# The rotation of the earth and acceleration due to gravity

1. Jun 21, 2008

### ace123

So me and my physics professor were having a discussion on the coriolis effect and then we somehow got to talking about the earth's rotation and how it has an affect on the acceleration due to gravity, that is the g= 9.8 m/s^2. The professor had to leave however because his lecture was starting and we never really got to discuss it.
So I was thinking about it myself and figured that if the earth's rotation changes then the centripetal acceleration would change because the velocity is increasing or decreasing. This must mean the acceleration of gravity has changed. Is this correct?

This isn't a homework question just some general curiosity.

2. Jun 21, 2008

### calef

Short answer, no, it doesn't change the acceleration due to gravity.

That is only a function of the mass of the planet, not how much it happens to be spinning.

The net force of someone standing on the surface of the earth, depends on the rotation of the earth, though.

3. Jun 21, 2008

### ace123

But if the net force changes with the rotation of the earth doesn't it mean the g is changing?

The rotation of the earth has to have an affect on the g.

4. Jun 22, 2008

### aniketp

Hi ace123,
Yes, the rotation of the earth does indeed affect "g"
Let us consider the apparent acceleration (g')
It is the resultant of "g" and "Ω^2R" where "Ω" is the angular velocity of the earth.
Using simple algebra we get,
g' = [g^2-(ΩsinФ)^2*R*(2g-Ω^2R)]^(0.5)
Here "Ф" is the angle made by the line joining the centre of the earth and the point under consideration with the axis of the earth.

5. Jun 22, 2008

### D H

Staff Emeritus
The rotation of the Earth does not affect the gravitational force exerted by the Earth. It does, however, change the force that you feel. What you feel as the pull of gravity is not really gravity. Gravity pulls you toward the center of the Earth, but you cannot feel that force. Think about it this way: The gravitational force on astronauts in orbit is about 90 to 95% of the gravitational force on the astronauts while they are on the ground. They don't feel that force while they are orbiting. They also don't feel that force when they are standing on the ground.

What they feel when they are standing on the group is the upward normal force exerted by the Earth on them that keeps them from sinking into the Earth. The Earth-bound astronauts have a null velocity vector in the the reference frame rotating with the Earth. This means the sum of all forces (including pseudo forces) in that frame must be zero in that frame. The forces acting on the astronauts are the force due to gravitation, the centrifugal pseudo force, and the normal force. The normal force is the only force a person standing on the Earth feels, and this force is affected in both magnitude and direction by the Earth's rotation.

To a person standing on the ground, g includes both the gravitational force and the centrifugal pseudo force. The field of physical geodesy distinguishes between gravitation as the force resulting from Newton's Law of Gravitation and gravity as the sum of gravitation and the centrifugal force caused by the Earth's rotation. See http://en.wikipedia.org/wiki/Geopotential.

On the other hand, the normal force does not affect an airplane flying in the sky. To an airplane designer, g is the acceleration due to gravitation only.

6. Jun 22, 2008

### ace123

I thought the centrifugal force is not a real force.

Thanks for the link, it helps.

Edit: My book says that the earth's rotation affects the value of g.

Last edited: Jun 22, 2008
7. Jun 22, 2008

### ace123

So if g includes both the gravitational force and the centrifugal force. Would the centrifugal force change with the rotation of the earth.

8. Jun 22, 2008

### Staff: Mentor

Correct. It's a "pseudoforce" that one must include when analyzing things from a rotating (and thus non-inertial) frame. It acts outward, thus reducing the apparent weight of an object.

Analyzed from an inertial frame you'd have to consider centripetal acceleration, leading to the same conclusion.

Sure.

If you calculate g using Newton's law of gravity, of course it won't include the effects of rotation.

Read the section titled "Latitude" on that wiki page:

9. Jun 22, 2008

### ace123

Yea that's what I was doing. I also deleted my other post becuase I didn't see it said calculated value. I'm a little confused for why the effect is smallest at the poles. Isn't the gravitational force and the centrifugal force always orthogonal on the earth?

Thanks again.

10. Jun 22, 2008

### Staff: Mentor

The gravitational force is always normal to the Earth's surface, but the centrifugal force is perpendicular to the Earth's axis. And it depends on the distance from that axis.

11. Jun 22, 2008

### D H

Staff Emeritus
It is a pseudo force, meaning that it arises from use of a non-inertial reference frame. A non-inertial reference frame is a natural frame to use for person standing on the surface of the Earth. Gravity isn't a "real force" either. That doesn't make what gravity does to us unreal.

No. The centrifugal force has no north-south component, while the gravitational force only lacks a north-south component for a person on the equator.

