3 comments on “Centrifugal Force at the Earth’s Surface

  1. Well, I did some of the numbers, anyway. I took your .026 m/s/s as a given. The calculations I made show not the derivation of the centrifugal acceleration, but rather how that centrifugal acceleration combines with the acceleration due to gravity to create a net effective acceleration.

  2. Hi Justin,

    You have a deluxe web site. It is truly commendable and remarkable.

    I found your site, because I was wondering today about the effects of centrifugal force on perceived weight. I didn’t do the numbers, but I set up the problem much as you did. However, in my mental model I included something which it seems you neglected. Let me explain:

    Acceleration is a vector. It has magnitude and direction. Vector arithmetic is not the same as normal (scalar) arithmetic. I think to do the problem correctly you need to do vector arithmetic to combine the two accelerations, as they do not act in the same direction, rather they are separated by an angle which I believe is the same as the latitude of your location–39 degrees in this case.

    Adding these two vectors boils down, I think, to finding the third side of a triangle which as two sides of .026 and 9.8, and an angle of (180-39) or 141 degrees. Do you remember the law of cosines? Me neither, but I looked it up (https://www.omnicalculator.com/math/triangle-side). I’ll let you check my math, but I got a vector sum of 9.777 m/s/s. subtracting that from 9.8 gives .023 m/s/s which is still a small fraction of 9.8, but it is significantly less than .026 m/s/s which was your answer for the change in net effective acceleration.

    I hope I didn’t make a math error. I think I have the right method however. Let me know what you think.

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