# What force balances centripetal force of Earth in inertial frame of reference?

I understand that in a non-inertial frame of reference rotating with the Earth, an object at rest has a weight that is equal to mg. This mg is the vector addition of the centrifugal force and the true gravitational vector. Thus mg is simply the addition of the vertical components of the true gravitational vector and the vertical component of the centrifugal force as the horizontal components cancel out (object at rest).

When I try to view this problem in an inertial frame of reference, however, I run into trouble. I am trying to balance the centripetal force with the true gravitational vector but it doesn't lead to the mg that I saw in the non-inertial frame of reference.

What is the best way to picture this scenario to come up with the same answer just different reference frames?

Thank you very much

• What is the "true gravitational vector"? An inertial frame of reference is simply one in free fall. In that system the floor will approach with an acceleration of +g and it will rotate with the Earth's rotation. Commented Sep 10, 2015 at 23:22
• In the non-rotating frame, the ground is accelerating away from you. So you need an additional force to keep up with it. Commented Sep 11, 2015 at 0:35

I understand that in a non-inertial frame of reference rotating with the Earth, an object at rest has a weight that is equal to mg. This mg is the vector addition of the centrifugal force and the true gravitational vector.

This part is correct. This is how g is defined.

Thus mg is simply the addition of the vertical components of the true gravitational vector and the vertical component of the centrifugal force as the horizontal components cancel out (object at rest).

This part is incorrect. The centrifugal force is a very small part of g. It is greatest at the equator, and even there, it is only about 0.0035 g. The centrifugal force does not cancel out the gravitational force.

You are ignoring that the ground is pushing up on you. In a frame rotating with the Earth, the correct expression for a person standing at rest on the surface of the Earth is $m\vec g + \vec N = 0$.

What about an inertial frame? The only forces acting on that person from the perspective of the inertial frame are the true forces, Newtonian gravitation and the normal force. Given that $m\vec g$ is Newtonian gravitation plus the centrifugal force, it's easy to calculate the Newtonian gravitational force acting on a person. The centrifugal force acting on the person is $-m\, \vec\omega \times (\vec\omega \times \vec r)$. The Newtonian gravitational force is thus $m\left(\vec g + \vec\omega \times (\vec\omega \times \vec r)\right)$. The net true force acting on the person is thus $$m\left(\vec g + \vec\omega \times (\vec\omega \times \vec r)\right) + \vec N = m\left(\vec g + \vec\omega \times (\vec\omega \times \vec r) - g\right) = m\,\vec\omega \times (\vec\omega \times \vec r)$$

This is not zero. It is in fact uniform circular motion, with the force always pointing at the Earth's rotation axis, and a rotation rate of one rotation per sidereal day. This is the behavior a person standing still on the surface of the rotating undergoes from the perspective of an inertial frame.

• This is very, very old, so I apologize for resurrecting it. I'm curious about your third paragraph, where you discuss "Newtonian gravitation and the normal force" and then immediately bring up the centrifugal force. Why does the centrifugal force show up at all in the inertial frame? "Given that mg is Newtonian gravitation plus the centrifugal force..." My understanding was that the centrifugal force doesn't show up in inertial reference frames, so what gives? Commented Jun 28 at 16:23
• @johnmorrison Geoscientists tend to use an Earth-fixed frame rather than an inertial frame. In particular, $g$ is always expressed in an Earth-fixed frame, which means that $g$ incorporates both gravitation acceleration and centrifugal acceleration. Commented Jun 30 at 8:49