Let us say a block of mass is placed on the surface of earth. Then while drawing the forces on that body, we say:

  1. Force $F = mg$ acting towards the center of Earth.
  2. Normal reaction $N$ offered by the surface of Earth.

If no other forces are acting, we say $F=N$. But what about the centrifugal force $m\omega^2R$ . Why don't we ever bring that into picture? What am I missing?

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    $\begingroup$ See also physics.stackexchange.com/q/8074/2451 , physics.stackexchange.com/q/299723/2451 $\endgroup$
    – Qmechanic
    May 12, 2011 at 5:24
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    $\begingroup$ ""Why don't we ever bring that into picture? What am I missing?"" You missed to read about the many cases where "that was brought into picture". $\endgroup$
    – Georg
    May 12, 2011 at 8:51
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    $\begingroup$ @Georg : Thanks for pointing that out. Now, I do want to read about those cases. Could you point me to some of those cases/examples/problems? $\endgroup$
    – claws
    May 12, 2011 at 9:03
  • $\begingroup$ Read any medium physics textbook on "gravity dependence on latitude". Any clockmaker knows why a pendulum at equator has to be somewhat shorter than at say, 50th parallel. $\endgroup$
    – Georg
    May 12, 2011 at 9:15

2 Answers 2


Because it's effect is smaller than the variation in $g$ due to earth's bulge (caused by the same centrifugal force) or the local geology - when you use $9.8m/s^2$ that's just an approximation.

The effect of the bulge and centrifugal force mean that $g$ at the equator is about 0.5% lower than $g$ at the poles

edit: velocity at equator $40,000 km / 24 h = 1666.7 km/h = 0.463 km/s$

'centrifugal g' = $(0.463 km/s)^2 )/ 6375 km = 0.03 m/s^2$ or 0.3% of 'g'

  • $\begingroup$ How is it smaller? $\omega$ is angular velocity of earth which is large and R is radius of earth which is also large. $\endgroup$
    – claws
    May 12, 2011 at 5:09
  • $\begingroup$ The magnitude of the force is smaller than the variation in 'g' - the centrifugal force is about 0.1 - 0.2% of 'g' $\endgroup$ May 12, 2011 at 5:12
  • $\begingroup$ Interestingly doing the numbers it's larger than I guessed $\endgroup$ May 12, 2011 at 5:18
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    $\begingroup$ ω is small - it's 2pi/(24*60*60) rad/s. It might help to think of it in F=mv^2/R terms, remember R is big $\endgroup$ May 12, 2011 at 5:26
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    $\begingroup$ @Floris - if you know a gold seller that will accept my calibration weight I can make money without 'g' ;-) $\endgroup$ Sep 1, 2017 at 20:52

I think that all the right physics is contained in Martin Beckett's answer and the comments on it, but I'd like to restate it in a way that may bring out what I think the key point is.

In practice, when we do experiments in a lab near Earth's surface, we use a value of $g$ that's been determined empirically at that location. For instance, we might determine it by dropping something in vacuum and measuring its acceleration with respect to our lab. That value of $g$ already includes the centrifugal contribution, so we don't need to (indeed we must not!) include it separately.

We often tell introductory physics students that $g$ is the "acceleration due to gravity," but strictly speaking we're telling a small lie when we do this: $g$ is really the acceleration due to gravity and inertial forces.

Of course, that lie is only a lie in the context of Newtonian mechanics: when we get to general relativity the distinction between gravity and inertial forces goes away anyway! The acceleration of a falling object in general relativity is most naturally thought of as being all inertial force: the falling object is moving along a geodesic, and the reason we see it as accelerating is that our lab is not in an inertial reference frame.


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