How Does the Coriolis Effect Influence River Water Levels at Different Banks?

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SUMMARY

The Coriolis effect causes a difference in water levels between the east and west banks of a river flowing northward. The height difference can be approximated by the formula 2Dwvsin(λ)/g, where D is the river width, v is the flow speed, w is the Earth's angular velocity, and λ is the latitude. A participant attempted to derive this relationship using a different formula but faced challenges in proving the equivalence. The discussion emphasizes the importance of correctly applying trigonometric identities and understanding the physical implications of the Coriolis effect on fluid dynamics.

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  • Understanding of fluid dynamics principles
  • Knowledge of the Coriolis effect and its implications
  • Familiarity with trigonometric functions and identities
  • Basic physics concepts related to motion and gravity
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  • Research the mathematical derivation of the Coriolis effect in fluid dynamics
  • Study the impact of latitude on the Coriolis effect using real-world examples
  • Explore the relationship between river flow dynamics and angular momentum
  • Learn about the applications of the Coriolis effect in meteorology and oceanography
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Students studying physics, environmental scientists, and engineers interested in fluid dynamics and the effects of Earth's rotation on water systems.

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Homework Statement




A river of width D flows Northward with speed v. Show that the water is lower at the west bank than at the east bank by approximately

2Dwvsinlamda/g

where w is the angular velocity of the Earth and lamda the latitude.

Homework Equations





The Attempt at a Solution



I get the difference in height as 2Dwvsin(lamda)cos(lamda)/sqrt(g^2 +(2wvsin(lamda)cos(lamda))^2)

So if this was the same as my answer it would mean that cos(lamda)/sqrt(g^2 +(2wvsin(lamda)cos(lamda))^2) = 1/g

Im just unsure how to prove this, it looks like a simple right angled triangle to show this by rearranging to show coslamda = sqrt(g^2 +(2wvsin(lamda)cos(lamda))^2)/g

Does this look right guys?
 
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Not really. You can't reduce your formula to the answer through any approximation that I can see. Why don't you just share your calculations with us?
 

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