# Gravitational Acceleration

1. Mar 7, 2006

### Milind_shyani

hello
The gravitational acceleration at the equator is the least and at the pole is the maximum due to the rotation of the earth on its axis.Why???

2. Mar 7, 2006

### Galileo

The earth is not precisely spherical, it is flattened at the poles because of the earth's rotation. This is simply because of the centrifugal force, objects at the equator have the greatest distance from the axis of rotation so are 'flung outwards' more than at the poles. So at the poles you are closer to the center of the earth and experience a greater gravitational acceleration.

3. Mar 7, 2006

### arildno

It should be borne in mind that "g" is generally meant to mean "effective gravitational acceleration" and includes the effects of a rotating Earth (which predicts a lower g at equator than by the poles), the deviatioric correction due to the non-spherical form of the Earth, as the two main correctional effects.

4. Mar 7, 2006

### BobG

Galileo is right - the oblateness of the Earth is the biggest factor in the difference in gravitational attraction.

However, the way the question is worded, I think it's addressing centripetal acceleration. Which point will have a larger linear velocity due to the rotation of the Earth: a point on the equator or a point on one of the poles?

Both the oblateness of the Earth and centripetal force contribute to reducing the net force, and the resulting acceleration, at the equator (with oblateness having nearly twice as much affect).

5. Mar 7, 2006

### arildno

I never said that Galileo was wrong..

6. Mar 7, 2006

### BobG

I didn't comment on your answer because I didn't understand it. :rofl:

7. Mar 7, 2006

### Staff: Mentor

To the OP -- just think about what happens as you spin the earth faster and faster....what force is acting on the person at the equator that is different from the person at the pole? Quiz question -- how fast to you have to spin the earth to spit off the person at the equator?

8. Mar 8, 2006

### andrevdh

Think of a girl on a merry-go-round. She will move at a larger speed when she is further away from the center, since the outer regions cover a larger distance in the same time. If the merry-go-round would spin faster and faster the girl would have to hold on for dear life or she would be flung off. She would conclude that some force is pulling her outwards from the center. This force we call the centrifugal force, it is not a real force. It results from her inertia - she wants to keep on moving in a straight line and have to hold on or press up against something in order to keep on the merry-go-round. The same happens with an object on the earth. The further it is away from the rotation axis of the earth the larger the centrifugal force it experiences. This leads to a decrease in the weight of the object - it wants to fly off the earth. Since the weight is related to the gravitational acceleration and mass (which stays the same under all circumstances) we conclude that gravity was somehow reduced by this effect.

9. Mar 8, 2006

### Hootenanny

Staff Emeritus
The effect of 'centrifugal' acceleration can be quanitfied by;
$$g_{a} = g - r\omega^2\cos^2\theta$$
where $g_{a}$ is the apparent force of gravity and $\theta$ is latitude. This assumes the earth is spherical, but is a good approximation for a non-spherical earth.