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 the 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 Initial and Final Acceleration
- During a solar eclipse, the Moon is positioned between the Earth and the Sun. The initial acceleration of the Earth toward the Sun (when the Moon is in line with the Sun) can be expressed as:
\[ a_i = a_s + a_m \]
where \( a_s \) is the acceleration due to the Sun, and \( a_m \) is the acceleration due to the Moon.

- When the Moon is on the other side of the Earth, the final acceleration can be expressed as:
\[ a_f = a_s - a_m \]

### Step 2: Calculate the Change in Acceleration
- The change in acceleration, \( \Delta a \), is given by:
\[ \Delta a = a_i - a_f \]
Substituting the expressions for \( a_i \) and \( a_f \):
\[ \Delta a = (a_s + a_m) - (a_s - a_m) = 2a_m \]

### Step 3: Calculate the Acceleration Due to the Moon
- The acceleration due to the Moon can be calculated using the formula for gravitational acceleration:
\[ a_m = \frac{G \cdot M_m}{r_m^2} \]
where:
- \( G = 6.67 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2 \) (gravitational constant)
- \( M_m = 7.36 \times 10^{22} \, \text{kg} \) (mass of the Moon)
- \( r_m = 3.8 \times 10^8 \, \text{m} \) (orbital radius of the Moon)

### Step 4: Substitute Values to Find \( a_m \)
- Plugging in the values:
\[ a_m = \frac{6.67 \times 10^{-11} \cdot 7.36 \times 10^{22}}{(3.8 \times 10^8)^2} \]

### Step 5: Calculate \( (3.8 \times 10^8)^2 \)
- Calculate the square of the radius:
\[ (3.8 \times 10^8)^2 = 14.44 \times 10^{16} \]

### Step 6: Calculate \( a_m \)
- Now substituting back into the equation for \( a_m \):
\[ a_m = \frac{6.67 \times 10^{-11} \cdot 7.36 \times 10^{22}}{14.44 \times 10^{16}} \]
- Performing the multiplication and division:
\[ a_m = \frac{49.09 \times 10^{11}}{14.44 \times 10^{16}} = 3.40 \times 10^{-5} \, \text{m/s}^2 \]

### Step 7: Calculate the Change in Acceleration
- Now, substituting \( a_m \) back to find \( \Delta a \):
\[ \Delta a = 2a_m = 2 \cdot 3.40 \times 10^{-5} = 6.73 \times 10^{-5} \, \text{m/s}^2 \]

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