Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
See also – Qmechanic May 12 '11 at 5:24
""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". – Georg May 12 '11 at 8:51
@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? – claws May 12 '11 at 9:03
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. – Georg May 12 '11 at 9:15
up vote 8 down vote accepted

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'

share|cite|improve this answer
How is it smaller? $\omega$ is angular velocity of earth which is large and R is radius of earth which is also large. – claws May 12 '11 at 5:09
The magnitude of the force is smaller than the variation in 'g' - the centrifugal force is about 0.1 - 0.2% of 'g' – Martin Beckett May 12 '11 at 5:12
Interestingly doing the numbers it's larger than I guessed – Martin Beckett May 12 '11 at 5:18
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 – Martin Beckett May 12 '11 at 5:26
Dear @Martin, you're wrong that the variation of $g$ due to the centrifugal force is smaller than the variation of $g$ because of the bulge, which is also caused by the same centrifugal force. Up to a factor of at most 2, they're the same. See – Luboš Motl May 12 '11 at 7:41

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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