# Little Gravity Question

1. Apr 9, 2005

### naren11

Because the rotation of the earth about it's own axis, a point on the equator experiences a centripetal acceleration of 0.034m/s(square). While a point at any of the poles experiences no such acceleration.

1) Show that the true weight of an object (the gravitational force of the object) at the equator is greator than it's apparent weight.

2) If the true weight of a person is 735 N, what is his apparent weight at the equator?

Thank you! :!!)

2. Apr 9, 2005

### quasar987

I love you too, but could you post what work you have done so far? If that includes no calculations, at least post a begining of reasoning or the reason why you don't know how to start this problem. Thx.

3. Apr 9, 2005

### Werg22

The centripal acceleration of the rotation is directed toward the center of the eart, and so for the gravity. So you have add up those two acceleration for an object located on the equator. As for an object on the pole, the centripetal acceleration is inexistent and thus only the gravity acts on it. Go figure.

Last edited: Apr 9, 2005
4. Apr 9, 2005

### Andrew Mason

$$W_{app} = \vec N = mg - F_c = W_{act} - F_c$$

Since $F_c$ is non-zero at the equator (ie it is .034m), apparent weight < actual weight.

You just have to work out the centripetal force from the centripetal acceleration.

AM

5. Apr 9, 2005

### HallsofIvy

Staff Emeritus
Welcome to "Physics Forum". Please read the "stickies" about homework help. We expect you to show what you have done and what you know about a problem so that we can give hints to help you without doing the problem for you.

In this problem, note that the "centripetal acceleration" requires force just to keep the person moving with the earth. It subtracts from the gravitational force, not adding to it!

6. Apr 9, 2005

### Werg22

But in that case the question is fallascious... And the force is provided by the earth magma and not gravity.

Last edited: Apr 9, 2005
7. Apr 9, 2005

### Andrew Mason

There are two forces acting on a mass on the surface of the earth (ignoring gravity of sun, moon, etc): earth gravity and the earth's normal force. These forces must sum to the net force, which is the mass multiplied by its actual acceleration.

$$\vec N + \vec F_g = m\vec a$$

What is the net acceleration of the mass? What is the Force of Gravity $F_g$? If you think about that you will be able to tell where the centripetal force comes from.

AM

8. Apr 9, 2005

### Werg22

Yes so Fn+Fg=centripetal force (Fg beeing opposite to Fn and thus being negative). This is just a physical consequence... And so the weight is greater and the equator compared to the one on the pole since the net force on the pole would be 0.

Last edited: Apr 9, 2005
9. Apr 9, 2005

### Andrew Mason

But the measured weight is the Normal force. For any place on the earth, the normal force is mg reduced by the radial component of the centripetal force.

So the measured weight (N) is less than mg at the equator. At the poles it is equal to mg.

AM

10. Apr 9, 2005

### Werg22

Is the earth rotation is due to the fact it is not a perfect sphere and thus gravity causes it to rotate?

11. Apr 9, 2005

### quasar987

Someone correct me if I'm wrong but I'd say earth's rotation is due to the time when the entire solar system was but dust star in rotation. This is when the rotationnal motion was aquired and there has been no reason for it to stop.

On the opposite, the earth is an ellipsoid BECAUSE it rotates. Because the earth is a rotational frame, there is a ficticious force oriented in the direction pointing away from the rotational axis that adds to that of gravity. The earth has shifted its shape in order that its surface be perpendicular in every point to this "corrected" force field. The result is an ellipsoid.

Last edited: Apr 9, 2005
12. Apr 9, 2005

### Andrew Mason

No need to use fictitious forces. You could say that the earth is an oblate spheroid because the earth's spin reduces gravitational pressure at the equator but not at the poles.

This is due to the fact that at the equator, part of the gravitational force(/unit area) is used to provide centripetal acceleration so there is less gravity available to provide pressure. This causes the earth to bulge slightly, causing the earth's surface to extend further from the earth's centre, thereby reducing the gravitational force further, causing a little more bulge, etc. (don't worry, it reaches a limit).

AM

13. Apr 10, 2005

### Werg22

So in fact gravity is responsable of keeping objects in balance with the earth rotation, and thus has to give a centripetal acceleration in order to keep this balance, I'm I wrong?