## Y2.00719178K (Mar 05 2007)

Message boards : Technical News : Y2.00719178K (Mar 05 2007)

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Metod, S56RKO
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Message 540273 - Posted: 3 Apr 2007, 6:19:31 UTC - in response to Message 540181.
Last modified: 3 Apr 2007, 6:27:54 UTC

If I recall correctly (it has been quite a while), as long as the center of mass is a particular distance away, and the object is spherical, it does not matter what size the sphere is as long as the radius of the sphere does not exceed the distance from the center of mass to the nearest point of the other object.

Spherical shells of different mass densities all look like a mass at a single point from outside of the outermost sphere.

Almost correct. If a spherical body's density is homogenous, then indeed it seems to be a single-point body regarding the direction of the gravitational force. However, the size of gravitational force is not exactly the same. One has to remember how it is properly calculated: it's a volume integral of (mass of a part of the body) divided by (square of distance between part of this body and the other body) multiplied by gravitational constant multiplied by mass of the other object. If the distance between two bodies is not much more than diameter of the larger one, then the distance between parts of bodies (when calculating gravitational force) varies quite a lot.
Example: when you stand on surface of Earth, then gravitational force of the cubic kilometre (or cubic mile if you like) you stand on is quite bigger than gravitational force of the cubic kilometre on the other side of planet (which tends to be some 6400 kilometres away).

If the body's density is not homogenous, then even direction of gravitational force is not the same as direction towards mass centre of the body when the distance between two bodies is not large enough. There are quite some places on Earth where direction of gravitational force deviates from direction towards mass centre of Earth. I'm not quite sure about the magnitude of deviation but I think it's in order of a couple of degrees.
[/edit]

I don't know what a plumb bob does at temperate latitudes. Does it hang exactly vertically?

Actually it does. This really depends on definition of horizontality, but mostly it is attributed to the direction of surface of water - eg. sea or lake. If you remember that water is also affected by the same forces then it's quite obvious that the shape of planet Earth (which' surface is 70% covered by water and all the rest is not really rigid sphere but rather small solid shelves floating on surface of a giant droplet of molten metals and minerals) reflects the equipotential surface of sum of all the forces - mainly gravity and centrifugal force. Extraterrestrial forces (such as Sun's gravity, Lunar gravity) average out over longer period of time due to rotation of Earth.
Metod ...
ID: 540273 ·
John McLeod VII
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Message 540741 - Posted: 4 Apr 2007, 3:57:52 UTC - in response to Message 540273.

If I recall correctly (it has been quite a while), as long as the center of mass is a particular distance away, and the object is spherical, it does not matter what size the sphere is as long as the radius of the sphere does not exceed the distance from the center of mass to the nearest point of the other object.

Spherical shells of different mass densities all look like a mass at a single point from outside of the outermost sphere.

Almost correct. If a spherical body's density is homogenous, then indeed it seems to be a single-point body regarding the direction of the gravitational force. However, the size of gravitational force is not exactly the same. One has to remember how it is properly calculated: it's a volume integral of (mass of a part of the body) divided by (square of distance between part of this body and the other body) multiplied by gravitational constant multiplied by mass of the other object. If the distance between two bodies is not much more than diameter of the larger one, then the distance between parts of bodies (when calculating gravitational force) varies quite a lot.
Example: when you stand on surface of Earth, then gravitational force of the cubic kilometre (or cubic mile if you like) you stand on is quite bigger than gravitational force of the cubic kilometre on the other side of planet (which tends to be some 6400 kilometres away).

If the body's density is not homogenous, then even direction of gravitational force is not the same as direction towards mass centre of the body when the distance between two bodies is not large enough. There are quite some places on Earth where direction of gravitational force deviates from direction towards mass centre of Earth. I'm not quite sure about the magnitude of deviation but I think it's in order of a couple of degrees.
[/edit]

If each spherical shell is of a consistent mass density, the direction of the gravitational force does not vary from a point at the center of the spherical shells. I agree that if the mass density is not symetrically distributed then the direction of the force would be changed.

