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A number of point masses each equal to 'm' are arranged along x-axis at $x = 1 m, x = 2 m,, x = 4 m,$ and so on. Another mass M is placed at the origin. Force experienced by 'M' due to infinite point masses is

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$\frac{4}{3} GMm$

$\frac{3}{4} GMm$

$\frac{GMm}{3}$

$GMm$

answer is A.

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Detailed Solution

Using principle of superposition,

$F = F_{1} + F_{2} + . . . . + F_{\infty}$

$\Rightarrow F = (\frac{{GMm}_{1}}{r_{1}^{2}} + \frac{{GMm}_{2}}{r_{2}^{2}} + \dots \infty)$

since, $m_{1} = m_{2} = m_{3} = . . . . . = m$

$\Rightarrow F = GMm (\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \dots \infty)$

$\Rightarrow F = GMm (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{4^{2}} + \dots \infty)$

$\Rightarrow F = GMm (\frac{1}{1} + \frac{1}{4} + \frac{1}{16} + \dots \infty)$

$\Rightarrow F = GMm (\frac{1}{1 - \frac{1}{4}}) = \frac{4}{3} GMm$

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