Masses of `1 kg` each are placed `1 m, 2 m, 4 m, 8 m`, ... from a point `P`. The gravitational field intensity at `P` due to these masses is

Options

A. `G`
B. `G`
C. `4G`
D. `4G//3`

Explanation

To find the gravitational field intensity at point P due to the masses placed at distances of \(1 m, 2 m, 4 m, 8 m, \ldots\), we will follow these steps:

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

### Step 2: Identify the Masses and Distances
We have masses of \(1 kg\) placed at distances:
- \(r_1 = 1 m\)
- \(r_2 = 2 m\)
- \(r_3 = 4 m\)
- \(r_4 = 8 m\)
- and so on...

### Step 3: Calculate the Gravitational Field Intensity from Each Mass
For each mass, we will calculate the gravitational field intensity at point P:

1. For \(m_1 = 1 kg\) at \(r_1 = 1 m\):
\[ E_1 = \frac{G \cdot 1}{1^2} = G \]

2. For \(m_2 = 1 kg\) at \(r_2 = 2 m\):
\[ E_2 = \frac{G \cdot 1}{2^2} = \frac{G}{4} \]

3. For \(m_3 = 1 kg\) at \(r_3 = 4 m\):
\[ E_3 = \frac{G \cdot 1}{4^2} = \frac{G}{16} \]

4. For \(m_4 = 1 kg\) at \(r_4 = 8 m\):
\[ E_4 = \frac{G \cdot 1}{8^2} = \frac{G}{64} \]

### Step 4: Sum the Gravitational Field Intensities
Since all the gravitational fields are in the same direction (towards point P), we can sum them up:
\[ E_{net} = E_1 + E_2 + E_3 + E_4 + \ldots \]
\[ E_{net} = G + \frac{G}{4} + \frac{G}{16} + \frac{G}{64} + \ldots \]

### Step 5: Recognize the Series
The series can be recognized as a geometric series where:
- First term \(a = G\)
- Common ratio \(r = \frac{1}{4}\)

### Step 6: Use the Formula for the Sum of an Infinite Geometric Series
The sum \(S\) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{G}{1 - \frac{1}{4}} = \frac{G}{\frac{3}{4}} = \frac{4G}{3} \]

### Final Answer
Thus, the gravitational field intensity at point P due to the masses is:
\[ E_{net} = \frac{4G}{3} \]