# Does Coriolis effect change the direction of an object moving parallel to equator, say along the tropic of Cancer?

It is usually said that the Coriolis force deflects a horizontally moving object to its right in the northern hemisphere and to the left in the southern hemisphere, and an object at the equator will be experiencing zero Coriolis force.

1. Now, what happens to an object rotating along with the Earth in a path parallel to the equator, along the tropic of Cancer? (ie, velocity of the object with respect to the rotational frame equals zero). Shouldn't the Coriolis force on the object be zero in that case?

2. Also, what happens if the object is moving eastwards along the tropic of Cancer, but with a non-zero relative velocity? What would be the direction of the Coriolis force? Would the object be deflected towards its right, ie, towards the equator, or would the object be experiencing a force in the direction of centrifugal force?

3. What if the object is moving eastwards along the equator, but with a non-zero relative velocity? What would be the direction of the Coriolis force? Would the object experience a force in the direction of centrifugal force?

It would appear you are asking about the following case: if an object has a velocity with respect to the Earth, in a direction parallel to the local latitude line, will there be a rotation-of-Earth effect?

For this situation let's use an airship. An airship is buoyant, it's propulsion can give it a considerable velocity relative to the Earth, and anything that tends to deflect the motion of the airshil will deflect it, because the airship has no friction with the ground.

An airship that is moving from west to east is circumnavigating the Earth's axis faster than the Earth itself. Compare that to a car going around a curve on a banked circuit.

(Early racing cars had very little grip. On a banked circuit the car is steered up the slope, so that the angle of the slope provides the required centripetal force. From the inside of the bend to the outside the slope is steeper and steeper.)

Conversely, an airship that is moving east to west is circumnavigating the Earth's axis slower than the Earth itself. So the airship will tend to slump down, towards the nearest pole.

About the shape of the Earth:
At the equator the radius of the Earth is 21 kilometers more than the radius at the poles. That is: to go from the equator to the poles is to go downhill. Imagine a planet with exactly the shape of the Earth, but not rotating. Then all fluid would flow to the poles. Over geologic time scale the Earth is deformable enough so that the Earth as a whole is always very close to hydrostatic equilibrium. The shape of the Earth is such that at every latitude the downhill slope is the slope that provides buoyant objects the required centripetal force to remain on the same latitude.

With the above in place, let's calculate how much sideways deflection we can expect, in proportion to velocity (along the local latitude line) with respect to the Earth.

I will use the uppercase letter $$\Omega$$ for the angular velocity of the Earth, and the lowercase letter $$\omega$$ for the angular velocity of the airship relative tot the Earth. I will use $$\omega_r$$ for this relative angular velocity. I will use $$R$$ for the radius of the Earth.

The expression for the required centripetal acceleration for circumnavigating motion is:

$$a = \omega^2 r \qquad (1)$$

The airship is moving west-to-east. To find the expected sideways acceleration we calculate the total required centripetal acceleration, and subtract the actually provided centripetal acceleration.

$$a = (\Omega + \omega_r)^2R - \Omega^2R \qquad (2)$$

$$a = \Omega^2R + 2 \Omega \omega_rR + \omega_r^2R - \Omega^2R \qquad (3)$$

$$a = 2\Omega \omega_rR + \omega_r^2R \qquad (4)$$

Now we convert the relative angular velocity of the airship to a relative linear velocity ( $$v=\omega r$$ )

$$a = 2\Omega v_r + v_r^2/R \qquad (5)$$

For the kind of velocity that an airship can reach the first term, $$2\Omega v_r$$, is much larger than the second term $$v_r^2/R$$

So:
Unless the crew of the airship takes countermeasures an airship moving along a local latitude line, with a velocity with respect to the Earth, will veer away from that course. The tendency to veer away is proportional to the velocity with respect to the Earth. The magnitude of the tendency to accelerate sideways will be $$2\Omega v_r$$. The direction of the veering is as follows: an airship moving west-to-east will swing wide, veering to the outside of the local latitude line. An airship moving east-to-west will veer to the inside of the local latitude line.

