The fLaw of Gravity

Hi, James

I'm no expert myself (probably even less than you) but one thing that is sure is that the earth's core must be more dense than the outer layers. If not because of gravity then at least from the fact that so many other layers are laying on 'top' of it. The more dense materials like iron etc. would have 'sinked' to the 'bottom' of the sphere because it is molten (or even just partially molten).
However, I agree with you that at a close distance from such a 'large' object the attractive force experienced cannot be deduced as been only located at the centre of the earth. The 'side' layer as seen from an observer must play some role, even if it is a small part. It is however far smaller than the downward force and thus not so obvious.
Another curious thing would be what happen if you move to the inside of the earth, like in a deep mine? If there are any miners reading this would they be able to confirm that the gravitational force 'inside' the earth might be slightly less. Most probably there should be but only very small.

Rudolf

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[James]: I've been wondering about the notion that a gravitational mass attracts other masses as if it's entire mass were at the geometric center of mass.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Every student of gravity wonders about this at some point, and most eventually check it out for themselves if it is not assigned as homework. It is not too difficult to set up the integrals to prove analytically and rigorously that any uniform-density spherical shell produces zero gravitational force anywhere inside and exactly an inverse-square-from-center gravitational force anywhere outside.

Of course, the same must be true for any sphere, which may be regarded as composed of numerous spherical shells. There is no need for different spherical shells to have the same density because the axiom is true for each uniform shell, regardless of its own density.

This is difficult to check numerically because, the closer the test point gets to the surface, the more the force on it is dominated by the nearest surface point, whose force is *not* the same as that of a uniform shell segment centered at the same place. This can be handled by a process called "regularization", which deals with the singularity problem as any point mass is approached indefinitely close, (causing its own force to approach infinity). -|Tom|-

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Rudolf]: Another curious thing would be what happen if you move to the inside of the earth, like in a deep mine?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

This result is also known theoretically and has been measured. Curiously, Earth's gravity gets stronger as we go deeper inside, down through the crust and well into the mantle. This is because the effect of our getting closer to the dense core is greater than the effect of loss of all attraction from the shells of matter above us. But eventually, at greater depths, the force diminishes, approaching zero at Earth's center. -|Tom|-

All this topic doesn't cost the strokes of keyboard, actually. Every doubter should just draw the force lines picture for a sphere and for a point mass to see they are identical at radii larger than the sphere radius. It's all just simple geometry and symmetry, a very palpable verity in itself, and used to be taught at elementary school...

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>

[Tom] Every student of gravity wonders about this at some point, and most eventually check it out for themselves if it is not assigned as homework. It is not too difficult to set up the integrals to prove analytically and rigorously that any uniform-density spherical shell produces zero gravitational force anywhere inside and exactly an inverse-square-from-center gravitational force anywhere outside.

<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Thanks, Tom. I've spent the day going over this problem thinking I must have overlooked something obvious. I finally found it. In the program above, I had used the entire (x,y,z) acceleration vector instead of just the x component. It then became a simple matter of multiplying the acceleration vector by the normalised x component and everything fell into place. The agreement between between the body centered and distributed mass calculations agreed to within 7 significant digits at ORBIT = 2. I believe this validates the method employed, though nobody seemed to pick up on what the actual problem was. For reference, here is the corrected program:



<pre id=code><font face=courier size=2 id=code>
Option Explicit

Const INCREMENT = 0.001
Const ORBIT = 1.001
Private Continue As Boolean

Private Sub Command1_Click()
Dim X As Double, Y As Double, Z As Double, YZ As Double
Dim MassIncr As Double
Dim Mass As Double
Dim Grav As Double
Dim X_Normal As Double

'Use a second button click to abort
If Continue = True Then Continue = False: Exit Sub
Continue = True

'Math time
MassIncr = INCREMENT ^ 3
For Z = -1 To 1.000001 Step INCREMENT
If Continue = False Then Exit Sub
For Y = -1 To 1.000001 Step INCREMENT
YZ = Y * Y + Z * Z
For X = -1 To 1.000001 Step INCREMENT
If X * X + YZ <= 1 Then
Mass = Mass + MassIncr
'*****Corrected code
X_Normal = (X + ORBIT) / Sqr((X + ORBIT) * (X + ORBIT) + YZ)
Grav = Grav + MassIncr / ((X + ORBIT) * (X + ORBIT) + YZ) * X_Normal
'*****End corrected code
End If
Next X
Next Y

'Show some math on Z increments
Label1.Caption = Z
Label2.Caption = Mass / (ORBIT * ORBIT)
Label3.Caption = Grav
If Mass <> 0 Then
Label4.Caption = Grav / (Mass / (ORBIT * ORBIT))
End If
DoEvents

Next Z

'Indicate we are done computing
Continue = False
Command1.Caption = "Done"
End Sub
</font id=code></pre id=code>

Now, for Agora, you didn't see the error in the program and you cited symmetry of force vectors as an equivalence of magnitudes. Baaad Agora, baaad...

James

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
Now, for Agora, you didn't see the error in the program and you cited symmetry of force vectors as an equivalence of magnitudes. Baaad Agora, baaad...
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

James,

I haven't even looked at your proggie. I don't need to. I'm well aware of the kind of error you've made just from the result. The error is typical in the particular exercize and quite laughable at that. The magnitude is represented by the density of force lines, the total number of force lines represents the charge or mass value. So beat it already!