ANS: Forces and Angular Motion at the North Pole and Equator

[SS2006]

Suppose the Earth is a perfect sphere with R = 6370 KM. If a person weighs exactly 600 N at the north pole, how much will the person weigh at the equator.? (Hint: The upward push of hte scale on the person is what the scale will read and is what we aer calling the weight in this case)
ANS: 579.9 N

this is on a angular motin in a plane work sheet, i got all the ones above but here I am blanked
i can't use energy equations, nor mv2/r, or can i? i don't know how ot attempt this.
or is it f = g m1 m2 / r2?
i just need a kickstart

 

[HallsofIvy]

No, you don't use energy equations, this problem has nothing to do with energy- only force. You don't NEED to use Gm₁m₂/r² because you are already told that the force due to gravity is 600 N. The difference between the north pole and the equator is the centrifugal "force" at the equator so you CAN use mv²/r.
Subtract the force necessary to hold the person in circular motion at the equator from the 600 N. That will be the reading on the scale.

 

[SS2006]

i need more ideas
dont know how to start
are you saing
mv2/r = 600 - t?
what do i use for v2
and m

 

[daniel_i_l]

The speed (v2) would be the velocity at the equator that would be the angular spin of the Earth around its axis times the radius. The mass would be the mass of the person. You can determine this from the 600N. Then just subtract the mv2/r from 600.

 

[mezarashi]

These are the equations from the force diagram you should be contemplating for the guy at the North Pole:

$$F_{grav} - F_{normal} = ma = 0$$

For the equator:

$$F_{grav} - F_{normal} = ma = m\frac{v^2}{r}$$

Fnormal is the reading on your scale. Does it make sense?