One for gravitation

is the answer to the last part = √((Ï€³R³)/2GM₀ln(2n/Ï€)) ?

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}}}

one thing celestine ,

you wrote

vdv=KdR/R²

but then we have

v²/2=k[1/x - 1/R ]

where v=-dx/dt

but you wrote

v²/2=k/R !!!

@ #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.

Actually, you cant. That series diverges. So you cannot determine the sum.

Help me find the sum _k=1Σ^n-1 Cosec(kπ/n) for n->∞

Rohan said it is easy-------- but I haven't been able to figure it out yet.

you mean each particle attracted towards the center by a mass nm?

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

but cel ... its not given n --> infinity ....... we have to try and do summation of that cosec thing which is wat i was trying to do ................

okay metal, what about the time?

I think the particles will all move towards the centre. At any instant, the force on one particle is given by Summation Of { Gm² /(4x² Sin(k[pi]/n) } ;{ k varying from 1 to n-1}; towards the centre, where x is the distance of the particles from the centre.
The particles will therefore move such that they always form the
n-gon, which keeps on becoming smaller and smaller.