Two masses 800 kg and 600kg are at a distance 0.25 m apart. Calculate the magnitude of the gravitational intensity at a point distant 0.20 m from the 800 kg and 0.15 m from the 600 kg mass. `G = 6.66 xx 10^(-11) Nm^(2) kg^(-2)`.

Explanation

To solve the problem of calculating the gravitational intensity at a point due to two masses, we can follow these steps:

### Step 1: Understand the Setup
We have two masses:
- Mass \( m_1 = 800 \, \text{kg} \)
- Mass \( m_2 = 600 \, \text{kg} \)

The distance between the two masses is \( d = 0.25 \, \text{m} \). We need to find the gravitational intensity at a point \( P \) which is \( 0.20 \, \text{m} \) from \( m_1 \) and \( 0.15 \, \text{m} \) from \( m_2 \).

### Step 2: Calculate Gravitational Intensity from Each Mass
The formula for gravitational intensity \( E \) due to a mass \( m \) at a distance \( r \) is given by:

\[ E = \frac{G \cdot m}{r^2} \]

Where \( G = 6.66 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).

#### For Mass \( m_1 \):
- Distance from point \( P \): \( r_1 = 0.20 \, \text{m} \)

Calculating \( E_1 \):

\[ E_1 = \frac{G \cdot m_1}{r_1^2} = \frac{6.66 \times 10^{-11} \cdot 800}{(0.20)^2} \]

Calculating \( (0.20)^2 = 0.04 \):

\[ E_1 = \frac{6.66 \times 10^{-11} \cdot 800}{0.04} = \frac{5.328 \times 10^{-8}}{0.04} = 1.332 \times 10^{-6} \, \text{N/kg} \]

#### For Mass \( m_2 \):
- Distance from point \( P \): \( r_2 = 0.15 \, \text{m} \)

Calculating \( E_2 \):

\[ E_2 = \frac{G \cdot m_2}{r_2^2} = \frac{6.66 \times 10^{-11} \cdot 600}{(0.15)^2} \]

Calculating \( (0.15)^2 = 0.0225 \):

\[ E_2 = \frac{6.66 \times 10^{-11} \cdot 600}{0.0225} = \frac{3.996 \times 10^{-8}}{0.0225} = 1.776 \times 10^{-6} \, \text{N/kg} \]

### Step 3: Determine the Resultant Gravitational Intensity
Since the point \( P \) is located such that the angles formed by the lines connecting \( P \) to \( m_1 \) and \( m_2 \) are perpendicular, we can use the Pythagorean theorem to find the resultant gravitational intensity \( E \):

\[ E = \sqrt{E_1^2 + E_2^2} \]

Substituting the values:

\[ E = \sqrt{(1.332 \times 10^{-6})^2 + (1.776 \times 10^{-6})^2} \]

Calculating \( E_1^2 \) and \( E_2^2 \):

\[ E_1^2 = 1.332^2 \times 10^{-12} = 1.77 \times 10^{-12} \]
\[ E_2^2 = 1.776^2 \times 10^{-12} = 3.15 \times 10^{-12} \]

Adding them:

\[ E = \sqrt{1.77 \times 10^{-12} + 3.15 \times 10^{-12}} = \sqrt{4.92 \times 10^{-12}} \approx 2.22 \times 10^{-6} \, \text{N/kg} \]

### Final Answer
The magnitude of the gravitational intensity at point \( P \) is approximately:

\[ E \approx 2.22 \times 10^{-6} \, \text{N/kg} \]