
#19
Jan913, 12:13 PM

PF Gold
P: 244

Vidar 



#20
Jan913, 12:49 PM

P: 503

"The instantaneous acceleration is independent from its mass."
Does this require treating it as a point mass? I'm thinking that spherical body "A" is placed at rest wrt body "B"... The center of A lies on the surface of a sphere "b" of equal radii from B... That surface "b" curves through A and we can distinguish the part of A's volume that is inside b and outside b... Half of A's volume is outside "b" when "b" approaches infinity, Majority of A's volume is outside b for finite radii from B, this proportion increases with A approaching B. How is the "point" mass A's actual changing distribution of acceleration treated here? 



#21
Jan913, 02:32 PM

HW Helper
P: 3,338





#22
Jan913, 02:38 PM

HW Helper
P: 3,338





#23
Jan913, 04:07 PM

P: 503

I understand the spherical mass being treated as a point mass, but what I'm asking is this:
For the spherical mass "A" and a body "B", look at the proportion of A's mass that is inside a shell defined so it's center is at B's center of mass and its radius is defined by the distance between the center of mass of B to the center of mass of A... that proportion of A's mass inside that shell changes with distance. When the distance is very large that proportion approaches 1/2 When the distance is decreased the proportion of A's mass inside the shell decreases. If we allow B to be a true point mass so that A's surface may approach all the way to contact with B's center of mass, then the shell radius becomes the radius of A itself. The volume of intersection of the shell and A becomes (pi*5r^3)/12 So the proportion of A inside the shell is ((pi*5r^3)/12)/(4/3(pi*r^3)) = 5/16 How does the inverse square relation account for this (I'm sure it does)? In other words, how does B's tidal effect of gravitational inverse square gradient across the volume of A match the changing distribution of A's mass that falls within and outside the sphere of radius Bcom to Acom? 



#24
Jan1013, 02:32 PM

HW Helper
P: 3,338

I don't see any significance. Sorry, maybe there is, but I can't see it.




#25
Jan1013, 04:59 PM

P: 503

The significance is that whereas the inverse square relation applies to true point masses, I'm not certain that it applies exactly to actual extended masses... the physical distribution of mass in an object stays fixed as the influence on that distribution varies with distance  I'm trying to see how the inverse square relation would match this or not.
I'm sure that this has been examined at some time in the history of gravitational development and that the answer is probably "yes, it matches perfectly", I just have not seen that demonstrated... 



#26
Jan1013, 05:30 PM

Mentor
P: 40,905





#27
Jan1013, 06:28 PM

P: 503

I'm not sure what you mean by "(for points external to the mass)"...
If you mean wrt another external mass' gravitation, then yes. The simplification of using a point mass for calculation assumes that the point mass will accelerate identically to the center of mass of an actual extended version of the body, or that this has been mathematically demonstrated. The simplification is so that the calculations far all the distributed location of mass in an extended body can be avoided. At first glance I'm not seeing the justification for this. For example, with changing distance between the mass sphere A and point body B, the distribution and proportion of A's mass within the radial distance to B is changing. I'm not seeing how this change is accounted for... basically I'm asking if the inverse square relation applies exactly for point masses, but has a small error factor when employed for extended masses, especially at close proximity. Maybe I'm not being clear because I have been using the example of a sphere. Take the example of a long rod... In a gravity free environment we measure and mark the center of mass of the rod. Now we put the rod in proximity to another object and place the orientation of the rod to be aligned radially to the other object. I can see how the measured center of mass would correspond to the point mass location applying the inverse square relation if the rod was a line (a "mass line") without width. It seems to me that as the rod's width is increased there will be a problem aligning the rod's center of mass with the location of the point mass acting gravitationally because the volume of the rod begins to present unbalanced proportions inside and outside the sphere of constant radius from the other object. I'm not seeing that discrepancy match the difference in acceleration across the length of the rod... Here is a specific example to see what I mean: Let the rod have a width. Call the end of the rod nearest to other body "N". That end N is a circle area on the end of the rod. For the distance from the other body to N, only the point in the center of the circle area at N is being influenced at the magnitude of the acceleration for that distance. The place on the rod at which the circle's area's circumference lies on the sphere about the other body occurs a little further up the rod. Call that circle's center M. Now repeat at the far end of the rod approaching the center. Call the center of the circle at the end of the rod "n" and the center of the circumference of the circle that is a section of the sphere about the other body when the sphere surface is at "n", call that point "m". Because the radius of curvature is different for the sphere centered about the other body for the two ends of the rod with width, NM > nm There seems to be a net displacement of the gravitational point mass location wrt the static center of mass. The longer the rod, the wider the rod, the more this presents itself. I was using spheres originally because they are the "fattest" volume in all directions. Does it help see what I'm pointing to using the rod as a simpler example? 



#28
Jan1113, 06:24 AM

HW Helper
P: 3,338

But luckily, for any two sphericallysymmetric mass distributions, we can apply the inverse square law to their centres of masses, and we will get the exact same answer as if we had done the whole calculation exactly. What I mean by 'exact' calculation, is that if we have some rigid body, in some external gravitational field [itex]\vec{g}[/itex] then [tex]\int \rho \vec{g} dV [/tex] Gives us the explicit formula to get force on a rigid body body due to an external gravitational field acting on that rigid body. I make the distinction 'external' gravitational field because there is also the gravitational field on parts of the rigid body due to other parts (of the same body), but since gravitational fields add up linearly, and we assume that the rigid body without the external gravitational field is not accelerating, then we can ignore any 'selfgravity'. Edit: wikipedia has a page on 'shell theorem', which seems to be related to this kind of stuff. 



#29
Jan1113, 01:29 PM

P: 503

Thanks BruceW, exactly what I was looking for!



Register to reply 
Related Discussions  
Subtracting centrifugal acceleration from acceleration caused by movement  General Physics  1  
acceleration and deacceleration using the 3axis acceleration data  Engineering Systems & Design  1  
Centripetal Acceleration and Tangential Acceleration problem driving me crazy!  Introductory Physics Homework  1  
Fluid Mechanics local acceleration and convective acceleration  Advanced Physics Homework  1 