Question

a large number of identical point masses m are placed along x axis at x=0, 1, 2, 4,........ the magnitude of gravitational force on mass at origin (x=0) will be

Answer 1

gravitational force between any two point mass is given by formula

$F = \frac{Gm_1m_2}{r^2}$

now here we need to find the gravitational force on mass placed at x = 0

$F = \frac{Gm^2}{1^2} + \frac{Gm^2}{2^2} + \frac{Gm^2}{4^2}+.........$

$F = Gm^2(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{2^4} + .........)$

$F = Gm^2(\frac{1}{1 - \frac{1}{4}})$

$F = \frac{4Gm^2}{3}$

so above is the net gravitational force on it

Answer 2

Answer:

F = 4Gm^2/3

Explanation:

Gravitational force between any two point mass is given by formula:

                 F = Gm1m2/r^2

Now here we need to find the gravitational force on mass placed at x = 0

F = Gm^2/1^2 + Gm^2/2^2 + Gm^2/4^2 +.......

F = Gm^2(1/1 + 1/2^2 + 1/2^4 +........)

F = Gm^2(1/1-1/4)

F = 4Gm^2/3

Hope this answer helps you.