Two bodies of mass 100kg and `10^(4)` kg are lying one meter apart. At what distance from 100 kg body will the intensity of gravitational field be zero

Options

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

Explanation

To solve the problem of finding the distance from the 100 kg body where the gravitational field intensity is zero, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Distance**:
- Let \( m_1 = 100 \, \text{kg} \) (mass of the first body).
- Let \( m_2 = 10^4 \, \text{kg} \) (mass of the second body).
- The distance between the two bodies is \( d = 1 \, \text{m} \).

2. **Define the Distance from the 100 kg Body**:
- Let \( x \) be the distance from the 100 kg body where the gravitational field intensity is zero.
- The distance from the 10,000 kg body to this point will then be \( 1 - x \).

3. **Write the Gravitational Field Intensity Expressions**:
- The gravitational field intensity \( E_1 \) due to the 100 kg mass at a distance \( x \) is given by:
\[ E_1 = \frac{G \cdot m_1}{x^2} \]
- The gravitational field intensity \( E_2 \) due to the 10,000 kg mass at a distance \( 1 - x \) is given by:
\[ E_2 = \frac{G \cdot m_2}{(1 - x)^2} \]

4. **Set the Gravitational Field Intensities Equal**:
- For the gravitational field intensity to be zero, we set \( E_1 = E_2 \):
\[ \frac{G \cdot 100}{x^2} = \frac{G \cdot 10^4}{(1 - x)^2} \]
- The \( G \) cancels out from both sides:
\[ \frac{100}{x^2} = \frac{10^4}{(1 - x)^2} \]

5. **Cross Multiply to Solve for x**:
- Cross multiplying gives:
\[ 100(1 - x)^2 = 10^4 x^2 \]
- Expanding the left side:
\[ 100(1 - 2x + x^2) = 10^4 x^2 \]
\[ 100 - 200x + 100x^2 = 10^4 x^2 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 100 - 200x + 100x^2 - 10^4 x^2 = 0 \]
\[ (100 - 10^4)x^2 - 200x + 100 = 0 \]
- Simplifying:
\[ -9999x^2 - 200x + 100 = 0 \]

7. **Using the Quadratic Formula**:
- Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = -9999, \quad b = -200, \quad c = 100 \]
\[ x = \frac{200 \pm \sqrt{(-200)^2 - 4 \cdot (-9999) \cdot 100}}{2 \cdot (-9999)} \]
\[ x = \frac{200 \pm \sqrt{40000 + 3999600}}{-19998} \]
\[ x = \frac{200 \pm \sqrt{4039600}}{-19998} \]

8. **Calculating the Values**:
- Calculate \( \sqrt{4039600} \approx 2009.9 \):
\[ x = \frac{200 \pm 2009.9}{-19998} \]
- Taking the positive root:
\[ x \approx \frac{2209.9}{-19998} \approx 0.1 \text{ m} \]

9. **Final Result**:
- The distance from the 100 kg body where the intensity of the gravitational field is zero is approximately:
\[ x \approx \frac{1}{11} \text{ m} \approx 0.0909 \text{ m} \]