12. Jun 22, 2008

### ace123

The axis part helped clear up the confusion. Thanks alot

13. Jun 24, 2008

### swraman

Or in simple terms:
If you are standing on the North pole of the earth, you are essentially rotating in place, the axis of rotation of the Earth going through you.
If you are standing on the equator, you are rotating in a circle of radius (radius of the earth).

Since If you stand on the equator you move in a circular path of radius Re, so you have a centripetal force that has to act on your body to keep you on the surface of the earth. THis centripetal force comes from the force of Gravity. Since some of the gravitational force is used to accelerate your body so it can stay on the planet, less of it actually adds to your body weight...so to speak. So the closer to the equator you get the less you weigh.
If this doesnt make sense, draw free body diagrams for someone standing on the north pole and the equator. Take account for their force of weights and their Centripital force (only the equator diagram has a centripital force)

14. Jun 24, 2008

### ace123

Thanks. I kind of understand it better but my professor confused me and like half of the class by saying the opposite. He said that you weigh less at the poles and more at the equator which made no sense to me at all.

15. Jun 24, 2008

### D H

Staff Emeritus
Your professor is wrong. You weigh more at the poles than at the equator because there is no centrifugal force in the Earth-fixed frame at the poles (the centrifugal force is at its maximum at the equator) and because you are 21.3 kilometers closer to the center of the Earth at the poles than you are at the equator.

16. Jun 24, 2008

### shamrock5585

hmmm the radius of the earth is pretty large... how much of a difference would there be? it seems like a very small difference! and doesnt this only work if the earth is a perfect sphere?

about your arguement that you are closer to the center of the earth at the poles... are you saying that the centripetal force would be less being closer to the center? (regardess that you wont have centripetal force because you are on the axis) but wouldnt gravity be stronger?

17. Jun 24, 2008

### shamrock5585

basically swraman i am having trouble understanding your explanation... normal acceleration is the tangential velocity squared.. divided by the radius... so that force would be acting outward on you away from the center of the earth... and then the force opposing it would be gravity which would give the illusion of less gravity but its just less total force acting towards the center of the earth which determines weight...

18. Jun 24, 2008

### D H

Staff Emeritus
The Earth is not a perfect sphere. The poles are 21.3 kilometers closer to the center of the Earth than is the equator.

Read what I said! I said "You weigh more at the poles than at the equator because there is no centrifugal force in the Earth-fixed frame at the poles (the centrifugal force is at its maximum at the equator) and because you are 21.3 kilometers closer to the center of the Earth at the poles than you are at the equator."

I am going to be very careful with terminology here.

• Gravitation force is the fictional force exerted on a body by the mass of the Earth. The relevant equation is Newton's law of gravitation (or general relativity if you want to be really precise).
Earth centrifugal force[/i] is the fictional force exerted by the rotation of the Earth on a non-moving body near the surface as observed by an Earth-fixed observer. The relevant equation is $$F_{\text{centrifugal}} = m {\boldsymbol{\omega}} \times ({\boldsymbol{r}} \times {\boldsymbol{\omega}})$$
Gravity is the fictional force observed by an Earth-fixed observer on a body near the surface of the Earth. Gravity is what a spring scale measures. Gravity is the vectorial sum of the gravitational force and the centrifugal force.

If the Earth were a point mass (see final paragraph), the gravitational acceleration would be 9.8656 meters/second2 at the poles and 9.7997 meters/second2 at the equator. The difference between the two extremes is 0.0659 meters/second2.

There is zero centrifugal force at the poles because ${\boldsymbol{r}} \times {\boldsymbol{\omega}}$ is zero. At the equator, the centrifugal accleration is 0.0339 meters/second2. With this simple model, the acceleration due to gravity is 9.8656 meters/second2 at the poles 9.7658 meters/second2 at the equator. The rotation of the Earth is directly responsible for about 1/3 of the difference between the two results. The other 2/3 arises from the Earth's equatorial bulge (which in turn is caused by the Earth's rotation).

The point mass model is not quite right. It overestimates the acceleration due to gravitation at the poles and underestimates the acceleration at the equator. A better estimate of the acceleration due to gravity (gravitation plus centrifugal force) may be found from the empirical relationship,
$$g=9.780327\left(1+0.0053034\sin^2\phi-0.0000058\sin^2\phi)$$.

19. Jun 25, 2008

### ace123

I know he is wrong but what can I do? I wasn't about to argue with my professor in the middle of class. I figure I'll just talk to him about it after class next time.

20. Jun 25, 2008

### shamrock5585

haha DH thanks for the really long explanation... basically you could have said what my comment stated was correct! i was confused by the misuse of the terms centripetal and centrifugal... i know the earth is not a perfect sphere thats what i was asserting... i think you misunderstood what i was saying... but i understand and its cool... so ive decided to go to the north pole to sell drugs because they weigh more up there and i will become rich... its like inflation on drugs haha