How Fast Must Earth Spin for Equatorial Weight to Be One Fifth of Polar Weight?

[Shauta]

Homework Statement

What must be the period of rotation of the Earth on its axis so that a person at the equator will have a reading on the scale that is approximately one fifth as much as he would at the North Pole?

(Given on the formula sheet)
Radius of Earth = 6.37 x 10^6 m

Homework Equations

Fc = m*ac
ac = (4*(pi^2)*r)/(T^2)
Fg = m*g

The Attempt at a Solution

I had tried making Force of gravity (Fg) equal to Centripetal Force (Fc), then solving for T. Though this gives me an answer that is close to the one that my teacher gave me, the process doesn't feel right. I then tried repeating the process, but first multiplying Fg by 1/5 (as stated in the question), however, this also doesn't give me the right answer.

 

[Simon Bridge]

Modelling the Earth as a uniform sphere and neglecting air pressure ;)

If the force of gravity is equal to the centripetal force, then you are orbiting the Earth ... how hard would you be pressing on the ground?

Start by drawing a free body diagram for the object being weighed.

 

[Shauta]

I think I understand what you are trying to say. If I'm orbiting the Earth, then I'm not exactly pressing on the ground right? But, now I'm confused, because wouldn't that make Fc equal zero then?

 

[Simon Bridge]

The centripetal force needed to keep you on a circular path is not the same thing as your weight, which, in turn, is not the same thing as how hard you press into the ground.

Do not confuse the weight on a set of scales with the force of gravity.
The second is the technical meaning of the word weight in physics and the first is the casual everyday meaning of the word weight.

In your free body diagram - you have a person standing on the scales on the surface of the Earth at the equator.
What are the physical forces acting on the person? Which direction do they act?

 

[HalfLostAndSearching]

I would approach this question by first understanding the concept of circular motion and how it relates to the Earth's rotation. The period of rotation of the Earth on its axis is known as the sidereal day, which is approximately 23 hours and 56 minutes. This means that it takes the Earth approximately 23 hours and 56 minutes to complete one full rotation on its axis.

To address the main question, we need to consider the centripetal force acting on an object at the equator and at the North Pole. At the equator, the centripetal force is equal to the force of gravity, which is given by the formula Fc = Fg = m*g, where m is the mass of the object and g is the acceleration due to gravity. At the North Pole, the centripetal force is equal to the force of gravity multiplied by the cosine of the latitude, given by Fc = Fg*cos(latitude) = m*g*cos(latitude).

To find the period of rotation that would result in a reading on the scale at the equator that is approximately one fifth of the reading at the North Pole, we can set up the following equation:

1/5 * m*g = m*g*cos(latitude)

Simplifying and solving for the period T, we get T = √(5/4) * √(r/g), where r is the radius of the Earth and g is the acceleration due to gravity. Substituting the given values, we get T = √(5/4) * 8.64 hours, which is approximately 10.4 hours.

Therefore, the period of rotation of the Earth on its axis must be approximately 10.4 hours for a person at the equator to have a reading on the scale that is one fifth of the reading at the North Pole. This can also be verified by dividing the sidereal day by 5, which gives us approximately 4.8 hours, and then adding this to the sidereal day, which gives us a total of approximately 10.4 hours.