Note that for this calculation it isn't necessary to know the mass of the airship. The required centripetal acceleration is provided by the Earth's gravity, and inertial mass and gravitational mass are equivalent.

Also, at any latitude the direction of the tendency to deflect is parallel to the plane of the equator. But at higher latitudes the Earth surface is at an angle to the plane of the equator. A local calculation must take that into account.

For an airship:
Maintaining altitude is managed by adjusting the buoyancy of the airship.
Maintaining course is managed by pointing the nose of the airship in the direction opposite to the tendency to veer away from the intended course.

In itself the rotation-of-Earth effect is a single effect, the direction is in the plane of the rotation. It's just that when you are flying an actual airship altitude and course are managed separately. So it's practical to think of the total rotation-of-Earth effect as decomposed in a vertical-to-the-local component and a horizontal-to-the-local component.

Meteorologists refer to the horizontal-to-the-local component as 'Coriolis effect'.

• //at any latitude the direction of the tendency to deflect is parallel to the plane of the equator// - Does this mean that the deflection of a plane moving westwards will be either upwards or downwards (vertical)? (I have edited my opening question) Mar 12 at 8:16
• @Curiouserandcuriouser See the wikipedia article about the Eötvös effect. (That wikipedia article was contributed by me. There have been edits since I wrote it; no significant changes, by the looks of it.) Mar 12 at 8:27
• Okay, I think it's clear now. If an object is moving at a relative velocity eastwards along the equator, the Coriolis force will be directed outwards (along the same direction of centrifugal force). If the object is moving eastwards at a higher latitude, say tropic of Cancer, the Coriolis force vector will be pointed in the same direction as that was the case while travelling along equator. But since the points on higher latitudes makes a non-zero angle with the equatorial plane, in effect, the object will be deflected towards the equator. Mar 12 at 12:42
• @Curiouserandcuriouser To emphasize: recognizing what happens in the case of east-to-west velocity (relative to the Earth) is by far the most important. From the equator to the poles the Earth's surface is downhill. That slope gives rise to a centripetal force. When an object has east-to-west velocity it is circumnavigating slower, so less centripetal force is required. But the downhill slope is the one for co-rotating, which means that for the east-to-west moving object there is a surplus of centripetal force. That surplus causes veering to the inside of the starting latitude. Mar 12 at 13:17

An object moving directly north or south at the equator will experience 0 Coriolis force. All other moving objects experience some Coriolis force. (For simplicity's sake, I've disregarded vertical motion here w.r.t. the ground).

More generally, the Coriolis force can be expressed as: $$\mathbf{F}=-2m(\mathbf{\Omega}\times\mathbf{v})$$ Here, $$\mathbf{\Omega}$$ is the angular velocity vector of the Earth (points along the axis of rotation), and $$\mathbf{v}$$ is the velocity of the object in the rotating frame of reference, i.e. with respect to the ground.

Note that this means an object travelling with no relative motion to the ground beneath it will have a velocity of 0 in the rotating reference frame.

• Okay, so what if an object is moving at eastwards velocity (relative to the rotating frame) along the tropic of Cancer (parallel to equator)? In that case, the Coriolis force won't be zero. In which direction would the Coriolis force be in that case? Would the object bend right (towards the equator), or does it act in the same direction of centrifugal force? Mar 12 at 7:41
• R.W. Bird provides a nice answer to your question below :) Mar 12 at 20:16

If the formula provided by DanDan101 is correct, then for an object at the equator moving north or south, the cross product will be zero. For motion to the west, the force is down. For an object moving north or south at a northern latitude, the force will be to the right. If the object is moving west, the force will be toward the axis of rotation. But this breaks into components; one toward the center of the earth, and the other to the north. An object moving horizontally in the northern hemisphere is always subject to a Coriolis deflection to the right. (In the southern hemisphere it is to the left.)