# Problem of the week - gravitation

• yoyigloria
In summary, the problem presented deals with the variation in apparent weight of a person at the equator and at the poles due to the rotation of the Earth. The solution involves using the formula for gravitational force and the radius of the Earth at the equator and poles. It is important to consider the Earth as a perfect sphere, as the distance from the center to any point on the surface is the same as the radius. The difference in weight is caused by the centripetal acceleration and force at the equator compared to the poles.
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!

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.

rock.freak667 said:
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

gh = go((re)/(h+re))^2

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

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

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

gh = go((re)/(h+re))^2

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

I think that is all that is required.

yoyigloria said:
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.

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.

## 1. What is the problem of the week - gravitation?

The problem of the week - gravitation is a weekly challenge or question related to the concept of gravitation, which is the force of attraction between two objects with mass.

## 2. How does gravitation work?

Gravitation is governed by Newton's Law of Universal Gravitation, which states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

## 3. What are some real-life examples of gravitation?

Some real-life examples of gravitation include the gravitational pull between the Earth and the moon, the orbit of planets around the sun, and the falling of objects towards the ground.

## 4. How is gravitation related to the concept of weight?

Gravitation is directly related to the concept of weight. Weight is the measure of the force of gravity on an object, and it is dependent on the mass of the object and the strength of the gravitational pull.

## 5. What are some applications of gravitation in modern technology?

Gravitation has many applications in modern technology, including satellite communication, space travel, and GPS navigation. Gravitational theories and equations are also used in fields such as astronomy, physics, and engineering.

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