Problem of the week - gravitation

[yoyigloria]

I received this problem this week and no one in my class has been able to solve it. We don't need to solve it exactly, but we should provide a conceptual answer. I would really appreciate it if anyone could help me out.
-The apparent weight of a person may vary in different latitudes due to the rotation of the earth. If the apparent weight is the force that the scale makes on the person, determine the variation in the apparent weight of a person at the equator and at one of the poles.

THANKS!

 

[rock.freak667]

In general the force that the scale reads is the gravitational force between the Earth and the person.

So use the formula for gravitational force and the radius of the Earth at the equator and the poles. That should give you the differing weights.

 

[Liquidxlax]

In general the force that the scale reads is the gravitational force between the Earth and the person.

So use the formula for gravitational force and the radius of the Earth at the equator and the poles. That should give you the differing weights.

9.780 m·s−2 at the equator to about 9.832 m·s−2

g_h = g_o((r_e)/(h+r_e))^2

gravity at a height = gravity standard (radius earth/(height + radius earth))^2

 

[yoyigloria]

I forgot to say that we are to regard the Earth as perfect sphere.

 

[rock.freak667]

9.780 m·s−2 at the equator to about 9.832 m·s−2

g_h = g_o((r_e)/(h+r_e))^2

gravity at a height = gravity standard (radius earth/(height + radius earth))^2

I think that is all that is required.

I forgot to say that we are to regard the Earth as perfect sphere.

If you regard it as a perfect sphere, then the distance from the center to any point on the surface is the same as the radius, meaning that the weight would not vary.

 

[yoyigloria]

Thanks to everyone for their replies.
i just found the answer. It has to do with the rotational axis of the Earth and the centripetal acceleration that is caused by a centripetal force. Apparently, it is zero at the poles and most pronounced at the equator causing the apparent weight and measured gravitational acceleration to be 0.35% smaller at the equator than at the pole.