A very large number of particle of same mass m are kept at horizontal distance of `1m, 2m, 4m, 8m` and so on from (0,0) point. The total gravitational at this point (0,0) is .

Options

A. `-8G`
B. `-3Gm`
C. `-4Gm`
D. `-2Gm`

Answer

To solve the problem of finding the total gravitational potential at the origin (0,0) due to a large number of particles placed at distances of 1m, 2m, 4m, 8m, and so on, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Gravitational Potential**:
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.

2. **Identifying the Distances**:
The particles are located at distances \( r_1 = 1m, r_2 = 2m, r_3 = 4m, r_4 = 8m, \ldots \). These distances can be expressed as \( r_n = 2^{n-1} \) for \( n = 1, 2, 3, \ldots \).

3. **Calculating the Total Gravitational Potential**:
The total gravitational potential \( V_{total} \) at the origin due to all these particles is the sum of the potentials from each particle:
\[ V_{total} = V_1 + V_2 + V_3 + V_4 + \ldots \]
This can be expressed as:
\[ V_{total} = -\frac{Gm}{1} - \frac{Gm}{2} - \frac{Gm}{4} - \frac{Gm}{8} - \ldots \]

4. **Factoring Out Common Terms**:
We can factor out \( -Gm \):
\[ V_{total} = -Gm \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \right) \]

5. **Recognizing the Series**:
The series inside the parentheses is a geometric series with the first term \( a = 1 \) and the common ratio \( r = \frac{1}{2} \). The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Applying this formula:
\[ S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

6. **Calculating the Total Gravitational Potential**:
Substituting the sum back into the equation for \( V_{total} \):
\[ V_{total} = -Gm \cdot 2 = -2Gm \]

### Final Answer:
Thus, the total gravitational potential at the point (0,0) is:
\[ \boxed{-2Gm} \]