A large number of identical point masses m ar | vedclass

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(B) Let $F_1, F_2, F_4, F_8, \dots$ be the gravitational forces exerted on the mass at the origin by the masses $m$ located at $x = 1, 2, 4, 8, \dots$

Vedclass · Accepted Answer

$\frac{4}{3} G m^2$

Vedclass · Answer

(B) Let $F_1, F_2, F_4, F_8, \dots$ be the gravitational forces exerted on the mass at the origin by the masses $m$ located at $x = 1, 2, 4, 8, \dots$ respectively.Using Newton's law of gravitation,$F = \frac{G m_1 m_2}{r^2}$,the forces are:$F_1 = \frac{G m^2}{1^2} = G m^2$$F_2 = \frac{G m^2}{2^2} = \frac{G m^2}{4}$$F_4 = \frac{G m^2}{4^2} = \frac{G m^2}{16}$$F_8 = \frac{G m^2}{8^2} = \frac{G m^2}{64}$The total force $F_{total}$ is the sum of these forces:$F_{total} = G m^2 \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots \right)$The term in the bracket is an infinite geometric progression ($G$.$P$.) with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.The sum of an infinite $G$.$P$. is given by $S = \frac{a}{1 - r}$.$S = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.Therefore,$F_{total} = G m^2 \left( \frac{4}{3} \right) = \frac{4}{3} G m^2$.

$A$ large number of identical point masses $m$ are placed along the $x$-axis at $x = 0, 1, 2, 4, 8, \dots$ units. The magnitude of the total gravitational force on the mass at the origin $(x = 0)$ will be:

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