I am studying accelerating reference frames and I have solved the famous problem of finding how the Coriolis force deflects the trajectory of a falling object on Earth, the answer being

$$ x_{east} = \frac{wgcos(\theta)}{3}\sqrt{\frac{8h^3}{g}}$$

where $\theta$ is the angle between the falling trajectory and the equator, and $w$ the angular velocity of the Earth.

My question, however, is this: Does the $x_{east}$ variable represent deflection from a fall path that already takes into account other effects such as the fact that the earth is not perfectly round, the centrifugal force, etc.?

In other words, is it correct to say that $x_{east}$ does not represent deflection from a path that would be directed straight towards the earth, but rather from a path that would be directed directly towards $\textbf g$, with $\textbf g$ being the local vector for gravity?

As a side note, is the effect of the centrifugal force, which (If I'm not mistaken should be oriented westwards), greater than that of the Coriolis force?

Can we simply use the formulas

$F_{cent} = -m\omega \times (\omega \times r)$


$F_{cor} = -2m\omega \times v$

to figure it out?

  • $\begingroup$ cosine theta? So it is zero at the equator? $\endgroup$
    – JEB
    Aug 17, 2019 at 3:21
  • $\begingroup$ @JEB Thanks for pointing it out. It is cos(theta) but then the angle is with the equator. My mistake. It's edited now. $\endgroup$ Aug 17, 2019 at 4:36

1 Answer 1


About centrifugal effect

Let's start at the Equator. Your distance to the Earth axis of rotation is the Earth radius (at the Equator), and your angular velocity is one revolution per day.

The required centripetal acceleration for that specific circumnavigating motion is about 0.034 m/s^2

That required centripetal acceleration goes at the expense of the amount of gravitational acceleration that a local gravity sensor will measure.

Note that since inertial mass and gravitational mass are equal any local gravity measurement can only give you the resultant gravity. That is: what the local gravity measurement shows is gravity with the locally required centripetal acceleration already subtracted.

By universal convention, it's just the practical thing to do, the value for gravity measurement is the measured value. In any book of tables of physical values, in any geophysics textbook, on Wikipedia, the value given is the measured value.

You can infer the value of the local true gravity by taking the measured gravity, and adding the local required centripetal acceleration that you have calculated.

Comparing Centrifugal effect and Coriolis effect.
To compare, calculate the locally required centripetal acceleration (0.034 m/s^2 at the Equator), and compare that with the value for the Coriolis effect.

On latitudes between poles and Equator

For simplicity let me take the case of 45 degrees latitude. Because of its rotation the Earth has an equatorial bulge. There is the distance of the poles to the Earth geometrical center, and there is the equatorial radius. The equatorial radius is about 20 kilometers more than the distance between poles and Earth geometrical center.

If you would have a celestial body with the exact shape of the Earth, but non-rotating, then on each hemisphere fluid would flow towards the nearest pole. On a celestial body with an equatorial bulge there is a downhill slope from Equator to pole. Of course, the slope is minute, but it's not zero.

At any latitude a local plumb line is perpendicular to the local surface.

A falling object will be subject to a Coriolis effect. Other than that it will, to a good approximation, fall parallel to the local plumb line. (It is an approximation. If you would need to know exactly then you would need to know in what way the approximation deviates from the actual motion.)

Again, the local gravity measurement gives you the local resultant gravity.

Using the Coriolis term to calculate the deflection is an approximation. The larger the height from which you drop the object the larger the deviation from exact calculation. (Again. If you would need to know exactly then you would need to know in what way the approximation deviates from the actual motion.)

Exhaustive calculation

The effect that you want to calculate is extremely small. Because of that a calculation that assumes a spherical Earth will not be accurate.

To approximate Earth gravity at a sufficient level of accuracy: McCullough's Formula:

For the shape of the Earth WSG84

The equation of motion (for rotating coordinate) system will then have McCullough's formula for gravitational acceleration, and both the centrifugal term and the coriolis term.

But here's the thing: the equation of motion for an exhaustive calculation is so complicated that the only option you have is to move to computer numerical analysis. But if you move to computer numerical analysis anyway then there is no longer any reason to use a rotating coordinate system. You can use the non-rotating coordinate system for the integration, and in a subsequent step transform the computed trajectory to any particular rotating coordinate system you want.

If you use a non-rotating coordinate system the implementation of the numerical analysis is simpler: no centrifugal term, no coriolis term; only McCullough's formula.


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