To find the resultant gravitational potential at the origin due to an infinite number of masses, each of 1 kg, 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-by-Step Solution:
1. **Understanding Gravitational Potential**:
The gravitational potential \( V \) at a point due to a mass \( m \) at a distance \( r \) is given by the formula:
\[
V = -\frac{Gm}{r}
\]
where \( G \) is the gravitational constant.
2. **Identifying the Positions**:
The masses are located at positions \( x = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \ldots \). This means we have masses at distances \( r = 1, 2, 4, 8, 16, \ldots \) from the origin.
3. **Calculating the Potential from Each Mass**:
The potential at the origin due to a mass at distance \( r \) is:
\[
V_r = -\frac{G \cdot 1}{r} = -\frac{G}{r}
\]
4. **Summing the Contributions**:
Since there are masses at both positive and negative distances, we can consider the contributions from positive distances and multiply by 2 (because the contributions from negative distances will be the same):
\[
V_{\text{total}} = 2 \left( -\frac{G}{1} - \frac{G}{2} - \frac{G}{4} - \frac{G}{8} - \ldots \right)
\]
5. **Recognizing 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 \( r = \frac{1}{2} \).
6. **Finding the Sum of the Series**:
The sum of an infinite geometric series is given by:
\[
S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{2}} = 2
\]
7. **Calculating the Total Potential**:
Substituting the sum back into the total potential:
\[
V_{\text{total}} = 2 \left( -G \cdot S \right) = 2 \left( -G \cdot 2 \right) = -4G
\]
8. **Final Result**:
The magnitude of the gravitational potential at the origin is:
\[
|V| = 4G
\]
### Conclusion:
The resultant gravitational potential at the origin due to the infinite number of masses is \( 4G \).
