# Calculating Weight at Earth's Equator: 600.0 N to 597.9 N

In summary, the Earth's rotation causes a difference in weight for a person at the north pole and at the equator. Using the equations for centripetal force and Newton's second law, one can calculate the person's weight at the equator based on their weight at the north pole and the Earth's angular velocity. The equation for angular velocity is 2π/T, where T is the period of rotation, which is 24 hours for the Earth. By finding the difference between the weight and the upward push of the scale, one can calculate the centripetal force and determine the person's weight at the equator.

## Homework Statement

Suppose the Earth is a perfect sphere with R=6370 km. if a person weighs exactly 600.0 N 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) ans: 597.9 N

Fg=Gm1m2/r^2
Fg=ma
Fc=Mv^2/r?

## The Attempt at a Solution

I have no real clue on how to solve such a question, i plugged in for Fg but the answer i seem to get is higher than that of 597.9 N. Need help for a solution please!

I will give you a hint...A person standing on the equator experiences a centripetal force towards the centre of the Earth .Work out what this is and draw a free body force diagram for the person.

Apply Newton's 2nd law. What's different about being on the equator compared to being on the north pole?

Think about the angular velocity at the north pole and the equator. Then calculate the centripetal acceleration at the equator and use this value to find the effective acceleration of gravity at the equator.

how could i determine the angular velocity though? i am not given a time

The time is 24 hours for one revolution! Check out this website for an explanation

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Gravity/AccOfGravity.html

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so i should use Fc=4pie^2Mr/ T^2 to solve for Force centripetal

Yes, use this equation but it's a little easier to find the effective acceleration of gravity at the equator, g - acentripetal, then multiply by the mass.

okay so it would be easier to just get the centripetal acceleration, then take gravity- acceleration centripetal then multiply by mass for F=ma right? but how would i get the acceleration centripetal when Ac=V^2/r and i don't have V?

V= Earth's angular velocity times the Earth's radius.

i am not familiar with Earth's angular velocity?

i am not familiar with Earth's angular velocity?

Really? You don't know how long a day is? Or you don't know how to calculate angular velocity from knowledge of the period of rotation?

don't know how to calculate angular velocity from knowledge of the period of rotation? I've never heard of the term for one so maybe that's why I am lost with what your saying

Angular frequency is what you really need for this problem; and it is defined by:

$$\omega\equiv2\pi f=\frac{2\pi}{T}=\frac{|v|}{r}$$

Angular velocity is a vector quantity with magnitude $\omega$ and direct determined by a cross product.

okay thank you i have gotten the correct answer, much appreciated

It takes 24 hours(one day)for one rotation of 360 degrees or 2pi radians.The centripetal force is m vsquared/r or mr omega squared,use the equation you are happiest with.At the poles the upward push of the scale (R) will equal the persons weight(600N).At the equator there must be a resultant force(the centripetal force)because the Earth is spinning and the person is accelerating towards its centre.It follows that R cannot be equal to the weight because the resultant force will then be zero and it further follows that weight minus R equals the centripetal force.From this you can calculate the recorded weight(R)

## 1. How is weight calculated at Earth's equator?

Weight at Earth's equator is calculated using the formula W = mg, where W is weight, m is mass, and g is the acceleration due to gravity. The value of g varies slightly at different points on Earth's surface due to factors such as altitude and latitude.

## 2. What is the weight at Earth's equator if it is given in Newtons?

The weight at Earth's equator can be calculated using the formula W = mg, where m is the mass in kilograms and g is the acceleration due to gravity in meters per second squared. If the weight is given in Newtons, it can be converted to kilograms by dividing by 9.8, the approximate value of g at the equator.

## 3. How does the weight at Earth's equator compare to the weight at the poles?

The weight at Earth's equator is slightly less than the weight at the poles due to the centrifugal force caused by Earth's rotation. This force is greater at the equator, causing objects to weigh slightly less compared to the poles.

## 4. Can the weight at Earth's equator change?

The weight at Earth's equator can change due to factors such as altitude, latitude, and the distribution of mass on Earth's surface. However, the change is typically very small and may not be noticeable in daily life.

## 5. Why is there a difference in weight at Earth's equator and poles?

The difference in weight at Earth's equator and poles is due to the centrifugal force caused by Earth's rotation. This force is greater at the equator, causing objects to weigh slightly less compared to the poles. Additionally, the shape of Earth plays a role, as the equator is slightly wider than the poles, resulting in a slightly larger distance from the center of the Earth at the equator compared to the poles.

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