If the earth is a point mass of `6 xx 10^(24)` kg revolving around the sun at a distance of `1.5 xx 10^(8)` km and in time ` T = 3.14 xx 10^(7)` s, then the angular momentum of the earth around the sun is

Options

A. `1.2 xx 10^(18) kgm^(2)//s`
B. `1.8 xx 10^(29) kgm^(2)//s`
C. `1.5 xx 10^(37) kgm^(2)//s `
D. `2.7 xx 10^(40) kgm^(2)//s`

Answer

To solve the problem of finding the angular momentum of the Earth around the Sun, we can follow these steps:

### Step 1: Identify the given values
- Mass of the Earth, \( m = 6 \times 10^{24} \) kg
- Distance from the Sun, \( r = 1.5 \times 10^{8} \) km
- Time period, \( T = 3.14 \times 10^{7} \) s

### Step 2: Convert distance from kilometers to meters
Since \( 1 \) km = \( 1000 \) m, we convert the distance:
\[ r = 1.5 \times 10^{8} \text{ km} = 1.5 \times 10^{8} \times 10^{3} \text{ m} = 1.5 \times 10^{11} \text{ m} \]

### Step 3: Calculate the angular velocity \( \omega \)
The angular velocity \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2 \times 3.14}{3.14 \times 10^{7}} = \frac{2}{10^{7}} = 2 \times 10^{-7} \text{ rad/s} \]

### Step 4: Calculate the angular momentum \( L \)
The angular momentum \( L \) can be calculated using the formula:
\[ L = m \cdot \omega \cdot r^2 \]
Substituting the values:
\[ L = (6 \times 10^{24}) \cdot (2 \times 10^{-7}) \cdot (1.5 \times 10^{11})^2 \]

### Step 5: Calculate \( r^2 \)
Calculating \( r^2 \):
\[ r^2 = (1.5 \times 10^{11})^2 = 2.25 \times 10^{22} \text{ m}^2 \]

### Step 6: Substitute \( r^2 \) into the angular momentum equation
Now substituting \( r^2 \) into the equation for \( L \):
\[ L = (6 \times 10^{24}) \cdot (2 \times 10^{-7}) \cdot (2.25 \times 10^{22}) \]

### Step 7: Perform the multiplication
Calculating:
\[ L = 6 \cdot 2 \cdot 2.25 \times 10^{24} \cdot 10^{-7} \cdot 10^{22} \]
\[ L = 27 \times 10^{39} \text{ kg m}^2/\text{s} \]

### Step 8: Adjust the exponent
\[ L = 2.7 \times 10^{40} \text{ kg m}^2/\text{s} \]

### Final Answer
The angular momentum of the Earth around the Sun is:
\[ L = 2.7 \times 10^{40} \text{ kg m}^2/\text{s} \]