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) "kg m"^(2)//s`
B. `1.8 xx 10^(29) "kg m"^(2)//s`
C. `1.5 xx 10^(37) "kg m"^(2)//s`
D. `2.7 xx 10^(40) "kg m"^(2)//s`

Explanation

To calculate the angular momentum of the Earth around the Sun, we can follow these steps:

### Step 1: Identify the given values
- Mass of the Earth, \( m = 6 \times 10^{24} \, \text{kg} \)
- Distance from the Sun, \( r = 1.5 \times 10^{8} \, \text{km} = 1.5 \times 10^{11} \, \text{m} \) (conversion from km to meters)
- Time period of revolution, \( T = 3.14 \times 10^{7} \, \text{s} \)

### Step 2: Calculate the moment of inertia (I)
The moment of inertia \( I \) of the Earth about the Sun can be calculated using the formula:
\[ I = m r^2 \]
Substituting the values:
\[ I = (6 \times 10^{24} \, \text{kg}) \times (1.5 \times 10^{11} \, \text{m})^2 \]
Calculating \( r^2 \):
\[ r^2 = (1.5 \times 10^{11})^2 = 2.25 \times 10^{22} \, \text{m}^2 \]
Now substituting back:
\[ I = 6 \times 10^{24} \times 2.25 \times 10^{22} = 13.5 \times 10^{46} \, \text{kg m}^2 \]

### Step 3: Calculate the angular velocity (\( \omega \))
The angular velocity \( \omega \) can be calculated using 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 \( L \) is given by:
\[ L = I \omega \]
Substituting the values we calculated:
\[ L = (13.5 \times 10^{46}) \times (2 \times 10^{-7}) = 27 \times 10^{39} \, \text{kg m}^2/\text{s} \]
This can be expressed as:
\[ L = 2.7 \times 10^{40} \, \text{kg m}^2/\text{s} \]

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