The change in the value of acceleration of earth toward sun, when the moon coomes from the position of solar eclipse to the position on the other side of earth in line with sun is :
(mass of moon `=7.36xx10^(22)` kg, orbital radius of moon `=3.8xx10^(8)` m.)

Options

A. `6.73xx10^(-2) m//s^(2)`
B. `6.73xx10^(-3) m//s^(2)`
C. `6.73xx10^(-4) m//s^(2)`
D. `6.73xx10^(-5) m//s^(2)`

Answer

To solve the problem of finding the change in the value of acceleration of Earth toward the Sun when the Moon moves from the position of a solar eclipse to the position on the other side of the Earth in line with the Sun, we can follow these steps:

### Step 1: Understand the Forces Acting on the Earth
When the Moon is positioned between the Earth and the Sun (solar eclipse), the gravitational force exerted by the Moon on the Earth (Fm) acts in the same direction as the gravitational force exerted by the Sun (Fs). When the Moon moves to the opposite side of the Earth, Fm acts in the opposite direction to Fs.

### Step 2: Write the Net Forces in Both Positions
1. **During Solar Eclipse:**
- Net force on Earth (F_net1) = Fs + Fm
2. **When Moon is on the other side:**
- Net force on Earth (F_net2) = Fs - Fm

### Step 3: Calculate the Change in Net Force
The change in net force (ΔF_net) when the Moon moves from one position to the other can be calculated as:
\[ \Delta F_{net} = F_{net1} - F_{net2} = (Fs + Fm) - (Fs - Fm) = 2Fm \]

### Step 4: Relate Change in Force to Change in Acceleration
Using Newton's second law, we can relate the change in net force to the change in acceleration:
\[ \Delta F_{net} = m \cdot (a_1 - a_2) = m \cdot \Delta a \]
Where \( m \) is the mass of the Earth, \( a_1 \) is the acceleration toward the Sun when the Moon is in the solar eclipse position, and \( a_2 \) is the acceleration when the Moon is on the opposite side.

### Step 5: Substitute for Change in Acceleration
From the previous step, we have:
\[ m \cdot \Delta a = 2Fm \]
Dividing both sides by \( m \):
\[ \Delta a = 2 \cdot \frac{F_m}{m} \]

### Step 6: Calculate the Gravitational Force Exerted by the Moon
The gravitational force exerted by the Moon on the Earth is given by:
\[ F_m = G \cdot \frac{M \cdot m}{r^2} \]
Where:
- \( G \) = gravitational constant \( = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) = mass of the Moon \( = 7.36 \times 10^{22} \, \text{kg} \)
- \( r \) = orbital radius of the Moon \( = 3.8 \times 10^{8} \, \text{m} \)

### Step 7: Substitute Values into the Equation
Substituting the values into the equation for \( \Delta a \):
\[ \Delta a = 2 \cdot \frac{G \cdot M}{r^2} \]
\[ \Delta a = 2 \cdot \frac{6.67 \times 10^{-11} \cdot 7.36 \times 10^{22}}{(3.8 \times 10^{8})^2} \]

### Step 8: Calculate the Result
Calculating the above expression:
1. Calculate \( (3.8 \times 10^{8})^2 = 1.444 \times 10^{17} \)
2. Calculate \( 6.67 \times 10^{-11} \cdot 7.36 \times 10^{22} = 4.905 \times 10^{12} \)
3. Substitute back into the equation:
\[ \Delta a = 2 \cdot \frac{4.905 \times 10^{12}}{1.444 \times 10^{17}} \approx 6.73 \times 10^{-5} \, \text{m/s}^2 \]

### Final Answer
The change in the value of acceleration of Earth toward the Sun is approximately:
\[ \Delta a \approx 6.73 \times 10^{-5} \, \text{m/s}^2 \]