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Assuming the radius of Earth as $6.36\cdot10^6\:\mathrm m$. Then i get for the centrifugal force of an object with the mass of $m=10,000\:\mathrm{kg}$:

$$r\cdot\omega^2 \cdot m = 6.36\:\mathrm m \cdot \left(\frac{1}{24\cdot3600\:\mathrm s} \right)^2\cdot 10,000\:\mathrm{kg} = 8.52\:\mathrm{N}$$

So when I want to know how big the gravitational force is on a mass $M$, which is the correct attempt?

  1. $F=m \cdot a = M \cdot g$ but here the centrifugal force is ignored. So the actual resulting force should be $M \cdot g - F_\mathrm{centrifugal}$.

or

  1. Is the value of $g$ actually given with the centrifugal force in mind? Would that mean that the gravitational force would actually be higher on a mass $M$ when the earth stops rotating?
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  • $\begingroup$ Read @Ben51 answer to correct your value of Earth rotationnal speed 😊 . $\endgroup$ – dan May 8 at 7:26
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You are correct that one would have to include the centrifugal force into the calculation.

However, note that $r\omega^2 = 8.54 \cdot 10^{-4}\frac{\mathrm{m}}{\mathrm{s}^2}$ is smaller than the lowest digit given on the standard approximate value of $g = 9.81 \frac{\mathrm{m}}{\mathrm{s}^2}$, so asking whether or not the centrifugal force is included in this is meaningless - the value would be the same either way at this level of precision.

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The centrifugal force does have an effect on the weight of objects on the Earth. It is about half the reason things are lighter at the equator than the poles, the other being that they are further from the center of the Earth, owing to its oblate shape (itself a result of the centrifugal force). The reason you got such a low estimate of the influence of the centrifugal force is a miscalculation of $\omega$. It is not 1/day but 2$\pi$/day. So the force, using your numbers, is 336 N rather than 8. If the value of g without the centrifugal force were 9.81 $m/s^2$, this would reduce it to about 9.78.

In practice, yes, the value of g used always includes not just a purely gravitational force, but also the centrifugal contribution. This is more convenient than considering the forces separately, and is possible because the centrifugal contribution is always the same at any given place (in contrast, for example, to the Coriolis force, which depends on the motion of the the object experiencing it). So, for example, at locations away from the equator and the poles, the direction of "gravity" (i.e. the direction a plumb bob points) is not toward the center of the Earth, but rather a bit toward the opposite pole from there, always perpendicular to the local surface of the geoid. The true gravitational force on a mass m points toward the center of the earth (or very nearly), the centrifugal force points outward from the Earth's axis, and mg is the sum of the two.

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    $\begingroup$ Hi Ben! Please note that this site supports Mathjax. It shouldn't take you too long to learn how to use it, and it will make your answers look much more professional and readable. Cheers! $\endgroup$ – AccidentalFourierTransform Jan 4 '18 at 17:21

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