- #1
physics4all
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- TL;DR
- I wondered what would happen if one tried to calculate Newtonian Gravity between two bodies but in many-to-many style (like all atoms to all atoms), i.e. completely ignoring
Newton’s Shell theorem. Did not find anything online about anybody trying that.
So, I tried it myself and got curious results
I couldn't get Latex text to render. Sorry if text looks unreadable...
[Mentor Update -- OP has updated the below text using LaTeX]
Thus, for two perfect spheres with radiuses r and s, densities Pr and Ps, which are positioned in space at distance d between their respective centers, expression for Universal Gravity I got looks this way
(1) ##F = GPsPrπ^2Gg ##
Where Gg is a function of r,s and d and looks like this:
(2) \begin{align}
Gg = \frac{1}{30d}(&\nonumber \\
&(4d^3sr+44dsr^3+44ds^3r)\nonumber \\
&+(-20d^2r^3+20s^2r^3-4r^5)(ln(d+s+r)+ln(d+s-r)-ln(d-s+r)-ln(d-s-r))\nonumber \\
&+(-20d^2s^3-4s^5+20s^3r^2)(ln(d+s+r)-ln(d+s-r)+ln(d-s+r)-ln(d-s-r))\nonumber \\
&+(d^5-10d^3s^2-10d^3r^2-15ds^4-15dr^4+30ds^2r^2)(ln(d+s+r)-ln(d+s-r)-ln(d-s+r)+ln(d-s-r)))
\end{align}
Took many steps to find the integral. Probably need to post somewhere separately…
Took me some time to find a precise enough calculator for natural logs. As expected, once plugging in numbers for all planets in Solar system, plus spheres Cavendish used in his experiment, Gg is always equal to
(3) ##\frac{16s^3r^3}{9d^2}##
Plugging into (3) into (1) results in standard looking
(4) ## F = \frac{GMm}{d^2} ##
Indeed, using Taylor series to model each natural log, and simplifying (2), yields a very dominant (3) with extra ignorable branches which get smaller and smaller with every Taylor iteration.
However, interesting is what happens when d is so big compared to s and r, that it can be considered ## \infty ## . Lets refactor logs as following:
\begin{align}
(5) &(ln(d+s+r)+ln(d+s-r)-ln(d-s+r)-ln(d-s-r)) =\nonumber \\
&= ln(d^2+s^2-r^2+2ds) - ln(d^2+s^2-r^2-2ds) =\nonumber \\
&= ln(d^2+s^2-r^2+2ds) - ln(d^2+s^2-r^2-2ds) + ln(d^2+s^2-r^2) - ln(d^2+s^2-r^2) =\nonumber \\
&= ln(\frac{d^2+s^2-r^2+2ds}{d^2+s^2-r^2}) - ln(\frac{d^2+s^2-r^2-2ds}{d^2+s^2-r^2}) =\nonumber \\
&= ln(1 + \frac{2ds}{d^2+s^2-r^2}) - ln(1 - \frac{2ds}{d^2+s^2-r^2})
\end{align}
Now, applying limit
\begin{align}
(6.1) lim_{d \rightarrow +\infty}{ ( ln(1 + \frac{2ds}{d^2+s^2-r^2}) - ln(1 - \frac{2ds}{d^2+s^2-r^2})) }= ln(e^\frac{2ds}{d^2+s^2-r^2}) - ln(e^\frac{-2ds}{d^2+s^2-r^2}) = \frac{4ds}{d^2+s^2-r^2} \nonumber
\end{align}
The same transformations for other log expressions
(6.2) ## lim_{d \rightarrow +\infty}{ (ln(d+s+r)-ln(d+s-r)+ln(d-s+r)-ln(d-s-r)) }= \frac{4dr}{d^2-s^2+r^2} ##
(6.3) ## lim_{d \rightarrow +\infty}{ (ln(d+s+r)-ln(d+s-r)-ln(d-s+r)+ln(d-s-r)) } = \frac{4sr}{d^2-s^2-r^2}##
Doing the same transformations to other logs and plugging everything into (1) and simplifying, produces (I skip intermediate steps, otherwise text seems to be too busy)
\begin{align}
(7) Gg = &\frac{ \frac{160ds^3r^3}{d^2 +s^2 -r^2}+\frac{160ds^3r^3}{d^2 -s^2 +r^2}- \frac{48ds^3r^3}{d^2 -s^2 -r^2}+\frac{96dsr^5}{d^2 -s^2 -r^2}-\frac{96dsr^5}{d^2 +s^2 -r^2}+\frac{96ds^5r}{d^2 -s^2 -r^2}-\frac{96ds^5r}{d^2 -s^2 +r^2} } {30d}
\end{align}
Whichin case of d >> s > r is roughly
(8) ##\frac{272s^3r^3}{30d^2} ##
(8) is 5.1 times bigger than (3), making Gravitational pull 5+ times stronger at a very long distance, even though still very small. So matter seems 5 times heavier. Pretty much the same ratio as Dark matter to baryonic matter that is needed to explain higher velocity.
