# Does the normal force really equal mg on earth? I think not...

## Main Question or Discussion Point

Dear community, I am very ashamed to have not noticed this result but well into my physics major, and I would like to share it with you. I am not sure it is right, which is my reason for posting it.

The normal force is classically referred as
N = mg (1)
but if that was really the case; it is, if the Normal force completely canceled the gravitational force; then there would be no centripetal acceleration which were to keep a body on Earth's surface accelerating downwards to stay in contact with the planet.

Indeed, I think it is more like
N = k(mg) (2)
where k is the fraction of the gravitational force which does not contribute to the centripetal acceleration, which then is canceled by the Normal force.

Thus this centripetal acceleration is

ac = mv2/R (3)
with R being the radius of Earth, i.e. 6.37 × 106 m, and v the tangential velocity of a body on the surface being 469 m/s,

ac = 0.035 m/s2 (4)

then if we have
mg = N + Fc
= k(mg) + mac

then
k = (g - ac)/g = 0.997 (5)

First, this seems to be completely negligible, but it is a reason not to solve N = mg with more than 3 decimals of precision if one does not account for the centripetal acceleration of the studied body.

Second, I think it is a mistake not to introduce the normal force as N=mg-mac, mainly because of its theoretical significance. I, as an undergraduate, wasn't aware of this and I think it deserves minimal exposition at least in basic physics books. It is nice to have N = mg, but the normal force DOES NOT equal the gravitational force. It does not cancel it.

Am I wrong on this? I looked up the wikipedia article of normal force on earth and it does not include the centripetal acceleration, which I think is wrong and misleading.

This is my first 'not very short' post; I am sorry for any errors! Please leave your comments.

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Orodruin
Staff Emeritus
Homework Helper
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First, this seems to be completely negligible, but it is a reason not to solve N = mg with more than 3 decimals of precision if one does not account for the centripetal acceleration of the studied body.
If you would do this you would first need to know g to that precision. Typically, any centripetal acceleration would be accounted for at that stage - it is just a matter of going to a non-inertial frame. Note that g is not the same all over the Earth, it also varies due to other effects.

• IntegralBeing
scottdave
Homework Helper
Yes, "centrifugal force" (the tendency to fly away from the center of rotation) does affect how fast an object accelerates downward. I remember learning that the value of g is specified, for different places on the Earth. There is more than just the centripetal force, which varies as you move around. The density of the Earth is hardly uniform. Different locations are at various altitudes (distance from center).

I think it is easier for scientists to experimentally determine the value of g for a location and publish it, then having to separately account for centripetal force.

• IntegralBeing
If you would do this you would first need to know g to that precision. Typically, any centripetal acceleration would be accounted for at that stage.
Imagine you measure some g accounting also for any centripetal acceleration. If you were to calculate the acceleration of free fall, this would then be the g which you measured, Being the free fall force just mg. But this means the g you measured is not the approximation of G*Mearth/Rearth2 but the approximation of G*Mearth/Rearth2 - vearth rotation2/Rearth.

In this way, then we cannot really say that mg is the gravitational force but just the 'free fall' force, can we?

I am really confused by this. It isn't as trivial as I thought

Orodruin
Staff Emeritus
Homework Helper
Gold Member
But this means the g you measured is not the approximation of G*Mearth/Rearth2 but the approximation of G*Mearth/Rearth2 - vearth rotation2/Rearth.
Clearly ##GM_\oplus/R_\oplus^2## is an approximation. It assumes that the Earth is a spherically symmetric non-rotating object - it isn't. It is a good approximation for most applications.

In this way, then we cannot really say that mg is the gravitational force but just the 'free fall' force, can we?

I am really confused by this. It isn't as trivial as I thought
What ##g## is depends on your reference frame. Gravitation is an inertial force and going to an accelerated frame is going to change the inertial force. In particular, if you go to a frame in free fall - there is no inertial force at all. There is nothing strange about that. When we talk about "g" it usually means something like "the inertial force acceleration in a frame at rest relative to the Earth at some place of the Earth's surface". Its value depends on the place chosen.

