An infinite number of point masses each equal to m are placed at x =1. x = 2, x = 4, x = 8 ……… What is the total gravitational potential at x = 0

The correct option is -2Gm.

Step-by-Step Explanation:

To find the total gravitational potential at the origin due to an infinite number of point masses, we can use the principle of superposition. According to this principle, the total gravitational potential at a point is the algebraic sum of the gravitational potentials produced by each individual mass at that point.

The gravitational potential $V$ at a distance $r$ from a point mass $m$ is given by the formula:
$V=-\frac{Gm}{r}$
where $G$ is the universal gravitational constant.

Here, identical point masses of mass $m$ are placed along the x-axis at distances:
$x=1$, $x=2$, $x=4$, $x=8$, and so on.

The total gravitational potential at the origin ($x=0$) is the sum of the potentials due to each of these masses:
$V=V_1+V_2+V_3+\cdots$
Substituting the distances into our potential formula:
$V=-\frac{Gm}{1}-\frac{Gm}{2}-\frac{Gm}{4}-\frac{Gm}{8}-\cdots$

We can factor out the common term $-Gm$ from the expression:
$V=-Gm\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \right)$

The series inside the parentheses is an infinite geometric progression (GP):
$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$
Here, the first term ($a$) is $1$ and the common ratio ($r$) is $\frac{1}{2}$.

The sum ($S$) of an infinite geometric progression with $\left|r\right|<1$ is calculated using the formula:
$S=\frac{a}{1-r}$

Substituting the values of $a$ and $r$ into the sum formula:
$S=\frac{1}{1-\frac{1}{2}}=\frac{1}{\left(\frac{1}{2}\right)}=2$

Now, we substitute this sum back into our expression for the total gravitational potential:
$V=-Gm\left(2\right)=-2Gm$

Thus, the total gravitational potential at $x=0$ is $-2Gm$.