Consider an infinite distribution of point ma | vedclass

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Question

(A) The gravitational force $F$ acting on the point mass at the origin due to the infinite distribution of point masses at $x = r, 2r, 4r, 8r, \ldots$

Vedclass · Accepted Answer

$\frac{4Gm^2}{3r^2}$

Vedclass · Answer

(A) The gravitational force $F$ acting on the point mass at the origin due to the infinite distribution of point masses at $x = r, 2r, 4r, 8r, \ldots$ is given by the sum of individual gravitational forces:$F = \frac{Gm^2}{r^2} + \frac{Gm^2}{(2r)^2} + \frac{Gm^2}{(4r)^2} + \frac{Gm^2}{(8r)^2} + \ldots$$F = \frac{Gm^2}{r^2} \left[ 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots ight]$This is an infinite geometric series with the first term $a = 1$ and common ratio $q = \frac{1}{4}$.The sum of an infinite geometric series is $S = \frac{a}{1-q}$.$S = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.Therefore,$F = \frac{Gm^2}{r^2} 	imes \frac{4}{3} = \frac{4Gm^2}{3r^2}$.

Consider an infinite distribution of point masses (each of mass $m$) placed on the $x$-axis as shown in the diagram. What is the gravitational force acting on the point mass placed at the origin?

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