The time period of revolution of moon around the earth is 28 days and radius of its orbit is `4xx10^(5)` km. If `G=6.67xx10^(-11) Nm^(2)//kg^(2)` then find the mass of the earth.

Answer

To find the mass of the Earth using the given information about the Moon's orbit, we can use the formula derived from Newton's law of gravitation and the centripetal force required for circular motion.

### Step-by-Step Solution:

1. **Convert the Time Period to Seconds:**
The time period \( T \) is given as 28 days. We need to convert this into seconds.
\[ T = 28 \text{ days} = 28 \times 24 \times 60 \times 60 \text{ seconds} \]
\[ T = 28 \times 86400 = 2419200 \text{ seconds} \]

2. **Convert the Radius to Meters:**
The radius \( r \) of the Moon's orbit is given as \( 4 \times 10^5 \) km. We need to convert this into meters.
\[ r = 4 \times 10^5 \text{ km} = 4 \times 10^5 \times 1000 \text{ m} = 4 \times 10^8 \text{ m} \]

3. **Use the Formula for Circular Motion:**
The centripetal force required for the Moon to stay in orbit is provided by the gravitational force between the Earth and the Moon. The formula is:
\[ F = \frac{G \cdot M \cdot m}{r^2} \]
Where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant \( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \),
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the Moon (which will cancel out),
- \( r \) is the radius of the orbit.

The centripetal force can also be expressed as:
\[ F = \frac{m \cdot v^2}{r} \]
Where \( v \) is the orbital speed of the Moon.

4. **Relate the Orbital Speed to the Time Period:**
The orbital speed \( v \) can be expressed in terms of the radius and the time period:
\[ v = \frac{2 \pi r}{T} \]

5. **Substituting for \( v \):**
Substitute \( v \) into the centripetal force equation:
\[ F = \frac{m \cdot \left(\frac{2 \pi r}{T}\right)^2}{r} = \frac{m \cdot 4 \pi^2 r}{T^2} \]

6. **Setting the Forces Equal:**
Now we set the gravitational force equal to the centripetal force:
\[ \frac{G \cdot M \cdot m}{r^2} = \frac{m \cdot 4 \pi^2 r}{T^2} \]

7. **Canceling \( m \) and Rearranging:**
Cancel \( m \) from both sides and rearrange to solve for \( M \):
\[ G \cdot M = \frac{4 \pi^2 r^3}{T^2} \]
\[ M = \frac{4 \pi^2 r^3}{G T^2} \]

8. **Substituting the Values:**
Now substitute the values of \( G \), \( r \), and \( T \):
\[ M = \frac{4 \pi^2 (4 \times 10^8)^3}{(6.67 \times 10^{-11}) (2419200)^2} \]

9. **Calculating \( M \):**
First calculate \( (4 \times 10^8)^3 \):
\[ (4 \times 10^8)^3 = 64 \times 10^{24} = 6.4 \times 10^{25} \]
Then calculate \( (2419200)^2 \):
\[ (2419200)^2 = 5850644480000 \approx 5.85 \times 10^{12} \]
Now substitute these values:
\[ M = \frac{4 \pi^2 (6.4 \times 10^{25})}{(6.67 \times 10^{-11}) (5.85 \times 10^{12})} \]

10. **Final Calculation:**
Calculate the final mass \( M \):
\[ M \approx \frac{4 \times 9.87 \times 6.4 \times 10^{25}}{6.67 \times 10^{-11} \times 5.85 \times 10^{12}} \approx 5.97 \times 10^{24} \text{ kg} \]

Thus, the mass of the Earth is approximately \( 5.97 \times 10^{24} \) kg.