Rotation of a sphere because of its gravitational field

1. Jan 3, 2012

ShayanJ

I wanna know how can I prove that a massive sphere rotates becuase of its gravitational field.I thought about it and did the following:
The gravitational force applied to a mass m which is on the axis prependicular to the plane of a hollow ring and passing through its center is as follows
${\it dF}={\frac {G{\it dM}\,mz}{ \left( {r}^{2}+{z}^{2} \right) ^{3/2}}}$
If I integrate the equation above from z=0 to z=2R I should get the force applied to a point on a hollow sphere and I get:
$F=1/8\,{\frac {GMm\theta\,\ln \left( z \right) \sqrt {2}}{\pi \,{R}^{2}}}$
$\theta$ is from 0 to $2 \pi$ and z from 0 to 2R.But as you can see,at z=0,it becomes infinity.So I wrote the taylor series of ln(z).But my calculations for $\omega$ didn't yeild 7.3 * 10^(-5) which is the angular velocity of earth.What's wrong?
thanks

2. Jan 3, 2012

Staff: Mentor

Why would you think that it would rotate? Explain what you mean.

3. Jan 3, 2012

ShayanJ

Earth rotates around itself and if you think,the only reason can be its own gravitational field.

4. Jan 3, 2012

Tea Jay

What about the fields of other bodies that can influence it, such as the moon or sun?

What about the motion of its crust or mantle?

5. Jan 3, 2012

6. Jan 3, 2012

D H

Staff Emeritus
No! The Earth rotates now because it was rotating in the past and because angular momentum is a conserved quantity.

Newton's first law, a body persists in its state of motion unless acted upon by an external force, is in a sense a statement regarding conservation of linear momentum. There's a direct analog of Newton's first law for rotational motion, a body persists in its state of rotation unless acted upon by an external torque.

External torques from the Moon, the Sun, and the other planets do act on the Earth. However, these external torques are small. Because they are small, they can be treated as being effectively zero for short periods of time (years and even decades are short periods of time in this context). Earth's angular momentum is more or less constant over the span of years or decades. It's only on the scale of centuries or longer (or on the scale of extremely precise measurements) that the Earth's angular momentum can not be viewed as being constant.

7. Jan 4, 2012

ShayanJ

This question came into my mind when our professor was teaching the central forces.
Looks like we just get excited when we learn sth new and stick everything to that new concept.

8. Jan 5, 2012

jetwaterluffy

No the only reason is the conservation of angular momentum. Although a similar effect does possibly exist: http://en.wikipedia.org/wiki/Gravitomagnetism but the effect is tiny.

9. Jan 5, 2012

Buckleymanor

So when the dust etc, was pulled together when it was forming by the gravity of the main body, angular momentum caused it to rotate.
Won't rain falling do the same.

10. Jan 5, 2012

D H

Staff Emeritus

Not really. First off, rain doesn't have much angular momentum. Much more importantly, whatever angular momentum rain does add to the Earth comes from the Earth+oceans+atmosphere system, not from outside. Rain does not add angular momentum to the Earth+oceans+atmosphere system.

There is an ongoing but small transfer of angular momentum from the Earth to the Moon. This is what causes the Earth's rotation rate to slowly decrease and the size of the Moon's orbit to slowly increase. However, this transfer is very slow. Over short spans of time (seasons, years, even decades), the angular momentum of the Earth+oceans+atmosphere is more or less constant. There is seasonal transfer of angular momentum between the Earth+oceans and the atmosphere, but this is cyclical and bounded. There is no secular change.

11. Jan 5, 2012

jetwaterluffy

It starts as a large cloud of dust spinning very slowly. As the size of it decreases, the spinning speed speeds up to compensate. This can't happen with rain, as rain doesn't decrease the size of the earth.

12. Jan 5, 2012

D H

Staff Emeritus
That's not a good model of how stars form, let alone planets. The Sun has 99.9% of the mass of the solar system, but only 1% of the angular momentum. This is the angular momentum problem.

Assuming the total angular momentum of the cloud with respect to its center of mass is non-zero, the formation of the protostar will act to flatten the cloud into a disc. With the formation of the disc, the material that falls into the protostar is that material that has little angular momentum. The formation of the disc solves the angular momentum problem.

Last edited: Jan 5, 2012
13. Jan 5, 2012

jetwaterluffy

Erm.. sorry can you rephrase that? Maybe I didn't get it when I was told, and subconsciously made something up to compensate, so I'd like you to explain it again for me. (the above post to me doesn't make sense.)

14. Jan 5, 2012

phinds

That is an excellent observation.

It's another way of saying the old saw "to a man with a hammer, all problems look like nails"

15. Jan 5, 2012

D H

Staff Emeritus
Starting over, then.
That is not a good model for how planets or stars form. It fails to explain why even though the Sun has 99.9% of the total mass of the solar system, it only has 1% of the solar system's total angular momentum. This is called the angular momentum problem.

A better model is the solar nebula disc model. The star and planets eventually form from some nebular cloud. Assume that the total angular momentum of this cloud with respect to the cloud's center of mass is non-zero. The star starts forming due to some uneven mass distribution within the cloud. As the protostar accumulates more mass, the increasing gravitational force will cause the nebular cloud to flatten out and form a rotating disc. The dust that has angular momentum with respect to this protostar will tend to end up somewhere on the disc. The dust that has near-zero angular momentum is what falls into the growing star. This nicely solves the angular momentum problem.

This spinning cloud does not explain planet formation, either. Planets form after the disc has formed. The orbital velocity of an orbiting object is $\sqrt{G(M+m)/r}$. A growing protoplanet will have a slightly greater orbital velocity than will the dust with which it is co-orbiting. The growing protoplanet will plow through this dust cloud, gathering mass and clearing the vicinity as it does. The growing protoplanet also migrates inward, encountering new dust. The density of the material is thus slightly greater in the direction of the protostar. From the perspective of the planet, it is encountering a wind of dust that is slightly denser toward the star.

This is a fairly good explanation of the formation of gas giants, but not quite so good for the rocky planets. There are still some unknowns left in explaining the formation of the rocky planets. The Earth, with its huge Moon, is particularly problematic. The current hypothesis is that a Mars-sized protoplanet collided with the proto Earth shortly after the formation of the solar system was more or less complete.

16. Jan 5, 2012

Buckleymanor

I imagined that the energy was provided from outside the Earth+oceans+atmosphere system by the Suns evaporation of water.
With regards to the explanation of the formation the Moon.If there was a large collision you would expect the Moon to be spinning because of it and at more of a rate than the Earth.
How come it has not been able to maintain it's spin from the impact you would expect it to be conserved.

17. Jan 5, 2012

Staff: Mentor

The moon is not spinning as fast as it once was because of tidal locking.
http://en.wikipedia.org/wiki/Tidal_locking

18. Jan 5, 2012

D H

Staff Emeritus
That makes no sense.

As Drakkith already noted, there's tidally locking to consider.

19. Jan 6, 2012

Buckleymanor

Interesting though it seems unclear if the Moon should speed up it's rotation if it's not spinning as fast as it was.

From the cited wikipedia article,
There seems to be a contradiction when B starts of rotating too slowly how does the Moon Know it started rotating at a faster speed than it is now.
For all purposes when it slows down it won't "know" if it started faster or that this was it's starting speed.

Last edited by a moderator: Jan 6, 2012
20. Jan 6, 2012

D H

Staff Emeritus
There's no contradiction, and the Moon doesn't need to "know" anything. (How can it?) It's all in the math, and that math is only concerned with what is happening now.