I've been asked this question twice now from our favourite science denying community, and I don't know how to parse it. I can do centrifugal force and gravity given mass, but have never tried to even see if there should be any deflection of gravity due to centrifugal force at a given latitude.

I'm guessing I'm being sold a bridge with his numbers, but can't tell for sure.

Here's the numbers he's given me:

  • Gravitational force (m/s^2) = 9.80576
  • Latitude (degrees) = 45
  • Velocity (mph) = 735.64
  • Centrifugal force (m/s^2) = 0.0239809
  • Gravity Direction Vector = (0.707107, 0.707107)
  • Centrifugal force Direction Vector = (1, 0)
  • Combined Force = (-6.90974,-6.93372)
  • Combined Force (relative to ground) = (0.016957,-9.7888)
  • Weight reduction (%) compared to North pole = 0.172929
  • Projectile drop (time=4.51 seconds)
  • Deflection south (cm) = 17.2454
  • Drop height 100 m.

    How on Earth do I go about double checking what he's telling me? I'll be attempting to share the answer on YT as a reference for handling this supposed "Gotcha!"

    From the CF he provides, I'm estimating an object at the surface would have a 712 g mass (from CalcTools), and a 17 cm drift for a 100 m drop for that kind of mass seems like he's off by a factor of 10 (from examples at different lats).


  • $\begingroup$ Please provide some textual context. What is the statement that you want to refute, and where is it exactly? Your link to a Youtube channel adds no useful info. $\endgroup$
    – Jan Doggen
    Sep 30, 2019 at 8:54
  • $\begingroup$ Excuse me. The channel is mine where I respond to gotchas, but this one's just a step too far for me. The claim is that an object at 45 deg N lat will deviate in a 100 m drop by 17 cm due to the vector addition from centrifugal force. He anticipates that this southwards vector would be measurable on an accelerometer, My belief is that he's fudged the numbers, after seeing an example of a cannonball in freefall at 32 deg N lat deviate by only 1.9 cm. My belief is that a typical accelerometer probably wouldn't detect that. $\endgroup$ Sep 30, 2019 at 8:59
  • $\begingroup$ Still: who is he and what is he claiming? Please take a step back and look at your question with a new look: will an outsider understand what you are asking here? $\endgroup$
    – Jan Doggen
    Sep 30, 2019 at 9:00
  • 1
    $\begingroup$ Not a full answer as I lack time, but: Centrifugal force does have an effect on apparent gravity, both in magnitude and direction. This is because the force due to gravity is towards the centre of the earth, but the centrifugal force is away from the earth's axis of rotation - so unless you're on the equator, the two vectors are not exactly opposite in direction. See earthscience.stackexchange.com/questions/14514/… However, this effect is small, and intuitively the numbers that you give (17cm drift on a 100m drop) sound v unlikely. $\endgroup$ Sep 30, 2019 at 18:56
  • 1
    $\begingroup$ @SemidiurnalSimon - The youtube poster was off by a factor of two. It's a 34 cm drift rather than 17 cm. The youtube poster intentionally chose a suboptimal definition of "up" (or "down") as being a line from the center of the Earth to the surface. Newton showed that the Earth must deviate a bit from being spherical over 300 years ago. That a spherical model is slightly substandard doesn't mean that the Earth must therefore be flat. That's a non sequitur. $\endgroup$ Oct 1, 2019 at 11:56

1 Answer 1


I'm guessing I'm being sold a bridge with his numbers, but can't tell for sure.

You're being sold a bridge with his numbers. If he had done the math correctly he would have obtained a deflection of 33.58 cm at 45° geocentric latitude rather than 17.245 cm. The youtube comments show a number of errors. Two key errors are that the commenter assumes that

  • Local vertical points directly toward the center of the Earth.
  • Gravitational acceleration due to the Earth's mass points in the same direction.

What those youtube comments do show is that the Earth is not quite a sphere. A closer approximation is an oblate spheroid. That said, his comments fall into the wronger than wrong category. Quoting Isaac Asimov,

When people thought the Earth was flat, they were wrong. When people thought the Earth was spherical, they were wrong. But if you think that thinking the Earth is spherical is just as wrong as thinking the Earth is flat, then your view is wronger than both of them put together.

Note that I qualified latitude in the above as geocentric. Geodesists have multiple concepts of latitude. Geocentric latitude is the angle between a line to the center of the Earth and the Earth's equatorial plane. Geodetic latitude is the angle between the normal to the reference ellipsoid and the Earth's equatorial plane. Yet another concept is astronomical latitude, which is the angle between true vertical and the Earth's equatorial plane.

The difference between geocentric and geodetic latitude is at its greatest at 45°. A geocentric latitude of 45° is equivalent to a geodetic latitude of 45° 11' 32.7". Converting that 11' 32.7" difference to radians and multiplying by 100 meters results in 33.58 cm. The difference between geodetic and astronomical latitude is called the deflection of the vertical (or vertical deflection). The deflection of the vertical is rather small, less than 2' (two minutes of arc) even in extremely mountainous terrain. Typical values are a few tens of arc seconds.

What this means is that an oblate spheroid is a better model of the Earth's shape than a sphere. That does not mean a flat Earth model is correct. The Earth's flattening (how squished the Earth is at the poles versus the at the equator) is rather small. The poles are 22 km closer to the center of the Earth than is the equator. Percentage-wise, that's a 0.345% difference.

The reason an oblate spheroid is a better model of the Earth's shape than a sphere is because the Earth is spinning. That spin does two things: It results in a centrifugal acceleration from the perspective of an observer fixed with respect to the rotating Earth, and it results in an equatorial bulge. If the Earth was a uniform density liquid, the spin would make the Earth naturally relax into an oblate spheroidal shape. "Straight down", the combined effects of gravitational and centrifugal acceleration (geodesists call this "gravity" to distinguish it from gravitation) would be exactly normal to the surface.

  • $\begingroup$ TBH, I'm shocked that it's as much as 17 or 34cm over a 100m drop. I expected an answer in the millimeters at most. But I'm prepared to accept that my intuition is wrong :-) $\endgroup$ Oct 1, 2019 at 15:11
  • $\begingroup$ @SemidiurnalSimon - It's a consequence of a bad definition of "down" (or "up"). Down is not toward the center of the Earth. It's whatever direction gravity pulls. The difference between "down" and "toward the center of the Earth" is at its greatest at 45° latitude. Except for a few regions (e.g., parts of the Himalaya), the difference between "down" and the inward normal to the reference ellipsoid is over than an order of magnitude smaller than the difference between geocentric and geodetic latitude. $\endgroup$ Oct 1, 2019 at 15:33
  • $\begingroup$ Yes, I understand the mechanism - I explained it in my own comment (albeit with less technical language, because I am not a geodesist). I'm just surprised that the effect is so large. That's all. $\endgroup$ Oct 1, 2019 at 15:38
  • $\begingroup$ Thanks. Despite the source of the question, it was a great learning opportunity. $\endgroup$ Oct 2, 2019 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.