# Does the gravity we feel equal G(M/r^2) - our centripetal acceleration

1. Apr 15, 2013

### D.Hayward

We all know that something travelling in cicular motion at a constant speed has a centripetal acceleration towards the axis of rotation. I suppose this means we (travelling around the axis of rotation of the earth) also require a centripetal acceleration. This can be provided by gravity which is calculated by a=G(M/r^2) to be 9.81 m/s^2. However a portion of this has to be 'used' just to stop us drifting into space, this should be equal to our centripetal acceleration. Therefore shouldnt we feel 9.81 m/s^2 minus our centripetal acceleration in everyday life and in physics practicals. Is this true?

2. Apr 15, 2013

### cepheid

Staff Emeritus
Welcome to PF,

Yes. Basically. You would not "weigh" as much at the equator as you would at the poles. However, you have to be very precise about what the word "weigh" means here. More precisely, the Earth would not have to push up on you with as much normal force in order to support you, and so you would "feel" less heavy (although I'm don't think the difference is discernible). Some people define "weight" as this, in which case yes, you would "weigh" less by that definition. Others define your weight as the force with which Earth's gravity pulls on you, which would not be different between the two locations.* Under this definition, your weight is the same, even if the surface of the Earth doesn't have to support as much of it.

*for a perfectly spherical Earth, that is. In reality, earth is not perfectly spherical and g varies with location on the Earth's surface.

http://curious.astro.cornell.edu/question.php?number=310 [Broken]

Last edited by a moderator: May 6, 2017