Here is a cute one. If you are INSIDE of a spherical shell that has mass, and the inside does not have mass, it does not matter where you are inside of the spherical shell, you will be in 0 Gravity. If you are on the outside of the spherical shell, you will be feeling the gravity of the sphere.

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ID: 540741 ·
RandyC

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Message 540826 - Posted: 4 Apr 2007, 11:29:29 UTC - in response to Message 540741.

If each spherical shell is of a consistent mass density, the direction of the gravitational force does not vary from a point at the center of the spherical shells. I agree that if the mass density is not symetrically distributed then the direction of the force would be changed.

Here is a cute one. If you are INSIDE of a spherical shell that has mass, and the inside does not have mass, it does not matter where you are inside of the spherical shell, you will be in 0 Gravity. If you are on the outside of the spherical shell, you will be feeling the gravity of the sphere.

I have to disagree with that one. Because of the distance-square(?) rule for gravity, the local gravity of the shell will exert a stronger force than the gravity of the shell farther away. Only when you travel to the center of the shell will you attain zero gravity.
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Metod, S56RKO
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Message 540852 - Posted: 4 Apr 2007, 13:03:55 UTC - in response to Message 540826.

Here is a cute one. If you are INSIDE of a spherical shell that has mass, and the inside does not have mass, it does not matter where you are inside of the spherical shell, you will be in 0 Gravity. If you are on the outside of the spherical shell, you will be feeling the gravity of the sphere.

I have to disagree with that one. Because of the distance-square(?) rule for gravity, the local gravity of the shell will exert a stronger force than the gravity of the shell farther away. Only when you travel to the center of the shell will you attain zero gravity.

No, John was right. Mass of sphere you see in some spherical angle is proportional to the size of surface, covered by the same spherical angle. Which is directly proportional to the distance between viewpoint and sphere. On the other hand, gravitational force of a mass is inversely proportional to the distance between two masses. In this case distances cancel out.

Humpf, I expressed my self rather worse. I have an excuse though: English is not my native language.
Therefore an example: if you take a part of sphere that can be seen under say 45Ëš angle horizontally and 45Ëš vertically, then the mass is proportional to
vertical_angle x horizontal_angle x distance
Now, if you are within a sphere and look at the closest 'wall', you'll see certain amount of sphere there. If you then look at the farthest 'wall' and the distance to the farthest part of sphere is, say, 4 times longer, then you'll see 4 times more sphere surface.

Metod ...
ID: 540852 ·
Odysseus
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Message 541270 - Posted: 5 Apr 2007, 7:25:38 UTC - in response to Message 540852.

Now, if you are within a sphere and look at the closest 'wall', you'll see certain amount of sphere there. If you then look at the farthest 'wall' and the distance to the farthest part of sphere is, say, 4 times longer, then you'll see 4 times more sphere surface.

Yes; another way to put it might be that when you move away from the centre, the increased attraction of the closer parts is exactly balanced by the increased area of the opposite side: the further from the centre you go, the smaller the â€˜near, strongâ€™ region becomes in comparison to the â€˜distant, weakâ€™ regions. A sphere is the only geometric form for which this is true, having in effect a â€˜square lawâ€™ of areas to precisely complement the â€˜inverse-square lawâ€™ of gravity.
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ML1
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Message 541325 - Posted: 5 Apr 2007, 11:11:59 UTC - in response to Message 541270.
Last modified: 5 Apr 2007, 11:12:35 UTC

Now, if you are within a [hollow] sphere...

... when you move away from the centre, the increased attraction of the closer parts is exactly balanced by the increased area of the opposite side: the further from the centre you go, the smaller the â€˜near, strongâ€™ region becomes in comparison to the â€˜distant, weakâ€™ regions. A sphere is the only geometric form for which this is true, having in effect a â€˜square lawâ€™ of areas to precisely complement the â€˜inverse-square lawâ€™ of gravity.

Thanks for a very good clear explanation.

Nicely avoids some mental/mathematical gymnastics working that one out!

Interesting.

Regards,
Martin

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Message boards : Technical News : Y2.00719178K (Mar 05 2007)

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