The moon takes about `27.3` days to revolve around the earth in a nearly circular orbit of radius `3.84xx10^5 km`. Calculate the mass of the earth from these data.

Explanation

To calculate the mass of the Earth using the given data about the Moon's orbit, we can follow these steps:

### Step 1: Convert the time period from days to seconds
The time period \( T \) of the Moon's revolution is given as \( 27.3 \) days. We need to convert this into seconds.

\[ T = 27.3 \, \text{days} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour} \]

Calculating this gives:

\[ T = 27.3 \times 24 \times 3600 = 2,358,720 \, \text{seconds} \]

### Step 2: Convert the radius from kilometers to meters
The radius \( r \) of the Moon's orbit is given as \( 3.84 \times 10^5 \) km. We need to convert this into meters.

\[ r = 3.84 \times 10^5 \, \text{km} \times 1000 \, \text{meters/km} = 3.84 \times 10^8 \, \text{meters} \]

### Step 3: Use the formula for the mass of the Earth
The formula relating the mass of the Earth \( M \), the radius of the orbit \( r \), the time period \( T \), and the gravitational constant \( G \) is given by:

\[ T^2 = \frac{4 \pi^2 r^3}{G M} \]

Rearranging this formula to solve for the mass of the Earth \( M \):

\[ M = \frac{4 \pi^2 r^3}{G T^2} \]

### Step 4: Substitute the known values
We know:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( r = 3.84 \times 10^8 \, \text{m} \)
- \( T = 2,358,720 \, \text{s} \)

Now substituting these values into the formula:

\[ M = \frac{4 \pi^2 (3.84 \times 10^8)^3}{(6.67 \times 10^{-11}) (2,358,720)^2} \]

### Step 5: Calculate the mass of the Earth
Calculating \( r^3 \):

\[ r^3 = (3.84 \times 10^8)^3 = 5.67 \times 10^{25} \, \text{m}^3 \]

Calculating \( T^2 \):

\[ T^2 = (2,358,720)^2 = 5.57 \times 10^{12} \, \text{s}^2 \]

Now substituting these into the equation for \( M \):

\[ M = \frac{4 \pi^2 (5.67 \times 10^{25})}{(6.67 \times 10^{-11}) (5.57 \times 10^{12})} \]

Calculating the numerator:

\[ 4 \pi^2 \approx 39.478 \]

Thus,

\[ M \approx \frac{39.478 \times 5.67 \times 10^{25}}{(6.67 \times 10^{-11}) (5.57 \times 10^{12})} \]

Calculating the denominator:

\[ (6.67 \times 10^{-11}) (5.57 \times 10^{12}) \approx 3.71 \times 10^{2} \]

Finally, calculating \( M \):

\[ M \approx \frac{39.478 \times 5.67 \times 10^{25}}{3.71 \times 10^{2}} \approx 6.023 \times 10^{24} \, \text{kg} \]

### Conclusion
The mass of the Earth is approximately \( 6.023 \times 10^{24} \, \text{kg} \).

---