# Earth’s Rotation and Centripetal Acceleration

*Most recent answer: 10/22/2007*

Q:

How does the rotation of the earth affect the weight of a stationary object at sea level?
Does an object of the same mass have a different weight at the poles than at the equator due to the centripetal force? Indeed, would we all weigh more if the earth were not rotating?

- Keith (age 47)

UK

- Keith (age 47)

UK

A:

Yup, the earth’s rotation makes the weight of objects a little less at
the equator. Gravity pulls down, but the object needs to accelerate in
the downwards direction in order to stay in a circular path around the
Earth’s rotational axis in order to stay on the Earth’s surface as it
turns. The centripetal acceleration is about 3.39 cm/sec^2 at the
equator (I’m getting this number from the CRC Handbook of Chemistry and
Physics), which is about 0.35% the acceleration of gravity at the
surface of the earth, g. There is an additional lightening factor, in
that the Earth bulges a little bit outwards at the equator because of
its rotation, making objects on the surface just a tad farther away
from the center, also making them lighter.

Yup, we’d all weigh just a little more if the Earth were not rotating.

Tom J.

Yup, we’d all weigh just a little more if the Earth were not rotating.

Tom J.

*(published on 10/22/2007)*

## Follow-Up #1: effect of Earth's spin on weights

Q:

Can you tell me how to get the component of the centrifugal acceleration due to the Earth's rotation that goes in the opposite direction to gravity? At the equator the outward centrifugal acceleration is simply subtracted from the inward gravitational acceleration, because they go in exactly opposite directions, so itï¿½s
(GM/r^2) - (Ï‰^2r[cos Ï†]) .
But further North they differ in direction more and more. The centrifugal force goes in a direction parallel to the plane of the equator, I want to work out the component of that which goes directly against gravity. It needs to be in the form of an acceleration, so it can be subtracted from GM/r^2. It seems to me that cos (Ï†), the cosine of the latitude should be multiplied in again, to get that component. Is that right? Thank you very much.

- David Martin (age 41)

Arizona

- David Martin (age 41)

Arizona

A:

You're right. The net effective downward force in the spinning frame on mass M is M*(g-ω^{2}*r*cos^{2}(φ)). In addition, there's a slight effective force toward the equator of M*ω^{2}*r*cos(φ)sin(φ). You can get this by taking that centrifugal force and breaking it into an upward and a horizontal component, with magnitudes proportional to cos(φ) and sin(φ) respectively.

Mike W.

*(published on 04/13/2014)*