Influence of Centrifugal force in 'g'

[Guiga]

Influence of "Centrifugal" force in 'g'

Hello everyone,

I was wondering one of these days if the 'centrifugal' force due to the rotation of the Earth in the Equator could cause any difference in the value of the gravitational acceleration; namely it would be supposedly less than the real value because we experience it pushing us outward, in opposite direction to gravity.

I know that if it has an implication it is probably very tiny compared to the gravitational field of the Earth but I just want to know if the logic is correct.

ps. I am aware that the centrifugal force doesn't exist but we experience it; the simple rotating bucket proves it.

Thanks in advance!

 

[zhermes]

You're absolutely right. It is a simple calculation to perform as-well: assume you're on the equator, find the centrifugal force using the radius of the Earth and the rotational velocity (e.g. using the length of a day).

 

[netheril96]

Yes.
And "g" values already include the contribution from centrifugal forces.

 

[Ken G]

What's more, the centrifugal force contributes to altering the local effective g in two ways-- one way is that if you interpret yourself as being in an inertial frame (when you aren't), you need to include the "centrifugal acceleration" directly into a reduction of the locally measured g, as mentioned above. But also, this modification alters the pressure balance inside the Earth and causes the equator to bulge out a little. That increases the distance to the center, so it actually reduces the purely gravitational contribution to g (though only a little bit, less than a percent). However, it's enough to affect things like track records, when they are measured to high precision.