# Centripetal acceleration at two different points on the earth

1. Dec 1, 2011

### hms.tech

1. The problem statement, all variables and given/known data
This is a question from my OCR A levels book.
City X is on the equator of the Earth whlie another city (City Y) is located latitude 52° N.
What is the centripetal acceleration on a piece of rock located in each city.

2. Relevant equations

data given : angular velocity of earth: 7.3*10^-5
radius of the earth : 6.4*10^6

a= r*w^2

3. The attempt at a solution

Acceleration at x = (6.4*10^6) * (7.3*10^-5)^2

i think that the acceleration at y will also be the same as that on X, regardless of the latitude or longitude(location on the earth) because they would have the same distance from the COM of the earth.

Whats the answer to this question ppl ... ? will the acceleration be the same at both locations ?

2. Dec 1, 2011

### Stonebridge

They are different.
If you think about it, a city at 90 latitude, at the pole, would have no centripetal acceleration as it would not be moving in a circle.
So this acceleration varies from the value you correctly calculated at the equator to zero at the pole.
Can you do a little trigonometry to work out how it depends on the angle?
The clue is in the fact it is zero at 90 degs and a maximum at 0 degrees.

3. Dec 1, 2011

### hms.tech

hhmm...alright, ur hint leads me to cos(theta) ...
so i think that at city Y the acceleration would be :
6.4*10^6 *cos(52) *(7.3*10^-5)^2

alright, now moving forward with my argument, if u say that the acceleration at the pole would be zero, then u are implying that a person at the pole would be wightless !!

4. Dec 1, 2011

### Stonebridge

I hope not!
At both places you will be attracted towards the centre of the Earth with a gravitational force (called your weight) that depends on the radius of the Earth.
The difference is that at the equator you are also moving in a circle around the centre of the Earth, while at the pole you are not. (At 52 degs N you are moving in a smaller radius circle.)
So at the equator you need a small centripetal force to move you in that circle. This force is provided by the gravitational attraction. The result is that your measured weight on a scale, for example, would be slightly less at the equator than that measured at the pole, where there is no centripetal force required.

5. Dec 1, 2011

### D H

Staff Emeritus
From the perspective of some person standing still on the surface of the earth, a rock sitting on the earth 52 latitude or at the equator is not accelerating at all. The rock is stationary. However, that is the wrong perspective for solving this problem. The earth is rotating at one revolution per day. You need to look at the problem from the perspective of an inertial frame; one in which the earth is rotating.

From this perspective, there are two forces acting on the rock, the downward gravitational force exerted by the earth as a whole, and the upward normal force exerted by the earth's surface. The rocks are fixed with respect to the rotating earth, so the rocks are undergoing uniform circular motion. That can only mean that the two forces acting on the rock don't quite balance. (If they did balance, Newton's first law says the motion would be ≪what≫?)

6. Dec 1, 2011

### hms.tech

maintain it's uniform velocity(ie go straight linearly tangent to the earth's surface)

Last edited: Dec 1, 2011
7. Dec 1, 2011

### hms.tech

thnx guys, i think it is abt 80% clear now...appreciate the help
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