The mass of the earth is `6xx10^(24)kg` and that of the moon is `7.4xx10^(22)kg`. The potential energy of the system is `-7.79xx10^(28)J`. The mean distance between the earth and moon is `(G=6.67xx10^(-11)Nm^(2)kg^(-2))`

Options

A. `3.8xx10^(8)m`
B. `3.37xx10^(6)m`
C. `7.6xx10^(4)m`
D. `1.9xx10^(2)m`

Explanation

To find the mean distance between the Earth and the Moon using the given potential energy of the system, we can use the formula for gravitational potential energy (U) between two masses:

\[ U = -\frac{G \cdot m_1 \cdot m_2}{r} \]

Where:
- \( U \) is the potential energy,
- \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)),
- \( m_1 \) is the mass of the Earth (\( 6 \times 10^{24} \, \text{kg} \)),
- \( m_2 \) is the mass of the Moon (\( 7.4 \times 10^{22} \, \text{kg} \)),
- \( r \) is the distance between the Earth and the Moon.

### Step 1: Write down the formula for potential energy

\[ U = -\frac{G \cdot m_1 \cdot m_2}{r} \]

### Step 2: Rearrange the formula to solve for \( r \)

\[ r = -\frac{G \cdot m_1 \cdot m_2}{U} \]

### Step 3: Substitute the known values into the equation

Given:
- \( U = -7.79 \times 10^{28} \, \text{J} \)
- \( m_1 = 6 \times 10^{24} \, \text{kg} \)
- \( m_2 = 7.4 \times 10^{22} \, \text{kg} \)
- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)

Substituting these values into the equation for \( r \):

\[ r = -\frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24}) \cdot (7.4 \times 10^{22})}{-7.79 \times 10^{28}} \]

### Step 4: Calculate the numerator

Calculating the numerator:

\[ 6.67 \times 10^{-11} \cdot 6 \times 10^{24} \cdot 7.4 \times 10^{22} \]

Calculating step by step:

1. \( 6.67 \times 6 = 40.02 \)
2. \( 40.02 \times 7.4 = 296.148 \)
3. Combine the powers of ten: \( 10^{-11} \times 10^{24} \times 10^{22} = 10^{35} \)

So, the numerator is:

\[ 296.148 \times 10^{35} \]

### Step 5: Calculate the denominator

The denominator is:

\[ -7.79 \times 10^{28} \]

### Step 6: Divide the numerator by the denominator

\[ r = \frac{296.148 \times 10^{35}}{7.79 \times 10^{28}} \]

Calculating:

1. Divide the coefficients: \( \frac{296.148}{7.79} \approx 38.0 \)
2. Subtract the powers of ten: \( 10^{35} / 10^{28} = 10^{7} \)

So:

\[ r \approx 38.0 \times 10^{7} \, \text{m} = 3.8 \times 10^{8} \, \text{m} \]

### Final Answer

The mean distance between the Earth and the Moon is approximately:

\[ r \approx 3.8 \times 10^{8} \, \text{meters} \]

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