What equal charges would have to be placed on earth and moon to neutralize their gravitational force of attraction?
Given that mass of earth `= 10^(25) kg` and mass of moon `= 10^(23) kg`

Options

A. `8.57xx10^(16)C`
B. `8.57xx10^(13)C`
C. `5.45xx10^(13)C`
D. `5.45xx10^(16)C`

Explanation

To solve the problem of determining the equal charges that need to be placed on the Earth and the Moon to neutralize their gravitational force of attraction, we will follow these steps:

### Step 1: Calculate the Gravitational Force
The gravitational force \( F_g \) between two masses is given by the formula:
\[ F_g = \frac{G \cdot M_1 \cdot M_2}{R^2} \]
where:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) (gravitational constant)
- \( M_1 = 10^{25} \, \text{kg} \) (mass of the Earth)
- \( M_2 = 10^{23} \, \text{kg} \) (mass of the Moon)
- \( R \) is the distance between the Earth and the Moon, approximately \( 3.84 \times 10^8 \, \text{m} \).

Substituting the values into the formula:
\[ F_g = \frac{(6.67 \times 10^{-11}) \cdot (10^{25}) \cdot (10^{23})}{(3.84 \times 10^8)^2} \]

### Step 2: Calculate the Distance Squared
Calculate \( R^2 \):
\[ R^2 = (3.84 \times 10^8)^2 = 1.47456 \times 10^{17} \, \text{m}^2 \]

### Step 3: Substitute and Calculate \( F_g \)
Now substitute \( R^2 \) back into the gravitational force equation:
\[ F_g = \frac{(6.67 \times 10^{-11}) \cdot (10^{25}) \cdot (10^{23})}{1.47456 \times 10^{17}} \]
Calculating the numerator:
\[ (6.67 \times 10^{-11}) \cdot (10^{25}) \cdot (10^{23}) = 6.67 \times 10^{37} \]
Now, substituting back:
\[ F_g = \frac{6.67 \times 10^{37}}{1.47456 \times 10^{17}} \approx 4.52 \times 10^{20} \, \text{N} \]

### Step 4: Set Up the Electrostatic Force Equation
The electrostatic force \( F_e \) between two equal charges \( Q \) is given by:
\[ F_e = \frac{k \cdot Q^2}{R^2} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 5: Set Gravitational Force Equal to Electrostatic Force
To neutralize the gravitational attraction, set \( F_g = F_e \):
\[ 4.52 \times 10^{20} = \frac{(9 \times 10^9) \cdot Q^2}{1.47456 \times 10^{17}} \]

### Step 6: Solve for \( Q^2 \)
Rearranging gives:
\[ Q^2 = \frac{4.52 \times 10^{20} \cdot 1.47456 \times 10^{17}}{9 \times 10^9} \]
Calculating the numerator:
\[ 4.52 \times 10^{20} \cdot 1.47456 \times 10^{17} \approx 6.66 \times 10^{37} \]
Now substituting:
\[ Q^2 = \frac{6.66 \times 10^{37}}{9 \times 10^9} \approx 7.4 \times 10^{27} \]

### Step 7: Take the Square Root to Find \( Q \)
Taking the square root:
\[ Q = \sqrt{7.4 \times 10^{27}} \approx 8.6 \times 10^{13} \, \text{C} \]

### Final Answer
The equal charges that need to be placed on the Earth and the Moon to neutralize their gravitational force of attraction is approximately:
\[ Q \approx 8.6 \times 10^{13} \, \text{C} \]