# Gravity at the equator

1. Dec 22, 2008

### Shackleford

Assuming a perfect sphere,

mg - w' = m (v^2/r)

g' = g - (v^2/r)

I understand there's a constant radial acceleration because we're not flying off the planet. Theoretically, the difference should be 0.0337 m/s^2, but isn't because the Earth "bulges" at the equator. Why is there a difference on a rotating perfect sphere? It's not quite connecting intuitively or physically.

2. Dec 22, 2008

### buffordboy23

It's because of centrifugal force:
http://en.wikipedia.org/wiki/Centrifugal_force

The centrifugal force exists when there is rotation in the observer's frame of reference, so the exact shape doesn't really matter, except to play a role in the magnitude of the effect, which is negligible for most purposes on earth. The bulge at the equator means that the mass of the earth is not spread uniformly (topography affects this too), so this creates variations in the local gravity.

Here is something else interesting to consider. Suppose your at some latitude. As the earth rotates on its axis, your motion traces out the circumference of the circle within a plane that is perpendicular to the earth's rotational axis. Now the centrifugal force vector lies in the same plane and is directed radially outward. Now consider the acceleration vector for gravity. It points toward the center of the earth. So when you drop an object, it actually deviates by a small angle from the direction of the gravitational acceleration vector--the result is negligible but it is there. Draw a picture and construct the resultant vector from the two aforementioned vectors and you can see this result.

3. Dec 22, 2008

### Dick

On a rotating perfect sphere on uniform density, it's pretty easy to calculate the difference between g at the poles and g at the equator. Just subtract the centripetal acceleration. If the planet deforms, it's not. I'm not sure what's confusing you. The bulge is going to change the gravitational field and the radius. How depends on the rigidity of the material.

4. Dec 22, 2008

### Shackleford

Centrifugal force is another fictitious force. The misconception is usually cleared up with recognition of Newton's Second Law.

5. Dec 22, 2008

### Dick

It's fictitious in the sense that it's not a 'force', it's an acceleration. It's still there. You still have to factor it into F=ma, because that contains both forces and accelerations.

6. Dec 22, 2008

### Shackleford

Why do you subtract the centripetal acceleration? How does the centripetal acceleration decrease the magnitude of gravitation at the equator? Obviously, I'm looking for more insight beyond the equations.

7. Dec 22, 2008

### buffordboy23

Although the centrifugal force (and even the coriolis force) are fictitious, they must be accounted for because we live on a rotating reference frame.

8. Dec 22, 2008

### Shackleford

So, it's more of an effective or observed gravity compared to it at a non-rotating location without the constant centripetal acceleration?

And to echo someone else's input, say, above the equation, an object's downward resultant vector would be slightly higher than it would be without the rotational consideration.

Last edited: Dec 22, 2008
9. Dec 22, 2008

### Dick

It doesn't decrease the magnitude of gravitation at the equator. That's the same. If the earth were spinning so fast that objects at the equator were being thrown off the earth then you could call that a negative g. I wouldn't, but that's my opinion. That's what they are talking about.

10. Dec 22, 2008

### Dick

exactly. You move part of the 'a' of F=ma to the 'F' side. For convenience. Since it's a fiction.

11. Dec 22, 2008

### Shackleford

Cool. I got it now. I like to understand the logic behind the equations. Thanks for your help, guys.