There are two bodies of masses 100 kg and 10000 kg separated by a distance 1 m . At what distance from the smaller body, the intensity of gravitational field will be zero

Options

A. `1/9 m`
B. `1/10 m`
C. `1/11 m`
D. `10/11 m`

Answer

To find the distance from the smaller body (100 kg) at which the intensity of the gravitational field is zero, we can follow these steps:

### Step 1: Understand the Gravitational Field
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: Set Up the Problem
We have two masses:
- \( m_1 = 100 \, \text{kg} \) (smaller body)
- \( m_2 = 10000 \, \text{kg} \) (larger body)

The distance between them is \( d = 1 \, \text{m} \).

Let \( x \) be the distance from the smaller body where the gravitational field intensity is zero. Therefore, the distance from the larger body will be \( (1 - x) \).

### Step 3: Write the Gravitational Field Equations
The gravitational field intensity due to the smaller body at distance \( x \) is:
\[ E_1 = \frac{G \cdot 100}{x^2} \]

The gravitational field intensity due to the larger body at distance \( (1 - x) \) is:
\[ E_2 = \frac{G \cdot 10000}{(1 - x)^2} \]

### Step 4: Set the Gravitational Fields Equal
For the intensity of the gravitational field to be zero, the magnitudes of the two fields must be equal:
\[ E_1 = E_2 \]
Thus,
\[ \frac{G \cdot 100}{x^2} = \frac{G \cdot 10000}{(1 - x)^2} \]

### Step 5: Cancel Out Common Terms
Since \( G \) is common in both sides, we can cancel it out:
\[ \frac{100}{x^2} = \frac{10000}{(1 - x)^2} \]

### Step 6: Cross Multiply
Cross multiplying gives us:
\[ 100 \cdot (1 - x)^2 = 10000 \cdot x^2 \]

### Step 7: Expand and Rearrange
Expanding the left side:
\[ 100 \cdot (1 - 2x + x^2) = 10000 \cdot x^2 \]
This simplifies to:
\[ 100 - 200x + 100x^2 = 10000x^2 \]

Rearranging gives:
\[ 100 - 200x + 100x^2 - 10000x^2 = 0 \]
\[ -9900x^2 - 200x + 100 = 0 \]

### Step 8: Simplify the Equation
Dividing the entire equation by 100:
\[ -99x^2 - 2x + 1 = 0 \]
Multiplying through by -1:
\[ 99x^2 + 2x - 1 = 0 \]

### Step 9: Use the Quadratic Formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 99 \), \( b = 2 \), and \( c = -1 \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 99 \cdot (-1)}}{2 \cdot 99} \]
\[ x = \frac{-2 \pm \sqrt{4 + 396}}{198} \]
\[ x = \frac{-2 \pm \sqrt{400}}{198} \]
\[ x = \frac{-2 \pm 20}{198} \]

Calculating the two possible values:
1. \( x = \frac{18}{198} = \frac{1}{11} \)
2. \( x = \frac{-22}{198} \) (not valid since distance cannot be negative)

### Final Answer
Thus, the distance from the smaller body where the gravitational field intensity is zero is:
\[ x = \frac{1}{11} \, \text{m} \]

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