Coriolis force, no latitude change

If you move an object without changing its latitude, according to $$\vec{F}_c=-2m\vec{\omega}\times\vec{v}$$ you could add or substract weight to the object itself, depending on the $$\vec{v}$$ direction. However... I can't visualize that intuitively, I mean, I've always thought that the coriolis effect had to do with the fact that the earth is rotating, but if you are, for example, moving on the same direction as earth's rotation, then shouldn't you be 'accelerating'?

Here is a visualisation:
Visualize an airship, neutrally buoyant, in stationary position above some point on the equator.

At the equator the Earth's rotation gives the Earth's surface a velocity of about 465 meters per second. The amount of centripetal force required to cause an object to move along a circular path with a radius of 6378 kilometer (the Earth's equatorial radius), at 465 m/s, is about 0.034 newton per kilogram of mass.

Thus objects at the equator have a little bit less weight than the weight that they would have if the Earth would be non-rotating. The effective gravity: true gravity minus the amount of centripetal force required to co-rotate with the Earth.

Now imagine the airship first cruising due east, trimmed to neutral buoyancy, and then the airship makes a U-turn.

When cruising at 25 m/s due east the total velocity becomes 465 + 25 = 490 m/s, which requires a centripetal force of about 0.375 newton (per kilogram of mass). Cruising at 25 m/s due West the total velocity is 465 - 25 = 440 m/s, requiring about 0.305 newton.

The crucial point is: when the airship is co-rotating with the rotating Earth it is already the case that providing required centripetal force goes at the expense of gravity. That is what sets up the difference between moving east or west.

A derivation for velocity constrained to a fixed latitude:

$$u$$ velocity with respect to the Earth, along latitude line
$$a_s$$ required centripetal acceleration when when stationary with respect to the Earth
$$a_u$$ required centripetal acceleration when moving at velocity u
$$a_r$$ resultant acceleration; the difference of the above two
$$\Omega$$ angular velocity of the Earth: one revolution per Sidereal day
$$\omega_r$$ angular velocity of the airship relative to the angular velocity of the Earth.
$$R$$ Radius of the Earth

As we know, the formula for required centripetal acceleration to sustain circular motion: $$a=\Omega^2R$$.

Hence for the resultant acceleration:

{\displaystyle {\begin{aligned}a_{r}&=a_{u}-a_{s}\\&=\left(\Omega +\omega _{r}\right)^{2}R-\Omega ^{2}R\\&=\Omega ^{2}R+2\Omega \omega _{r}R+\omega _{r}^{2}R-\Omega ^{2}R\\&=2\Omega \omega _{r}R+\omega _{r}^{2}R\\&=2\Omega u+{\frac {u^{2}}{R}}\\\end{aligned}}}

This expression gives the change in weight as the airship makes a U-turn (no latitude change).

• That's a really good explanation, thank you VERY much. Apr 10 '20 at 14:29

Imagine standing on the equator and throwing something straight up. At ground level the sideways velocity due to rotation is $$2\pi R_\oplus/P$$ where $$P$$ is the length of the day. Of course, you and the object are both moving at this speed, so you only see the vertical component. But at height $$h$$ the object still has that sideways speed, while the atmosphere around it has velocity $$2\pi (R_\oplus + h)/P$$, a difference of $$2\pi h/P$$. Hence, from the perspective of the air moving with the Earth at that altitude, the object is moving sideways.

This becomes extra relevant if the object is a parcel of air lofted up by convection: it will tend to drift westwards. Were it instead descending it would have too much lateral speed, and drift eastwards.