One for gravitation

sir with reference to #18

i had missed a pi
so for n=1000 ratio is close to pi

for n= something very great wont it tend to pi

in this can we have an approximation ??

The case n\gg 1 is interesting. As the number n of point masses increase, M_n increases even if the total mass of the system is fixed at M₀, i.e. m=M₀/n. So seemingly from our last result the time of collapse should tend to zero. That is, the more finely a given amount of matter is spread around a circle, the shorter the time it takes for it to collapse under its own gravitational attraction. However, there is no point in examining a continuous matter distribution spread an arbitrary thin line; the extent of the matter in the transverse, i.e. radial direction cannot be neglected.

To finally conclude the post.
From the symmetry of the configuration and the initial conditions, it is easy to see that all bodies will fall towards the center of the n-gon with the same non-uniform acceleration. The structure maintains its initial shape (but the size reduces)
The net force on any one particle, say the n-th one due to all the others, is directed towards the center of the circle in which the polygon is inscribed, and is given by (as in #3)
F=G\dfrac{m^2}{r^2}\sum_{k=1}^{n-1}\dfrac{1}{4\sin(k\pi/n)}
when the distance of each is r from the center.
This force is identical to the gravitation attraction of fixed body at the center whose mass is
M_n=\dfrac{m}{4}\sum_{k=1}^{n-1}\dfrac{1}{\sin(k\pi/n)}
The values of the masses M_n (in units of m) can be evaluated for all n as
M₂=0.25,
M₃=.58
M₄=.96
......
M₁₀=3.86
......
However, there is no closed form. (in fact the summation cannot be written as an integral, since it diverges).
The time T of the collapse from an initial distance R onto a central mass M_n can be considered as half the period T_e for a degenerate elliptical orbit of semi-major axis R/2. The period T_c of a circular orbit of radius R can be calculated directly from the dynamics of circular motion as
T_c=\sqrt{\left(\dfrac{4\pi^2}{GM_n}\right)R^3}
But according to Kepler's third law,
\left(\dfrac{T_e}{T_c}\right)^2=\left(\dfrac{R/2}{R}\right)^3
Thus, finally, we obtain the time of collapse as
\boxed{T=\pi\sqrt{\dfrac{R^3}{8GM_n}}}

sorry i dint mention that

R is changing and i took R initial ∞ to make things simpler

@ #18 yes i missed a pi (typing hurriedly rest is correct)

Celestine and rohan:
How did you get the transition
\sum_{k=1}^{n-1}\csc\left(\dfrac{k\pi}{n}\right)\longrightarrow n \int_{\pi/n}^{(n-1)\pi/n}\csc\theta \ \mathrm{d}\theta
as n→∞?
In fact there is a vast difference. For n = 1000, the left side is approximately 4477.59 while the right side evaluates to 12912.3.

maybe .////

reason why sir posted n>>1 instead of n→∞ is that at n→∞ , T =0

so however wen n>>1 the abv aproximations yield a definite ans

even i was wasting time finding summation of cosec series but cudnt reach that

see second part of Q hes asking when n>>1 i jus took it infinity to make ans simpler

and i dont know a method to simplify that cosec series

post it if possible