Coriolis Effect on Fast Projectile

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SUMMARY

The discussion centers on the Coriolis effect's influence on a projectile fired nearly horizontally toward the east. Participants clarify that the Coriolis acceleration, represented as \(\vec{a}_C = -2\vec{\omega} \times \vec{v}_0\), results in a southward deflection of the projectile due to the Earth's rotation. The formula for calculating this deflection is derived as \(\omega D^2 \sin\lambda / v_0\), where \(\omega\) is the angular velocity of the Earth, \(\lambda\) is the latitude, and \(D\) is the distance traveled.

PREREQUISITES
  • Understanding of the Coriolis effect in physics
  • Familiarity with projectile motion principles
  • Knowledge of angular velocity and its implications
  • Basic trigonometry, particularly sine functions
NEXT STEPS
  • Study the mathematical derivation of the Coriolis effect on projectiles
  • Explore the impact of latitude on projectile motion
  • Learn about angular momentum and its relation to the Coriolis effect
  • Investigate real-world applications of the Coriolis effect in ballistics
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Physicists, engineers, and students studying mechanics, particularly those interested in the effects of Earth's rotation on projectile motion.

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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?
 
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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.
 
Gokul43201 said:
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.
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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