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 follow these steps:

### Step 1: Understand the relationship between gravitational force and centripetal force
The gravitational force acting on the Moon due to the Earth provides the necessary centripetal force for the Moon's circular motion. The equations for these forces are:

1. Gravitational Force:
\[ F_g = \frac{G \cdot M_e \cdot m}{r^2} \]
where \( G \) is the gravitational constant, \( M_e \) is the mass of the Earth, \( m \) is the mass of the Moon, and \( r \) is the distance between the centers of the Earth and the Moon.

2. Centripetal Force:
\[ F_c = \frac{m \cdot v^2}{r} \]
where \( v \) is the orbital velocity of the Moon.

### Step 2: Set the gravitational force equal to the centripetal force
Since these two forces are equal, we can set them equal to each other:
\[ \frac{G \cdot M_e \cdot m}{r^2} = \frac{m \cdot v^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G \cdot M_e}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \):
\[ \frac{G \cdot M_e}{r} = v^2 \]

### Step 3: Relate orbital velocity to the time period
The orbital velocity \( v \) can also be expressed in terms of the time period \( T \):
\[ v = \frac{2 \pi r}{T} \]
Substituting this into the previous equation gives:
\[ \frac{G \cdot M_e}{r} = \left(\frac{2 \pi r}{T}\right)^2 \]
Squaring both sides:
\[ \frac{G^2 \cdot M_e^2}{r^2} = \frac{4 \pi^2 r^2}{T^2} \]

### Step 4: Rearranging to find \( M_e \)
Rearranging the equation to isolate \( M_e \):
\[ M_e = \frac{4 \pi^2 r^3}{G T^2} \]

### Step 5: Substitute the known values
1. Convert the radius \( r \) from kilometers to meters:
\[ r = 4 \times 10^5 \text{ km} = 4 \times 10^8 \text{ m} \]
2. Convert the time period \( T \) from days to seconds:
\[ T = 28 \text{ days} = 28 \times 24 \times 3600 = 2419200 \text{ seconds} \]
3. Substitute \( G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \), \( r = 4 \times 10^8 \text{ m} \), and \( T = 2419200 \text{ s} \) into the equation for \( M_e \):
\[ M_e = \frac{4 \pi^2 (4 \times 10^8)^3}{(6.67 \times 10^{-11}) (2419200)^2} \]

### Step 6: Calculate \( M_e \)
Now, we can calculate \( M_e \):
1. Calculate \( (4 \times 10^8)^3 \):
\[ (4 \times 10^8)^3 = 64 \times 10^{24} = 6.4 \times 10^{25} \]
2. Calculate \( (2419200)^2 \):
\[ (2419200)^2 = 5857697280000 \]
3. Substitute these values back into the equation:
\[ M_e = \frac{4 \cdot (3.14)^2 \cdot (6.4 \times 10^{25})}{(6.67 \times 10^{-11}) \cdot (5857697280000)} \]
4. Calculate the final value to find \( M_e \).

After performing the calculations, we find that:
\[ M_e \approx 6.4 \times 10^{24} \text{ kg} \]

### Final Answer:
The mass of the Earth is approximately \( 6.4 \times 10^{24} \text{ kg} \).