# Gravity And Earth's Rotation

1. Oct 16, 2009

### Warmacblu

If the Earth rotated faster about its axis, how would the value of gravity change?

2. Oct 16, 2009

### DavidSnider

Faster = less gravity.

Last edited: Oct 16, 2009
3. Oct 16, 2009

### Nabeshin

Pop quiz: how fast does the earth have to rotate for there to be zero gravity?

4. Oct 16, 2009

### Warmacblu

How come? Is it like those theme park rides where you spin in a circle while standing and you feel like you are floating upwards?

5. Oct 16, 2009

### Nabeshin

Not quite, but something similar, yes. I like to think of it this way, in terms of inertial reference frames:

In order to be in circular motion there is a required centripetal force. In the case of us standing on the earth the only force there is is that of gravity. For a given rotational velocity, a force mv^2/r is required to keep us in that motion. Of course, this force is provided by gravity. In a sense, the gravitational force is "split" into partly providing for your rotational motion, and partly for the normal weight component. The larger the rotational motion, the more of the force is "allocated" towards the rotational part, so the less there is to be perceived as weight (keep in mind the gravitational force is fixed).

It's not exactly a rigorous way of thinking about it, but I hope it makes sense to you.

6. Oct 18, 2009

### Warmacblu

Does this mean that less is perceived as weight or the actual value of g, let us say 10 m/s2, actually decreases?

7. Oct 18, 2009

### pixel01

Approximately the angular velocity of the ISS rotation.

8. Oct 18, 2009

### HallsofIvy

Strictly speaking, the force of gravity is not changed by rotation. The weight of an object, which is the difference between the gravitational force on the object and the "centrifugal force" due to the rotation changes.

An object rotating at distance r and angular velocity $\omega$ feels a "centrifugal force" of $mr\omega^2$ (that is really the centripetal force necessary to keep the object in orbit). If the gravitational force is mg, there would be "zero gravity" when $mr\omega^2= mg$ or $\omega= \sqrt{g/r}$.

9. Oct 18, 2009

### Warmacblu

So the force we feel when spinning at a high angular velocity is the centrifugal force and not any change in gravity?

10. Oct 18, 2009

### D H

Staff Emeritus
Actually, neither. You can't feel gravity and you can't feel centrifugal force.

You can feel the ground pushing up on you, however.

11. Oct 19, 2009

### pixel01

I do not think you feel the ground pushing up, just like the feeling of astronauts when on board the ISS.

12. Oct 19, 2009

### Ich

D H is talking about General Relativity, while HallsofIvy uses Newtonian terms. Both are right, conserning the respective point of view, but as GR is our standard theory today, and as it's a question about the nature of forces, I'd subscribe to D H's answer.

13. Oct 19, 2009

### Wallace

Actually DH is correct from a Newtonian perspective as well. Whether you are standing on the Earth surface or orbiting in the ISS, from a Newtonian perspective you are receiving the same acceleration due to gravity (roughly speaking, the increased distance from the Earth in the ISS makes it a little smaller) but the main difference is that in the ISS case, the station is accelerating freely with you, hence there is not contact force between you and the floor, whereas in the case of standing on the Earth the ground is not moving, because its acceleration is balance by the contact force of the material below it. You 'feel' gravity only because you are trying to accelerate but the ground is pushing back.

14. Oct 19, 2009

### HallsofIvy

What are you referring to here? If the earth were rotating fast enough that you felt "no gravity", then, yes, of course, you "do not feel the ground pushing up". But at any rotation less than that (as our normal rotation) you definitely feel the grouhd pushing up. That's what "weight" is.

15. Oct 19, 2009

### D H

Staff Emeritus
Wallace already replied to this, but seeing how you misrepresented what I wrote, I need to answer as well. What I wrote is correct from either a Newtonian or a GR perspective.

That one cannot feel centrifugal force is easy to deal with: Which centrifugal force? From the perspective of an Earth-centered, Earth-fixed frame a person standing on the equator is stationary. This is a rotating frame, so there is a centrifugal acceleration of 3.3915 cm/s^2 acting on the person. From the perspective of an Earth-centered frame that is rotating about the Earth's rotation axis, but at 17 times the Earth rotation rate, the centrifugal acceleration on this person at the equator becomes 9.8 m/s2 -- equal but opposite to the gravitational acceleration. (The downward coriolis force is rather large in this frame.) From the perspective of a non-rotating Earth-centered frame, there is no centrifugal force. All three frames (Earth-centered Earth-fixed, hyper-rotating, and inertial) will agree on the gravitational and normal forces exerted on the person.

That one cannot feel gravitational force from a Newtonian perspective: (1) We do not know how to create a gravity shield, and (2) the gradient of the Earth's gravitational force over the span of a person's body is incredibly small. The latter would not be the case for an astronaut in low orbit around a neutron star.

16. Oct 19, 2009

### D H

Staff Emeritus
That is all you feel. Even though the gravitational acceleration on astronauts on board the ISS is about 90% of that on the surface of the Earth, astronauts on board the ISS feel weightless.

17. Oct 19, 2009

### Ich

I misread Warmacblu's statement, missed the "feel"-part of it.

18. Oct 19, 2009

### pixel01

In fact I can only imagine the 'feel' (I do not know any of this 4rum ever been on board the ISS). Anyway that is not difficult to imagine and I have once experience a free fall when in a plane for about at least 5 seconds.
In the surface of the earth, the gravity is stronger than when you are up 360km, sure, but then suppose the earth rotate a bit faster (not too fast that you are thrown away) and you will be in the same condition as an astronaut in space.

19. Oct 19, 2009

### D H

Staff Emeritus
If the Earth rotated once per 1.40699 hours (as opposed to once per 23.93447 hours), the normal force needed to keep an object at the equator stationary with respect to the rotating Earth would be zero. A person at the equator on such a hyper-rotating Earth would feel weightless. They would feel weightless precisely because the normal force is zero.

20. Oct 19, 2009

### Warmacblu

So let me get this straight. As the Earth spins faster, the value of g (9.8m/s^2) remains the same? And it feels like it's decreasing?

21. Oct 23, 2009

### Moonglum

Since g is defined as the force of gravity (given by Newtons law of Gravitation) divided by the mass of the test object, the value of g has no dependance on the rotational speed. Therefore your statement is correct!

22. Oct 23, 2009

### D H

Staff Emeritus
That is wrong in two ways. Firstly, g is defined as 9.80665 m/s2, exactly. Secondly, that definition does reflect the Earth's rotation rate. We live on a rotating Earth. Suppose you time the fall of an object initially at rest with respect to the rotating Earth in vacuum (this is exactly what a gravimeter does). The acceleration of the toward the Earth most definitely does depend on the Earth's rotation rate.

23. Oct 24, 2009

### qraal

The Earth wouldn't need to rotate as fast as the ISS is orbitting for the gravity to cancel out because it would undergo significant plastic deformation before becoming unstable at the equator. It would deform into an oblate spheroid and start losing mass at the equator once the force of gravity and the centrifugal force were balanced (i.e. weight = 0.) I can't remember the exact rotation rate, but it depends strongly on the density-radius profile of the Earth's interior. In some simulations of the Earth-Moon separating from each other just after the Big Whack they're rotating once every 6 hours and that might be close to the mass-loss limit.