Infinite number of masses, each of mass m, are placed along a straight line at distances of r, 2r, 4r, 8r, etc. from a reference point O.Then the gravitational field intensity at point O will be

Options

A. `(5Gm)/(4r^(2))`
B. `(4Gm)/(3r^(2))`
C. `(3Gm)/(2r^(2))`
D. `(2Gm)/(r^(2))`

Explanation

To find the gravitational field intensity at point O due to an infinite number of masses placed along a straight line at distances of r, 2r, 4r, 8r, etc., we can follow these steps:

### Step 1: Understand the Gravitational Field Intensity
The gravitational field intensity (E) due to a point mass (m) at a distance (d) is given by the formula:
\[ E = \frac{Gm}{d^2} \]
where G is the gravitational constant.

### Step 2: Identify the Distances of the Masses
The masses are located at distances:
- First mass at distance \( r \)
- Second mass at distance \( 2r \)
- Third mass at distance \( 4r \)
- Fourth mass at distance \( 8r \)
- And so on...

### Step 3: Calculate the Gravitational Field Intensity from Each Mass
For each mass, we can calculate the gravitational field intensity at point O:
1. From the first mass at distance \( r \):
\[ E_1 = \frac{Gm}{r^2} \]
2. From the second mass at distance \( 2r \):
\[ E_2 = \frac{Gm}{(2r)^2} = \frac{Gm}{4r^2} \]
3. From the third mass at distance \( 4r \):
\[ E_3 = \frac{Gm}{(4r)^2} = \frac{Gm}{16r^2} \]
4. From the fourth mass at distance \( 8r \):
\[ E_4 = \frac{Gm}{(8r)^2} = \frac{Gm}{64r^2} \]
5. Continuing this pattern, the nth mass at distance \( 2^{n-1}r \) will contribute:
\[ E_n = \frac{Gm}{(2^{n-1}r)^2} = \frac{Gm}{4^{n-1}r^2} \]

### Step 4: Sum the Gravitational Field Intensities
The total gravitational field intensity at point O is the sum of the contributions from all masses:
\[ E_{total} = E_1 + E_2 + E_3 + E_4 + \ldots \]
\[ E_{total} = \frac{Gm}{r^2} + \frac{Gm}{4r^2} + \frac{Gm}{16r^2} + \frac{Gm}{64r^2} + \ldots \]

### Step 5: Factor Out Common Terms
We can factor out \( \frac{Gm}{r^2} \):
\[ E_{total} = \frac{Gm}{r^2} \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots \right) \]

### Step 6: Recognize the Series
The series inside the parentheses is a geometric series with:
- First term \( a = 1 \)
- Common ratio \( r = \frac{1}{4} \)

The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Step 7: Substitute Back to Find Total Intensity
Now substituting back into the equation for \( E_{total} \):
\[ E_{total} = \frac{Gm}{r^2} \cdot \frac{4}{3} = \frac{4Gm}{3r^2} \]

### Final Answer
Thus, the gravitational field intensity at point O is:
\[ E_{total} = \frac{4Gm}{3r^2} \]