Question:

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

Updated On: Jun 6, 2022
  • $ G/2 $
  • $ G $
  • $ 2G $
  • $ 4G $
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The Correct Option is C

Solution and Explanation

Gravitational potential
$ V=GM\left( \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}}+...... \right) $
$ =G\times 1\left( \frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.... \right) $
$ =G\left( \frac{1}{1-1/2} \right) $
$ \left( \because sum\,of\,GP=\frac{a}{1-r} \right) $
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

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,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

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

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].