Gravitational Force and weight at the equator

[MiniJo]

Homework Statement

Suppose the Earth is a perfect sphere with R=6370 km. If a person weighs exactly 600.0N at the north pole, how much will the person weigh at the equator? (Hint: the upward push of the scale on the person is what the scale will read and is what we are calling the weight in this case.)

Homework Equations

Fg = Gm1m2 / r^2
Fg = mg

The Attempt at a Solution

This was my belief: that since in the question the Earth is presumed to be a perfect sphere, it means the radius will be constant at all points of the surface. Therefore, shouldn't the weight stay the same? But the answer to the question is 597.9 N, not 600.0 N and I just don't understand how.

I have a feeling it might have something to do with the normal force, but I'm not sure.

 

[Office_Shredder]

Are you supposed to take into account the rotation of the Earth around its north south pole axis?

 

[MiniJo]

I doubt that we have to take that into consideration. Our teacher never touched upon that yet.

 

[nrqed]

I doubt that we have to take that into consideration. Our teacher never touched upon that yet.

I am sure that you must have to take into account the rotation of the Earth, otherwise the answer would be trivially the same weight. Have you covered uniform circular rotation, in which case the net force is [itex] F_c = \frac{m v^2}{ r} [/itex] ?

Then just apply this and use the fact that the net force is equal to a normal force upward minus mg . Solve for the normal force.

Patrick