You can launch from any latitude, you can launch at any angle (eastward, westward etc), you can launch at any elevation, and you can specify the nozzle velocity (zero for release to free motion, high speed for launch with a gun.)

    If memory serves me the textbook 'Classical dynamics' by (authors Stephen Thornton and Jerry Marion) offers several discussions.

    If an object is dropped from a high tower, why does it not land at the base of the tower? Answer: (simplified version that neglects air resistance effects). Once the object is released it is effectively in orbital motion. (An orbit that will be shortlived, as it intersects the Earth's surface.)
    There is Kepler's law for orbital motion: when an orbiting object is pulled closer to its primary its speed increases, (in such a way that equal areas are swept out in equal intervals of time.)
    An extreme example is the orbit of Halley's comet: during the part of the orbit that the comet is falling to the center of the solar system the comet is gaining speed all the time. Halley's comet sweeps around the Sun, and then starts climbing outward again. Moving against the pull of the Sun the comet is losing velocity all the time.

    Returning to the example of releasing an object from the top of a high tower: the Earth's gravity is pulling the object in, so the angular velocity of the object increases. The Java applet I mentioned is also designed to illustrate the 'object released from high tower' scenario.

    In this particular case the fastest calculation strategy to obtain the motion with respect to the inertial coordinate system is to apply conservation of angular momentum.

    Last edited by a moderator: May 4, 2017
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