Now, orbital velocity based on (1) I believe would be (need to remove density of orbiting body and its density)
(9) ##V = \sqrt{\frac{GPsπGgd}{\frac{4}{3}πr^3}} ##
Lets take derivative of (9) (ignoring G, Ps, and π for simplicity) from above:
\begin{align}
(10)
V’=&\frac{((5d^4+(-30s^2-30r^2)d^2-15s^4+30r^2s^2-15r^4)ln(\frac{(d-s-r)(d+s+r)}{(d-s+r)(d+s-r)})\nonumber \\
-40s^3dln(\frac{(d-s+r)(d+s+r)}{(d-s-r)(d+s-r)})\nonumber \\
-40r^3dln(\frac{(d+s-r)(d+s+r)}{(d-s-r)(d-s+r)})\nonumber \\
+20rsd^2+60rs^3+60r^3s)}\nonumber \\
&{4\sqrt{10}r^\frac{3}{2}\sqrt{(d^5-10s^2d^3-10r^2d^3-15s^4d+30r^2s^2d-15r^4d)ln(\frac{(d-s-r)(d+s+r)}{(d-s+r)(d+s-r)})\nonumber \\
+(-20s^3d^2-4s^5+20r^2s^3)ln(\frac{(d-s+r)(d+s+r)}{(d-s-r)(d+s-r)})\nonumber \\
+(-20r^3d^2+20r^3s^2-4r^5)ln(\frac{(d+s-r)(d+s+r)}{(d-s-r)(d-s+r)})\nonumber \\
+(4rsd^3+44rs^3d+44r^3sd)}}
\end{align}
Using Taylor series, produces exactly the same negative acceleration as 1\nonumber \\sqrt(d) would do (plus some small ignorable branches):
(11) ## V’ = -\frac{0.6\sqrt(s^3)}{d\sqrt(d)} ##
However, using the same log transformations (5), at very large distance d, acceleration flips to positive and becomes a very small
(12) ##V’ = \frac{5\sqrt(s^3)}{3d\sqrt(d)} ##
In other words, at very large distances, Gravitational pull becomes more efficient and makes a smaller body to rotate around a much bigger body with pretty much constant velocity.
Granted, Universal Gravity is largely abandoned in favor of General Relativity, but still interesting.
P.S. After I wrote this text,, an idea came to see what happens at distances very close to a massive body, i.e. d~(s+r). Turns out in this case (1) is bigger than (4) too. Exactly as Einstein predicted. Another coincidence…
[Mentor Update -- OP has updated the below text using LaTeX]
Thus, for two perfect spheres with radiuses r and s, densities Pr and Ps, which are positioned in space at distance d between their respective centers, expression for Universal Gravity I got looks this way
(1) ##F = GPsPrπ^2Gg ##
Where Gg is a function of r,s and d and looks like this:
(2) \begin{align}
Gg = \frac{1}{30d}(&\nonumber \\
&(4d^3sr+44dsr^3+44ds^3r)\nonumber \\
&+(-20d^2r^3+20s^2r^3-4r^5)(ln(d+s+r)+ln(d+s-r)-ln(d-s+r)-ln(d-s-r))\nonumber \\
&+(-20d^2s^3-4s^5+20s^3r^2)(ln(d+s+r)-ln(d+s-r)+ln(d-s+r)-ln(d-s-r))\nonumber \\
&+(d^5-10d^3s^2-10d^3r^2-15ds^4-15dr^4+30ds^2r^2)(ln(d+s+r)-ln(d+s-r)-ln(d-s+r)+ln(d-s-r)))
\end{align}
Took many steps to find the integral. Probably need to post somewhere separately…
Took me some time to find a precise enough calculator for natural logs. As expected, once plugging in numbers for all planets in Solar system, plus spheres Cavendish used in his experiment, Gg is always equal to
(3) ##\frac{16s^3r^3}{9d^2}##
Plugging into (3) into (1) results in standard looking
(4) ## F = \frac{GMm}{d^2} ##
Indeed, using Taylor series to model each natural log, and simplifying (2), yields a very dominant (3) with extra ignorable branches which get smaller and smaller with every Taylor iteration.
