- #1
Zula110100100
- 253
- 0
Is there one out there? Do we have any reason to believe we can treat other objects like point masses as well?
I ask because if you consider line-world, and there was a 4m segment with uniform density 3kg/m located with it's left end at (3), the center of mass would be at (5), and I am located at (0), the acceleration from gravity is:
[tex]A_{g} = \frac{Gm}{r^2{}}[/tex]
[tex]A_{g} = \frac{12G}{25} = .48G[/tex]
Now I cut the object in half, so I have two 2m segments, each still a uniform 3kg/m, and it hasn't moved, but the center of mass for the nearer segment is now at (4), and the other is at (6):
[tex]A_{g} = \frac{Gm_{1}}{r_{1}^2{}} + \frac{Gm_{2}}{r_{2}^2{}}[/tex]
[tex]A_{g} = \frac{6G}{16} + \frac{6G}{36} = .375G + .167G = .542G[/tex]
So then I tried to do it with an integral(not 100% sure I am doing it right) and this is what I got
[tex]\Delta A_{g} = \frac{G\rho \Delta x}{x^{2}}[/tex]
[tex]A_{g} = G\rho \int \frac{1}{x^{2}}dx[/tex]
[tex]A_{g} = G\rho \int^{7}_{3} \frac{1}{x^{2}}dx[/tex]
[tex]A_{g} = 3G(-\frac{1}{7}+\frac{1}{3}) = 3G(-\frac{3}{21}+\frac{7}{21}) = 3G(\frac{4}{21}) = \frac{12G}{21}\equiv .571G[/tex]
That's almost a fifth more than the original estimation, so one could assume if point mass fails in 1d, it could fail in some higher dimensional configuration. Unless there is an error above, or some reason why it works in 3d and not 1d...Anyone know this?
I ask because if you consider line-world, and there was a 4m segment with uniform density 3kg/m located with it's left end at (3), the center of mass would be at (5), and I am located at (0), the acceleration from gravity is:
[tex]A_{g} = \frac{Gm}{r^2{}}[/tex]
[tex]A_{g} = \frac{12G}{25} = .48G[/tex]
Now I cut the object in half, so I have two 2m segments, each still a uniform 3kg/m, and it hasn't moved, but the center of mass for the nearer segment is now at (4), and the other is at (6):
[tex]A_{g} = \frac{Gm_{1}}{r_{1}^2{}} + \frac{Gm_{2}}{r_{2}^2{}}[/tex]
[tex]A_{g} = \frac{6G}{16} + \frac{6G}{36} = .375G + .167G = .542G[/tex]
So then I tried to do it with an integral(not 100% sure I am doing it right) and this is what I got
[tex]\Delta A_{g} = \frac{G\rho \Delta x}{x^{2}}[/tex]
[tex]A_{g} = G\rho \int \frac{1}{x^{2}}dx[/tex]
[tex]A_{g} = G\rho \int^{7}_{3} \frac{1}{x^{2}}dx[/tex]
[tex]A_{g} = 3G(-\frac{1}{7}+\frac{1}{3}) = 3G(-\frac{3}{21}+\frac{7}{21}) = 3G(\frac{4}{21}) = \frac{12G}{21}\equiv .571G[/tex]
That's almost a fifth more than the original estimation, so one could assume if point mass fails in 1d, it could fail in some higher dimensional configuration. Unless there is an error above, or some reason why it works in 3d and not 1d...Anyone know this?