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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?

Qmechanic
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claws
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    See also https://physics.stackexchange.com/q/8074/2451 , https://physics.stackexchange.com/q/299723/2451 – Qmechanic May 12 '11 at 05:24
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    ""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 08:51
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    @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 09:03
  • 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 09:15

2 Answers2

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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'

Piper McCorkle
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Martin Beckett
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  • 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 05:09
  • 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 05:12
  • Interestingly doing the numbers it's larger than I guessed – Martin Beckett May 12 '11 at 05:18
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    ω is small - it's 2pi/(246060) 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 05:26
  • 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 http://physics.stackexchange.com/q/8074/ – Luboš Motl May 12 '11 at 07:41
  • @Lubos - yes, I worked it out. But it's still small enough that if you are using 9.8 for'g' you don't need to care – Martin Beckett May 12 '11 at 12:32
  • Yes, sure, unless you of course buy gold for a million of dollars and using a different weight, they steal $3000 from you. ;-) – Luboš Motl May 12 '11 at 12:49
  • @Luboš - or my new diet that just moves people form Oslo to Mexico city – Martin Beckett May 12 '11 at 19:46
  • If you buy gold, you bring a calibration weight with you and use a balance. That removes the effect of $g$ from your purchase... – Floris Sep 01 '17 at 20:04
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    @Floris - if you know a gold seller that will accept my calibration weight I can make money without 'g' ;-) – Martin Beckett Sep 01 '17 at 20:52
  • @MartinBeckett presumably if you want to buy enough gold that it matters, you would want to confirm the calibration of the supplier's scale - and if your measurements disagree, walk away. See also this – Floris Sep 02 '17 at 01:00
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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.

Ted Bunn
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