Infinite number of masses, each of mass 3 kg are placed along the y-axis at y = 1 m, 3 m, 9 m 27 m ... The magnitude of resultant gravitational potential in terms of gravitatinal constant at the origin (y = 0) is

Options

A. 4.5 G unit
B. 6 G unit
C. 3 G unit
D. 9 unit

Answer

To solve the problem of finding the resultant gravitational potential at the origin due to an infinite number of masses placed along the y-axis, we can follow these steps:

### Step 1: Identify the positions of the masses
The masses are placed at positions along the y-axis at \( y = 1 \, m, 3 \, m, 9 \, m, 27 \, m, \ldots \). These positions can be expressed as \( y_n = 3^{n-1} \) for \( n = 1, 2, 3, \ldots \).

### Step 2: Write the formula for 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, \( m \) is the mass, and \( r \) is the distance from the mass to the point where we are calculating the potential.

### Step 3: Calculate the potential at the origin due to each mass
For each mass of \( 3 \, kg \) located at \( y_n \):
- The distance from the origin (0,0) to the mass at \( y_n \) is \( r_n = y_n = 3^{n-1} \).
- Therefore, the potential at the origin due to the mass at \( y_n \) is:
\[ V_n = -\frac{G \cdot 3}{3^{n-1}} = -\frac{G \cdot 3}{3^{n-1}} = -G \cdot 3^{1-n} \]

### Step 4: Sum the potentials from all masses
The total potential \( V \) at the origin is the sum of the potentials due to all the masses:
\[ V = \sum_{n=1}^{\infty} V_n = \sum_{n=1}^{\infty} -G \cdot 3^{1-n} \]
This can be rewritten as:
\[ V = -G \cdot 3 \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^{n-1} \]

### Step 5: Recognize the series as a geometric series
The series \( \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^{n-1} \) is a geometric series with the first term \( a = 1 \) and common ratio \( r = \frac{1}{3} \). The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \]

### Step 6: Substitute back into the potential expression
Now substituting the sum back into the potential expression:
\[ V = -G \cdot 3 \cdot \frac{3}{2} = -\frac{9G}{2} \]

### Step 7: Take the magnitude of the potential
Since the question asks for the magnitude of the gravitational potential:
\[ |V| = \frac{9G}{2} \]

### Final Answer
Thus, the magnitude of the resultant gravitational potential at the origin is:
\[ \frac{9G}{2} \]