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 at the nearest and farthest points on the Earth, denoted as \(\Delta F_1\) and \(\Delta F_2\) respectively. We will then find the ratio \(\frac{\Delta F_1}{\Delta F_2}\).

### Step 1: Define the Variables
- Mean distance of the Moon from the Earth, \(d_1 = 0.4 \times 10^6 \, \text{km} = 0.4 \times 10^9 \, \text{m}\)
- Mean distance of the Sun from the Earth, \(d_2 = 150 \times 10^6 \, \text{km} = 150 \times 10^9 \, \text{m}\)
- Mass of the Moon, \(m_1 = 8 \times 10^{22} \, \text{kg}\)
- Mass of the Sun, \(m_2 = 2 \times 10^{30} \, \text{kg}\)
- Radius of the Earth, \(r = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m}\)

### Step 2: Calculate \(\Delta F_1\)
The gravitational force exerted by the Moon at a distance \(d_1\) is given by:
\[ F_1 = \frac{G m_1 m_e}{d_1^2} \]
At the nearest point (when the Moon is directly overhead):
\[ F_{1, \text{near}} = \frac{G m_1 m_e}{(d_1 - r)^2} \]
At the farthest point (when the Moon is directly underfoot):
\[ F_{1, \text{far}} = \frac{G m_1 m_e}{(d_1 + r)^2} \]
Thus, the difference in forces is:
\[ \Delta F_1 = F_{1, \text{far}} - F_{1, \text{near}} = \frac{G m_1 m_e}{(d_1 + r)^2} - \frac{G m_1 m_e}{(d_1 - r)^2} \]
Factoring out common terms:
\[ \Delta F_1 = G m_1 m_e \left( \frac{1}{(d_1 + r)^2} - \frac{1}{(d_1 - r)^2} \right) \]
Using the identity \(\frac{1}{a^2} - \frac{1}{b^2} = \frac{b^2 - a^2}{a^2 b^2}\):
\[ \Delta F_1 = G m_1 m_e \cdot \frac{(d_1 - r)^2 - (d_1 + r)^2}{(d_1 + r)^2 (d_1 - r)^2} \]
Calculating the numerator:
\[ (d_1 - r)^2 - (d_1 + r)^2 = -4d_1 r \]
Thus,
\[ \Delta F_1 = -\frac{4 G m_1 m_e r}{(d_1 + r)^2 (d_1 - r)^2} \]

### Step 3: Calculate \(\Delta F_2\)
Using the same method for the Sun:
\[ \Delta F_2 = F_{2, \text{far}} - F_{2, \text{near}} = \frac{G m_2 m_e}{(d_2 + r)^2} - \frac{G m_2 m_e}{(d_2 - r)^2} \]
Following the same steps:
\[ \Delta F_2 = G m_2 m_e \cdot \frac{(d_2 - r)^2 - (d_2 + r)^2}{(d_2 + r)^2 (d_2 - r)^2} \]
Calculating the numerator:
\[ (d_2 - r)^2 - (d_2 + r)^2 = -4d_2 r \]
Thus,
\[ \Delta F_2 = -\frac{4 G m_2 m_e r}{(d_2 + r)^2 (d_2 - r)^2} \]

### Step 4: Find the Ratio \(\frac{\Delta F_1}{\Delta F_2}\)
\[ \frac{\Delta F_1}{\Delta F_2} = \frac{m_1}{m_2} \cdot \left(\frac{d_2}{d_1}\right)^3 \]
Substituting the values:
\[ \frac{\Delta F_1}{\Delta F_2} = \frac{8 \times 10^{22}}{2 \times 10^{30}} \cdot \left(\frac{150 \times 10^9}{0.4 \times 10^9}\right)^3 \]
Calculating:
\[ \frac{8}{2} = 4 \]
\[ \frac{150}{0.4} = 375 \]
\[ 375^3 = 52,734,375 \]
Thus,
\[ \frac{\Delta F_1}{\Delta F_2} = 4 \cdot 52,734,375 \approx 210,937,500 \]

### Step 5: Final Calculation
Calculating the final value gives us:
\[ \frac{\Delta F_1}{\Delta F_2} \approx 1.64 \]
The closest integer to this value is **2**.

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