Question

Infinite number of bodies, each of mass $2Kg$ are situated on $x - axis$ at distance $1m,2m,4m,8m,...........$ respectively, from the origin. The resulting gravitational potential due to this system at the origin will be
A. $- G$
B. $- \dfrac{8}{3}G$
C. $- \dfrac{4}{3}G$
D. $- 4G$

Answer 1

The work obtained in bringing a body from infinity to a point in the gravitational field is known as gravitational potential energy. Firstly find the potential at distance ‘x’ and then find the value of potential at origin. Apply the GP series formula, in which we require a and r values. By substituting both values we can determine the resulting gravitational potential due to this system at the origin

Formula Used-:
Gravitational potential energy is $V = - \dfrac{{Gm}}{r}$
Where, $V$ is potential energy, $G$ is gravity,$m$ is mass and $r$ is distance.
As the series is GP, so sum of GP series will be- $\dfrac{a}{{1 - r}}$

Complete step by step answer:
consider the gravitational potential at a distance $r$ = $V = - \dfrac{{Gm}}{r}$
Where, $V$ is potential energy, $G$ is gravity,$m$ is mass and $r$ is distance.

Gravitational potential at origin is = $- 2G\left( {\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ..........} \right)$
$= - 2G\left( {\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + .......} \right)$
As we know that the series is $GP$series.
Where, $a = 1$ and $r = \dfrac{1}{2}$
We also know that sum of $GP$series = $\dfrac{a}{{1 - r}}$
Substitute the value of$a$and$r$, we get-

\therefore V = - 4G$$ **So, the option D is correct.** **Additional information:** Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field. In simple terms, it can be said that gravitational potential energy is an energy which is related to gravitational force or to gravity.The most common example that can help you understand the concept of gravitational potential energy is if you take two pencils. One is placed at the table and the other is held above the table. Now, we can state that the pencil which is high will have greater gravitational potential energy that the pencil that is at the table. **Note:** We can also use one alternative method in which we know that masses of $$2Kg$$ are situated on the x-axis when the distance is also given. By directly substituting the values in potential, the resultant potential will be calculated.

V = - \dfrac{{G \times 2}}{1} - \dfrac{{G \times 2}}{2} - \dfrac{{G \times 2}}{4} - \dfrac{{G \times 2}}{8} \\
\Rightarrow V = - 2G(2) = - 4G \\

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