What is the centripetal acceleration if you are moving with $100\frac{km}{h}$ on the equator if you are are moving a) east b) west?

Shouldn't the acceleration be the same given with $a_{cp}=\frac{v^2}{r}=\omega^2r$. Has the angular speed of earth $\omega_e=\frac{2\pi}{r}$ any influence on the cetripetal acceleration while moving east or west?

Calculate the Coriolison acceleration ($\vec{a_c}=\frac{\vec{F_c}}{m}$value and direction) if you move with $100\frac{km}{h}$ a) east b) west c) north d) south.

Using $\vec{F_c}=-2m(\vec{\omega}\times\vec{v})$ we get that $\vec{a_c}=-2(\vec{\omega}\times\vec{v})$. It's not that hard to calculate the value for a) and b) because $\vec{\omega}$ and $\vec{v}$ are perpendicular. In a) we have that the direction is perpendicular away from the surface of the earth and in b) perpendicular to the center of the earth. But what happens in c) and d)? The angle changes so the value changes also over time... The direction stays the same so thats not a problem to calculate. Can it even be calculated or?


closed as off-topic by David Hammen, AccidentalFourierTransform, ACuriousMind, CuriousOne, user36790 May 19 '16 at 3:22

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  • $\begingroup$ You need to show some work when you ask a homework question at this site. We do not do your homework for you. We do however help you overcome your misunderstandings. But that means that you have shown us some work. $\endgroup$ – David Hammen May 18 '16 at 18:51
  • $\begingroup$ @DavidHammen I have been googling for answers and I used the theory we had learned but I couldn't get a clear answer. I always try my best so I don't have to rely on someone here. But if I really can't get anywhere I ask for help so I can understand where my errors and problems are in the way I am thinking. $\endgroup$ – HeatTheIce May 18 '16 at 19:12
  • $\begingroup$ I think in its current form the question shows effort, and has a conceptual question in it. I vote to re-open. $\endgroup$ – Floris May 22 '16 at 1:54

For the first part of your question, you have to realize that your net velocity (the one that you plug into the expression for centripetal force) is the vector sum of the surface velocity and your velocity relative to the surface. If you were running West as fast as the earth turns East, you would "stay in place" and the sun would appear to stop moving in the sky. At that point there would be no centripetal force relative to the earth (although there would still be some relative to the sun...)

As for Coriolis acceleration; you have the expression. When you travel due North or South on the equator, the cross product is zero: there is no Coriolis force. Move away from the equator, and a force appears that is proportional to the sine of the latitude. The direction of the force changes as you cross the equator - if you look at global wind patterns, you will see that the two are related.

  • $\begingroup$ So in the first part, if I am going east then I will have that $\omega=\omega_e+\omega_v=\frac{2\pi}{T}+\frac{v}{r}$ and if I am traveling to the west we would have $\omega=\omega_e-\omega_v=\frac{2\pi}{T}-\frac{v}{r}$. I wasn't sure if I should take the relative angle angle velocity or just the one from the vehicle, bcs the earth rotates "without" acceleration. $\endgroup$ – HeatTheIce May 18 '16 at 20:59
  • $\begingroup$ An object on the surface of the earth is accelerating (to stay attached to the earth). Yes, your expressions for $\omega$ look correct. Viachle = vehicle? $\endgroup$ – Floris May 18 '16 at 21:02
  • $\begingroup$ Why is the cross produkt equal to zero if you travel due to North or South? There is still an angle between the velocity vector and the angle velocity vector which is $\neq 0$. Then there should still exist a Coriolis force... And sorry for the mistake, sometimes I do make dumb mistakes when I write late at night. $\endgroup$ – HeatTheIce May 18 '16 at 22:17
  • $\begingroup$ The vector pointing due North at the equator is parallel to the axis of rotation of the earth. Why do you think there is an angle between them? $\endgroup$ – Floris May 18 '16 at 22:56
  • $\begingroup$ At the start there is no Angle, but as I am getting closer to the North Pole the angle is getting bigger until it reaches 90°. At the equator it's easy, bcs the angle is always 90°, but here it changes over time. $\endgroup$ – HeatTheIce May 18 '16 at 23:05

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