Question:

Infinite number of masses, each $1 \,kg$ are placed along the $x-$ axis at $x=\pm 1 m, \pm 2 m, \pm 4 m, \pm 8 m, \pm 16 \,m, \ldots$ the magnitude of the resultant gravitational potential in terms of gravitational constant $G$ at the origin $(x=0)$ is

Updated On: Jun 18, 2022
  • G/2
  • G
  • -2G
  • 4G
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The Correct Option is D

Solution and Explanation

Given each mass $= 1\,kg$
and are placed in the following way
we know that $V = \frac{GM}{r}$
where $m = $mass , $r = $ distance from origin
Now applying formula.
gravitational potential : $2\left(-\frac{G \times 1}{1}-\frac{G \times 1}{2} \cdots\right)$
$=2\left[\frac{G \times 1}{1}+\frac{G \times 1}{2}+\frac{G \times 1}{4}+\cdots\right]$
$=-2 G\left[1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots\right]$
$=-2 G\left[\frac{1}{1-\frac{1}{2}}\right]$
$=-2 G(2)=-4 G$
in $ \infty $ G.P
$ S_{\infty}=\frac{a}{1-r}$
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Concepts Used:

Newtons Law of Gravitation

Gravitational Force

Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.

Newton’s Law of Gravitation:

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
  • Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)

By combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]