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## 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

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

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

with R being the radius of Earth, i.e. 6.37 × 10

then if we have

then

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-ma

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.

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

**a**(3)_{c}= mv^{2}/Rwith R being the radius of Earth, i.e. 6.37 × 10

^{6}m, and v the tangential velocity of a body on the surface being 469 m/s,**a**(4)_{c}= 0.035 m/s^{2}then if we have

**mg = N + F**

= k(mg) + ma_{c}= k(mg) + ma

_{c}then

**k = (g - a**(5)_{c})/g = 0.997First, 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-ma

_{c}, 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.