However, interesting is what happens when d is so big compared to s and r, that it can be considered ## \infty ## . Lets refactor logs as following:
\begin{align}
(5) &(ln(d+s+r)+ln(d+s-r)-ln(d-s+r)-ln(d-s-r)) =\nonumber \\
&= ln(d^2+s^2-r^2+2ds) - ln(d^2+s^2-r^2-2ds) =\nonumber \\
&= ln(d^2+s^2-r^2+2ds) - ln(d^2+s^2-r^2-2ds) + ln(d^2+s^2-r^2) - ln(d^2+s^2-r^2) =\nonumber \\
&= ln(\frac{d^2+s^2-r^2+2ds}{d^2+s^2-r^2}) - ln(\frac{d^2+s^2-r^2-2ds}{d^2+s^2-r^2}) =\nonumber \\
&= ln(1 + \frac{2ds}{d^2+s^2-r^2}) - ln(1 - \frac{2ds}{d^2+s^2-r^2})
\end{align}
Now, applying limit
\begin{align}
(6.1) lim_{d \rightarrow +\infty}{ ( ln(1 + \frac{2ds}{d^2+s^2-r^2}) - ln(1 - \frac{2ds}{d^2+s^2-r^2})) }= ln(e^\frac{2ds}{d^2+s^2-r^2}) - ln(e^\frac{-2ds}{d^2+s^2-r^2}) = \frac{4ds}{d^2+s^2-r^2} \nonumber
\end{align}
The same transformations for other log expressions
(6.2) ## lim_{d \rightarrow +\infty}{ (ln(d+s+r)-ln(d+s-r)+ln(d-s+r)-ln(d-s-r)) }= \frac{4dr}{d^2-s^2+r^2} ##
(6.3) ## lim_{d \rightarrow +\infty}{ (ln(d+s+r)-ln(d+s-r)-ln(d-s+r)+ln(d-s-r)) } = \frac{4sr}{d^2-s^2-r^2}##
Doing the same transformations to other logs and plugging everything into (1) and simplifying, produces (I skip intermediate steps, otherwise text seems to be too busy)
\begin{align}
(7) Gg = &\frac{ \frac{160ds^3r^3}{d^2 +s^2 -r^2}+\frac{160ds^3r^3}{d^2 -s^2 +r^2}- \frac{48ds^3r^3}{d^2 -s^2 -r^2}+\frac{96dsr^5}{d^2 -s^2 -r^2}-\frac{96dsr^5}{d^2 +s^2 -r^2}+\frac{96ds^5r}{d^2 -s^2 -r^2}-\frac{96ds^5r}{d^2 -s^2 +r^2} } {30d}
\end{align}
Whichin case of d >> s > r is roughly
(8) ##\frac{272s^3r^3}{30d^2} ##
(8) is 5.1 times bigger than (3), making Gravitational pull 5+ times stronger at a very long distance, even though still very small. So matter seems 5 times heavier. Pretty much the same ratio as Dark matter to baryonic matter that is needed to explain higher velocity.
Now, orbital velocity based on (1) I believe would be (need to remove density of orbiting body and its density)
(9) ##V = \sqrt{\frac{GPsπGgd}{\frac{4}{3}πr^3}} ##
Lets take derivative of (9) (ignoring G, Ps, and π for simplicity) from above:
\begin{align}
(10)
V’=&\frac{((5d^4+(-30s^2-30r^2)d^2-15s^4+30r^2s^2-15r^4)ln(\frac{(d-s-r)(d+s+r)}{(d-s+r)(d+s-r)})\nonumber \\
-40s^3dln(\frac{(d-s+r)(d+s+r)}{(d-s-r)(d+s-r)})\nonumber \\
-40r^3dln(\frac{(d+s-r)(d+s+r)}{(d-s-r)(d-s+r)})\nonumber \\
+20rsd^2+60rs^3+60r^3s)}\nonumber \\
&{4\sqrt{10}r^\frac{3}{2}\sqrt{(d^5-10s^2d^3-10r^2d^3-15s^4d+30r^2s^2d-15r^4d)ln(\frac{(d-s-r)(d+s+r)}{(d-s+r)(d+s-r)})\nonumber \\
+(-20s^3d^2-4s^5+20r^2s^3)ln(\frac{(d-s+r)(d+s+r)}{(d-s-r)(d+s-r)})\nonumber \\
+(-20r^3d^2+20r^3s^2-4r^5)ln(\frac{(d+s-r)(d+s+r)}{(d-s-r)(d-s+r)})\nonumber \\
+(4rsd^3+44rs^3d+44r^3sd)}}
\end{align}
Using Taylor series, produces exactly the same negative acceleration as 1\nonumber \\sqrt(d) would do (plus some small ignorable branches):
(11) ## V’ = -\frac{0.6\sqrt(s^3)}{d\sqrt(d)} ##
However, using the same log transformations (5), at very large distance d, acceleration flips to positive and becomes a very small
(12) ##V’ = \frac{5\sqrt(s^3)}{3d\sqrt(d)} ##
In other words, at very large distances, Gravitational pull becomes more efficient and makes a smaller body to rotate around a much bigger body with pretty much constant velocity.
Granted, Universal Gravity is largely abandoned in favor of General Relativity, but still interesting.
P.S. After I wrote this text,, an idea came to see what happens at distances very close to a massive body, i.e. d~(s+r). Turns out in this case (1) is bigger than (4) too. Exactly as Einstein predicted. Another coincidence…
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