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First of all I am defining x axis in east of object and y to be in the north of object . Let us suppose a body is falling from a height; then after falling a certain height indeed gain some velocity in the direction which I am defining in the local height of the observer in z direction. indeed in a rest frame outside Earth it would not have velocity in y or x direction but in a frame standing still on earth (perhaps moving with earth) it will gain a velocity in the east direction and y direction so when I determine the coriolis force for a certain moment then to determine the coriolis force while doing the matrix of 2$m\omega \times v$should I put $\vec v$ is equal to $\dot x\hat{i} + \dot y \hat{j} + \dot z \hat{k}$ as in the in the earth rotating frame it has velocity in the x and y directions but in the rest frame outside earth it never has velocity in x or y direction. So should I put $\vec v$ =$0\hat{i} +0\hat{j} + \dot z\hat{k}$ ? Show my exact question is that what should I take as the velocity to detect the velocity as the perception as in perception of a rest frame outside Earth or should I take the velocity of a observer who is in the earth frame who can't perceive his own rotation so sees the ball shifting.

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  • $\begingroup$ I think this could be a good question but it's not exactly clear what you're asking. Can you clean up your question a little bit? $\endgroup$ – Zack Hutchens Dec 14 '18 at 3:00
  • $\begingroup$ @ZackHutchens when an object falls down in earth which is a rotating frame then it would have some velocity in z direction or in local height of object but after subsequently falling it gains some velocity in x and y direction. So while determining coriolis force. $\2m \omega \times v$ should the v represented as earth standing observers perception or in rest frane perception outside earth. I think you may be able to understand the question now and you are welcome to suggest edits to clarify the question. $\endgroup$ – Nobody recognizeable Dec 14 '18 at 3:05
  • $\begingroup$ @ZackHutchens is it clear now ? $\endgroup$ – Nobody recognizeable Dec 14 '18 at 3:11
  • $\begingroup$ I think so. I posted an answer. If it does not help, then you can ask questions and I will try again. $\endgroup$ – Zack Hutchens Dec 14 '18 at 3:22
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The equivalent of the Newton's second law in the noninertial frame is expressed as $$m\ddot{\vec r} = \vec{F}' - 2m \vec{\omega} \times\vec{v}.$$ Here everything is written in terms of the rotating coordinate system. $F'$ is the actual net force on the system, but it is written to be a function of the rotating coordinates. It also absorbed the centrifugal force, so in your case $\vec{F}' = m\vec{g}$ but $|\vec{g}| \neq 9.8$ N/kg.

So, the point here is that $\vec{v}$ is taken to be in the rotating coordinates -- is the viewpoint of an observer fixed on the Earth.

Thus for the initial velocity you have $\vec{v}_0 = \dot{z}\hat{k}$. As the object begins to fall, it will become deflected from the Coriolis effect, and this velocity will in accordance with $\vec{v} \times \vec{\omega}$. However for the Earth, $\omega$ is quite small at $\sim 10^{-4}$ s$^{-1}$. Thus, to a first approximation: $$m\ddot{\vec r} = \vec{F}' - 2m \vec{\omega} \times\vec{v}_0,$$ which will allow you to calculate the longitudinal deflection (in the x-direction).

As given in Goldstein p. 174, if you have that text, the corresponding inertial frame is a coordinate system which is stationary with respect to "local stars".

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  • $\begingroup$ Ok so i see you are saying $\vec v$ should be taken in perception of the rest frame observer outside earth .( Probably stars' can be regarded as rest frame as their rotation is negligible. ) . Please justify this comment. $\endgroup$ – Nobody recognizeable Dec 14 '18 at 3:29
  • $\begingroup$ It's a good approximation to neglect the deflection $v$ because otherwise it's a really hard differential equation to solve. And yes, the local stars are a good approximation to an inertial coordinate system because their rotation is negligible. $\endgroup$ – Zack Hutchens Dec 14 '18 at 3:34
  • $\begingroup$ if v is taken in perception of outside earth observer (who is in a rest frame) then it has velocity in z direction only ; it never really deflects in other direction. So you'd never need to estimate or neglect anything. There's the problem. $\endgroup$ – Nobody recognizeable Dec 14 '18 at 3:39
  • $\begingroup$ Correct. This may help: youtube.com/watch?v=49JwbrXcPjc $\endgroup$ – Zack Hutchens Dec 14 '18 at 3:42
  • $\begingroup$ neglect the previous point. I think i got it. You are saying $\vec v$ is taken in rotating frame. But the invariance is shown with stars frame. $\endgroup$ – Nobody recognizeable Dec 14 '18 at 3:49

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