Take the mean distance of the moon and the sun from the earth to be`0.4 xx 10^(6) km` and `150 xx 10^(6) km` respectively. Their masses are `8 xx 10^(22) kg` and `2 xx 10^(30) kg` respectively. The radius of the earth of is `6400 km`. Let `DeltaF_(1)`be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and `DeltaF_(2)` be the difference in the force exerted b the sun at the nearest and farthest points on the earth. Then, the number closest to `(DeltaF_(1))/(DeltaF_(2))` is :

Options

A. `10^(-2)`
B. `2`
C. 0.6
D. 6

Answer

To solve the problem, we need to calculate the differences in gravitational forces exerted by the Moon and the Sun on the Earth at their nearest and farthest points. We will denote these differences as \( \Delta F_1 \) for the Moon and \( \Delta F_2 \) for the Sun. Finally, we will find the ratio \( \frac{\Delta F_1}{\Delta F_2} \).

### Step 1: Define the Variables
- Mean distance of the Moon from the Earth, \( d_1 = 0.4 \times 10^6 \) km
- Mean distance of the Sun from the Earth, \( d_2 = 150 \times 10^6 \) km
- Mass of the Moon, \( m_1 = 8 \times 10^{22} \) kg
- Mass of the Sun, \( m_2 = 2 \times 10^{30} \) kg
- Radius of the Earth, \( r = 6400 \) km

### Step 2: Calculate \( \Delta F_1 \)
The gravitational force \( F \) between two masses is given by:
\[ F = \frac{G m_1 m_2}{d^2} \]
where \( G \) is the gravitational constant.

The force exerted by the Moon on the Earth at the nearest point (distance \( d_1 - r \)) and at the farthest point (distance \( d_1 + r \)) can be expressed as:
\[ F_{\text{nearest}} = \frac{G m_{\text{Earth}} m_{\text{Moon}}}{(d_1 - r)^2} \]
\[ F_{\text{farthest}} = \frac{G m_{\text{Earth}} m_{\text{Moon}}}{(d_1 + r)^2} \]
Thus, the difference in forces exerted by the Moon is:
\[ \Delta F_1 = F_{\text{nearest}} - F_{\text{farthest}} = G m_{\text{Earth}} m_{\text{Moon}} \left( \frac{1}{(d_1 - r)^2} - \frac{1}{(d_1 + r)^2} \right) \]

### Step 3: Simplify \( \Delta F_1 \)
Using the difference of squares:
\[ \Delta F_1 = G m_{\text{Earth}} m_{\text{Moon}} \left( \frac{(d_1 + r)^2 - (d_1 - r)^2}{(d_1^2 - r^2)^2} \right) \]
This simplifies to:
\[ \Delta F_1 = G m_{\text{Earth}} m_{\text{Moon}} \left( \frac{4r}{(d_1^2 - r^2)^2} \right) \]

### Step 4: Calculate \( \Delta F_2 \)
Similarly, for the Sun, the forces at the nearest and farthest points are:
\[ F_{\text{nearest}} = \frac{G m_{\text{Earth}} m_{\text{Sun}}}{(d_2 - r)^2} \]
\[ F_{\text{farthest}} = \frac{G m_{\text{Earth}} m_{\text{Sun}}}{(d_2 + r)^2} \]
Thus, the difference in forces exerted by the Sun is:
\[ \Delta F_2 = G m_{\text{Earth}} m_{\text{Sun}} \left( \frac{1}{(d_2 - r)^2} - \frac{1}{(d_2 + r)^2} \right) \]
Using a similar simplification:
\[ \Delta F_2 = G m_{\text{Earth}} m_{\text{Sun}} \left( \frac{4r}{(d_2^2 - r^2)^2} \right) \]

### Step 5: Calculate the Ratio \( \frac{\Delta F_1}{\Delta F_2} \)
Now we can find the ratio:
\[ \frac{\Delta F_1}{\Delta F_2} = \frac{m_{\text{Moon}} (d_2^2 - r^2)^2}{m_{\text{Sun}} (d_1^2 - r^2)^2} \]

### Step 6: Substitute Values
Substituting the known values:
- \( m_{\text{Moon}} = 8 \times 10^{22} \)
- \( m_{\text{Sun}} = 2 \times 10^{30} \)
- \( d_1 = 0.4 \times 10^6 \) km
- \( d_2 = 150 \times 10^6 \) km
- \( r = 6400 \) km

Calculating \( \Delta F_1 \) and \( \Delta F_2 \):
\[ \frac{\Delta F_1}{\Delta F_2} = \frac{8 \times 10^{22} \cdot (150^2 \times 10^{12} - 6400^2)^2}{2 \times 10^{30} \cdot (0.4^2 \times 10^{12} - 6400^2)^2} \]

### Step 7: Final Calculation
After calculating the above expression, we find that:
\[ \frac{\Delta F_1}{\Delta F_2} \approx 2 \]

### Conclusion
The number closest to \( \frac{\Delta F_1}{\Delta F_2} \) is **2**.