Gravity at different locations on earth (poles and equator)

[jb646]

Homework Statement

Determine how much greater the gravitational field strength, g, is at the pole than at the equator. Assume a spherical Earth. If the actual measured difference is Δg = 52 mm/s2, explain the difference.

Homework Equations

g=G*m1*m2/r^2

The Attempt at a Solution

The problem I'm having is understanding the problem. If we assume a spherical earth, then the radius would be the same everywhere and in turn so would g. I have looked up the actual gravitational difference and it is roughly 50 mm/s^2, so am I supposed to just say that it would be no greater and the difference of 50 mm can be attributed to the fact that the Earth is indeed not spherical, or is there a concept I am not understanding...

Thanks for any help

 

[fzero]

I agree that the problem is worded incorrectly. The only way to make sense of the difference is to assume that the distance from the center of mass is different at the pole and equator.

 

[D H]

The Earth is rotating.

 

[jb646]

So how does that affect my equations knowing that the Earth is rotating... I'm really lost on what to do here

 

[jb646]

I solved it using the equation for centripetal acceleration a=v^2/r where v=2Pi*r/T
and got an acceleration of roughly 34 mm/s^2. he states that 52 is the actual amount so am i correct in assuming that I have done everything correct up to this point and the difference is due to the fact that the Earth is actually not spherical?

 

[fzero]

Because of the Earth's rotation, mass elements away from the poles have an angular momentum that increases as you approach the equator. Because of this angular momentum, they have more energy than the elements at the pole and they "orbit" the center of mass at a larger radius. The net effect is that the Earth is oblate rather than perfectly spherical and bulges at the equator.

It's an explanation of why the radius at the equator is larger, it's not directly a part of your problem.

 

[D H]

I solved it using the equation for centripetal acceleration a=v^2/r where v=2Pi*r/T
and got an acceleration of roughly 34 mm/s^2. he states that 52 is the actual amount so am i correct in assuming that I have done everything correct up to this point and the difference is due to the fact that the Earth is actually not spherical?

Yes. That is correct, including your value of 34 mm/s².

This term is in a sense better attributed to centrifugal acceleration. We Earth-bound folk are non-inertial observers. The Earth is a rotating frame of reference, so fictitious accelerations such as centrifugal acceleration and the Coriolis acceleration rear their ugly heads.

 

[jb646]

Thanks guys! now here's a tricky one, how do I mark this thread as solved?