Infinite number of masses, each of `1 kg`, are placed along the x-axis at `x = +- 1m, +- 2m, +-4m, +- 8m, +- 16m`.. The gravitational of the resultant gravitational potential in term of gravitaitonal constant `G` at the origin `(x = 0)` is

Options

A. `G//2`
B. `G`
C. `2G`
D. `4G`

Explanation

To find the resultant gravitational potential at the origin due to an infinite number of 1 kg masses placed along the x-axis at positions \( x = \pm 1m, \pm 2m, \pm 4m, \pm 8m, \pm 16m, \ldots \), we can follow these steps:

### Step 1: Understand the Gravitational Potential Formula
The gravitational potential \( V \) due to a mass \( m \) at a distance \( r \) is given by the formula:
\[ V = -\frac{Gm}{r} \]
where \( G \) is the gravitational constant.

### Step 2: Calculate the Potential from Each Mass
Since the masses are placed symmetrically about the origin, we can calculate the potential contributions from the positive and negative sides separately. The distances from the origin for the masses at \( x = 1, 2, 4, 8, 16, \ldots \) are \( 1, 2, 4, 8, 16, \ldots \).

For each mass at position \( x \):
- At \( x = 1m \): \( V_1 = -\frac{G \cdot 1}{1} = -G \)
- At \( x = 2m \): \( V_2 = -\frac{G \cdot 1}{2} = -\frac{G}{2} \)
- At \( x = 4m \): \( V_3 = -\frac{G \cdot 1}{4} = -\frac{G}{4} \)
- At \( x = 8m \): \( V_4 = -\frac{G \cdot 1}{8} = -\frac{G}{8} \)
- And so on...

### Step 3: Consider the Negative Side
The contributions from the negative side will be the same in magnitude but negative in direction:
- At \( x = -1m \): \( V_{-1} = -G \)
- At \( x = -2m \): \( V_{-2} = -\frac{G}{2} \)
- At \( x = -4m \): \( V_{-3} = -\frac{G}{4} \)
- At \( x = -8m \): \( V_{-4} = -\frac{G}{8} \)
- And so on...

### Step 4: Combine the Contributions
Since the potential is a scalar quantity, we can add the contributions from both sides:
\[ V_{\text{total}} = 2 \left( -G - \frac{G}{2} - \frac{G}{4} - \frac{G}{8} - \ldots \right) \]

### Step 5: Recognize the Series
The series inside the parentheses is a geometric series:
\[ S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \]
This series has a first term \( a = 1 \) and a common ratio \( r = \frac{1}{2} \). The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{2}} = 2 \]

### Step 6: Calculate the Total Potential
Substituting back into our equation for \( V_{\text{total}} \):
\[ V_{\text{total}} = 2 \left( -G \cdot 2 \right) = -4G \]

### Step 7: Find the Magnitude
Since the question asks for the magnitude of the gravitational potential at the origin:
\[ \text{Magnitude} = |V_{\text{total}}| = 4G \]

### Final Answer
The magnitude of the resultant gravitational potential at the origin is:
\[ \boxed{4G} \]