# Is the normal force smaller than gravitational force due to earth's rotation? [duplicate]

In physics class we learn that the normal or contact force exactly equals the force of earth's gravity pulling you to the center of the Earth and thus cancel each other out. I thought that the normal force does not exactly equal the force of gravity on an object.

The earth is rotating around its axis. That means that, for example, a book lying on a table has no constant velocity since its direction of motion is constantly changing with the circular motion around the Earth's axis. The centripetal force due to this rotation, for a book with mass of 1 kg is given by: $F=\frac{mv^2}{r}$. The speed of the rotation is 465.1 m/s and the radius 6,378,000 m. So this should give a force of around 0.034 N. This means there is a net force.

I thought this would mean that the normal force does not exactly cancel out the gravitational force. Is this true? For example does the normal force on a book, with a gravitational force (on the equator) of $9.78 m/s^2$ and thus a force of $9.78 N$ equal $9.78-0.034=9.746 N$ (or is it that if one talks about the gravitational acceleration, this effect has been taken in account? That is to say, when we talk about gravitational acceleration do we mean an experimental number, so that this effect has been taken in account, or the theoretical acceleration without the effect)? Could this effect be compared with the effect that you are being pushed to the side when a car makes a curve and that the book is like this ´tilt´ a bit upwards?

Yeah, you are right in assuming that the net force of gravity is not g at all points on earth. Due to rotation, the force is less at the equator than at the poles by a factor of the centripetal acceleration outwards $\times cos\theta$. Moreover, the earth is not exactly a sphere but rather an ellipsoid. So, the value of g also varies from place to place due to the distance from the center.