To calculate the resultant gravitational potential at the origin due to an infinite number of masses of 2 kg each placed along the x-axis at positions \( x = \pm 1m, \pm 2m, \pm 4m, \pm 8m, \ldots \), we can follow these steps:
### Step 1: Understanding Gravitational Potential
The gravitational potential \( V \) at a point due to a mass \( m \) is given by the formula:
\[
V = -\frac{Gm}{r}
\]
where \( G \) is the universal gravitational constant and \( r \) is the distance from the mass to the point where we are calculating the potential.
### Step 2: Calculate Contribution from Each Mass
For each mass of 2 kg located at \( x = \pm 1m, \pm 2m, \pm 4m, \pm 8m, \ldots \), we will calculate the gravitational potential at the origin (0,0).
1. **For \( x = \pm 1m \)**:
- Distance \( r = 1m \)
- Contribution from each mass:
\[
V_{1} = -\frac{G \cdot 2}{1} = -2G
\]
- Total contribution from both masses:
\[
V_{1, total} = -2G - 2G = -4G
\]
2. **For \( x = \pm 2m \)**:
- Distance \( r = 2m \)
- Contribution from each mass:
\[
V_{2} = -\frac{G \cdot 2}{2} = -G
\]
- Total contribution from both masses:
\[
V_{2, total} = -G - G = -2G
\]
3. **For \( x = \pm 4m \)**:
- Distance \( r = 4m \)
- Contribution from each mass:
\[
V_{3} = -\frac{G \cdot 2}{4} = -\frac{G}{2}
\]
- Total contribution from both masses:
\[
V_{3, total} = -\frac{G}{2} - \frac{G}{2} = -G
\]
4. **For \( x = \pm 8m \)**:
- Distance \( r = 8m \)
- Contribution from each mass:
\[
V_{4} = -\frac{G \cdot 2}{8} = -\frac{G}{4}
\]
- Total contribution from both masses:
\[
V_{4, total} = -\frac{G}{4} - \frac{G}{4} = -\frac{G}{2}
\]
### Step 3: Summing All Contributions
Now we need to sum the contributions from all the masses:
\[
V_{total} = V_{1, total} + V_{2, total} + V_{3, total} + V_{4, total} + \ldots
\]
This can be expressed as:
\[
V_{total} = -4G - 2G - G - \frac{G}{2} - \frac{G}{4} - \ldots
\]
### Step 4: Recognizing the Series
The series can be recognized as a geometric series:
\[
V_{total} = -2G \left( 2 + 1 + \frac{1}{2} + \frac{1}{4} + \ldots \right)
\]
The series inside the parentheses is a geometric series with the first term \( a = 2 \) and common ratio \( r = \frac{1}{2} \).
### Step 5: Sum of the Infinite Geometric Series
The sum \( S \) of an infinite geometric series is given by:
\[
S = \frac{a}{1 - r}
\]
Substituting \( a = 2 \) and \( r = \frac{1}{2} \):
\[
S = \frac{2}{1 - \frac{1}{2}} = \frac{2}{\frac{1}{2}} = 4
\]
### Step 6: Final Calculation
Substituting back into the potential equation:
\[
V_{total} = -2G \cdot 4 = -8G
\]
### Final Answer
Thus, the resultant gravitational potential at the origin is:
\[
\boxed{-8G}
\]
