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. `7.60xx10^(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 follow these steps:

### Step 1: Write the formula for gravitational potential energy
The gravitational potential energy (U) of a two-body system can be expressed as:

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

where:
- \( U \) is the potential energy,
- \( G \) is the gravitational constant,
- \( m_1 \) is the mass of the first body (Earth),
- \( m_2 \) is the mass of the second body (Moon),
- \( r \) is the distance between the two bodies.

### Step 2: Substitute the known values into the equation
Given:
- Mass of Earth, \( m_1 = 6 \times 10^{24} \, \text{kg} \)
- Mass of Moon, \( m_2 = 7.4 \times 10^{22} \, \text{kg} \)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
- Potential energy, \( U = -7.79 \times 10^{28} \, \text{J} \)

Substituting these values into the potential energy formula gives:

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

### Step 3: Rearrange the equation to solve for \( r \)
We can rearrange the equation to isolate \( r \):

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

### Step 4: Calculate the numerator
Calculating the numerator:

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

Calculating \( 6.67 \times 6 \times 7.4 \):

\[ 6.67 \times 6 = 40.02 \]
\[ 40.02 \times 7.4 \approx 296.148 \]

So the numerator becomes:

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

### Step 5: Calculate the denominator
The denominator is:

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

### Step 6: Divide the numerator by the denominator
Now we can find \( r \):

\[ r = \frac{296.148 \times 10^{35}}{7.79 \times 10^{28}} \approx 38.01 \times 10^{7} \]

### Step 7: Calculate \( r \)
Calculating \( r \):

\[ r \approx 3.801 \times 10^{8} \, \text{m} \]

### Step 8: Take the square root to find the distance
Since we need the distance \( r \):

\[ r \approx 1.9 \times 10^{2} \, \text{m} \]

Thus, the mean distance between the Earth and the Moon is approximately:

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

### Final Answer
The mean distance between the Earth and the Moon is approximately \( 3.8 \times 10^{8} \, \text{m} \).

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