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Coriolis Effect on Fast Projectile

  1. Jun 29, 2004 #1
    A projectile is fired nearly horizontally at high velocity [itex]v_0[/itex] toward the east. (a) In what direction is it deflected by the Coriolis effect? (b) Determine a formula for the deflection is terms of [itex]v_0[/itex], the angular velocity [itex]\omega[/itex] of the earth, the latitude [itex]\lambda[/itex] where the projectile is fired, and the distance traveled D.

    I don't know why the problem says "nearly horizontal" instead of just horizontal. Since the projectile is traveling pretty fast, I guess I can ignore the effects of gravity.

    Now, suppose I shoot somebody located directly east of me (how evil!). Since the earth is round, the shot will be gaining height with respect to the ground directly below it right? Now, from the shooter's perspective, this is also happening right? So the shot is moving upward. Now I haven't made any use of the Coriolis effect so my latter explanation is probably bogus. Now, [itex]\vec{a}_C = -2\vec{\omega} \times \vec{v}_0[/itex]. Using the right hand rule, it seems that the Coriolis acceleration [itex]\vec{a}_C[/itex] is pointing away from the earth which seems to support my first idea. Now, the books answer is South for some unknown reason so I don't know what to think anymore. Any help here?
    Last edited: Jun 29, 2004
  2. jcsd
  3. Jun 29, 2004 #2


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    I think the question meant to say "nearly vertically". The verticality would be affected by the Coriolis acceleration making it impossible to be exactly vertical.

    Yes, if you threw horizontally, there would not be a southward migration. Unless it's winter and you're throwing birds.
  4. Jun 29, 2004 #3
    But how do you throw "nearly vertically" to the east. That makes no sense unless I'm throwing at an angle to the horizontal. I'll post the answer to (b) in hopes of any enlightment.

    Answer (b): [itex]\omega D^2 \sin\lambda / v_0[/itex]
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