• IntegralBeing
sophiecentaur
Gold Member
then if we have
mg = N + Fc
= k(mg) + mac

then
k = (g - ac)/g = 0.997 (5)
You have assumed that the two forces act along the same direction but that's only the case at the equator. The forces are VECTORS. There is No centripetal acceleration at the poles and the direction is in the plane of the circle of latitude, anywhere else.
As other people have mentioned, there are many other potential factors involved so where do you stop.? He who rides a tiger cannot dismount.

• IntegralBeing
Orodruin
Staff Emeritus
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As other people have mentioned, there are many other potential factors involved so where do you stop.?
The answer to this should be pretty evident. You stop when you have reached your desired precision. For most applications it is not necessary to know g to sub-% precision.

sophiecentaur
Gold Member
The answer to this should be pretty evident. You stop when you have reached your desired precision. For most applications it is not necessary to know g to sub-% precision.
It can b e relevant when buying and selling valuable substances like gold but only just. There are apocryphal tales of crooked dealers buying gold at the equator and selling it at high latitudes but I do wonder. (Plus, a Balance works correctly everywhere and that's what's used in the trade)

Orodruin
Staff Emeritus
Homework Helper
Gold Member
There are apocryphal tales of crooked dealers buying gold at the equator and selling it at high latitudes but I do wonder. (Plus, a Balance works correctly everywhere and that's what's used in the trade)
The obvious counter measure to this is to not use a spring-based displacement scale but a balance or spring-based period scale. If I was a crooked dealer I would find it much simpler to tamper with the scale calibration/counter weights so it is likely hearsay.

• sophiecentaur
IntegralBeing
• Any centripetal force you experience is due to the spin of the earth.
• The earth spins about the polar axis.
• The rotation at the poles is zero and this increases to a maximum as you move to the equator.
Now imagine your situation at the equator. The forces acting on you are your own weight (W) and the reaction (N), but these are not equal in size because you are accelerating towards the earths centre. N is smaller than W and by an amount which provides the centripetal force (C) which is needed to make you accelerate. You can write:

W-N=C

The difference is very small and you can look it up. As you move to the poles the centripetal force required gets smaller and N gets closer to W. At the poles W=N.

• IntegralBeing
jtbell
Mentor
I, as an undergraduate, wasn't aware of this
When I was an undergraduate about 45 years ago, it was at least mentioned in our introductory physics course. I'm pretty sure we did exercises involving "apparent gravity" versus "actual gravity" at the equator at least. Also, I remember using a simplified formula to calculate a "standard" value for (apparent) g at our location for laboratory exercises, taking latitude into account. It may have been one of the formulas mentioned in the Wikipedia article:

https://en.wikipedia.org/wiki/Gravity_of_Earth

Lord Jestocost
.....and I think it deserves minimal exposition at least in basic physics books.....
I think that the topic is addressed in the textbook "Physics, Parts I and II" by D. Halliday and R. Resnick (John Wiley & Sons, Inc.).

Nugatory
Mentor
When I was an undergraduate about 45 years ago, it was at least mentioned in our introductory physics course. I'm pretty sure we did exercises involving "apparent gravity" versus "actual gravity" at the equator at least.
I had the same experience at about the same general time.... and also much emphasis on developing the ability to easily and quickly decide which effects will be significant and which can be ignored in any particular problem.

I will add my voice. Yes this happens. Yes you weigh 0.05% less at the equator than at the poles, and yes in my experience this is pointed out in freshman physics. It is also pointed out that the earth bulges at the equator for this reason, and to add an even less significant effect, that means you are further from the center of the earth when at sea level at the equator than at the poles and so weigh even a tiny bit less than less.

Looks to me like at the poles and along the equator the orientation of G is perpendicularly normal to a tangent plane at the Earth's surface, but at latitudes* between the poles and equator these orientations' distal radial projections increasingly approach being parallel to the equatorial plain. A tall enough structure** would either have to curve with altitude to maintain alignment (longitudinal compression through a curved line) or remain straight (sustaining differential orientation of "normal" with altitude as bending stress, requiring increasingly slanted upper floors to maintain local perpendicular normal, etc.)...

* - wonder if the deviation from geometric normal occurs at a particular N/S latitudes?
**- wonder how high before these things need accounting in the architectural design?