To find the point between the Earth and the Moon where the gravitational field intensity is zero, we can use the concept of gravitational field intensity (or gravitational field strength). 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, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
### Step-by-Step Solution
1. **Define the Variables**:
- Let \( d \) be the distance between the Earth and the Moon, \( d = 3.85 \times 10^8 \, \text{m} \).
- Let \( m_e = 6.0 \times 10^{24} \, \text{kg} \) (mass of the Earth).
- Let \( m_m = 7.26 \times 10^{22} \, \text{kg} \) (mass of the Moon).
- Let \( x \) be the distance from the Earth to the point where the gravitational field is zero.
2. **Set Up the Equation**:
- The gravitational field intensity due to the Earth at distance \( x \) is:
\[
E_e = \frac{G \cdot m_e}{x^2}
\]
- The gravitational field intensity due to the Moon at distance \( d - x \) is:
\[
E_m = \frac{G \cdot m_m}{(d - x)^2}
\]
- At the point where the gravitational field intensity is zero, the magnitudes of these two fields must be equal:
\[
E_e = E_m
\]
- Therefore, we have:
\[
\frac{G \cdot m_e}{x^2} = \frac{G \cdot m_m}{(d - x)^2}
\]
3. **Cancel \( G \)**:
- Since \( G \) is a common factor, we can cancel it from both sides:
\[
\frac{m_e}{x^2} = \frac{m_m}{(d - x)^2}
\]
4. **Cross Multiply**:
- Cross multiplying gives us:
\[
m_e \cdot (d - x)^2 = m_m \cdot x^2
\]
5. **Expand and Rearrange**:
- Expanding the left side:
\[
m_e \cdot (d^2 - 2dx + x^2) = m_m \cdot x^2
\]
- Rearranging gives:
\[
m_e \cdot d^2 - 2m_e \cdot dx + (m_e - m_m) \cdot x^2 = 0
\]
6. **Substitute Known Values**:
- Substitute \( m_e = 6.0 \times 10^{24} \, \text{kg} \), \( m_m = 7.26 \times 10^{22} \, \text{kg} \), and \( d = 3.85 \times 10^8 \, \text{m} \) into the equation:
\[
6.0 \times 10^{24} \cdot (3.85 \times 10^8)^2 - 2 \cdot 6.0 \times 10^{24} \cdot 3.85 \times 10^8 \cdot x + (6.0 \times 10^{24} - 7.26 \times 10^{22}) \cdot x^2 = 0
\]
7. **Solve the Quadratic Equation**:
- This is a quadratic equation in the form \( Ax^2 + Bx + C = 0 \). Use the quadratic formula:
\[
x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]
- Calculate the coefficients \( A \), \( B \), and \( C \) from the expanded equation.
8. **Calculate the Roots**:
- Find the value of \( x \) that satisfies the equation, which will give the distance from the Earth to the point where the gravitational field intensity is zero.
### Final Answer
After performing the calculations, you will find the distance \( x \) from the Earth to the point where the gravitational field intensity is zero.
