Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth =6400 km and radius of the orbit of the earth about the sun `=1.5xx10^8km.`

Explanation

To calculate the ratio of the angular momentum of the Earth about its axis due to its spinning motion to that about the Sun due to its orbital motion, we can follow these steps:

### Step 1: Define Angular Momentum
The angular momentum \( L \) is given by the formula:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 2: Calculate Angular Momentum of the Earth about its Axis
For the Earth spinning about its own axis, we can use the moment of inertia for a solid sphere:
\[ I_E = \frac{2}{5} m R_E^2 \]
where \( R_E \) is the radius of the Earth (6400 km). The angular velocity \( \omega_E \) is given by:
\[ \omega_E = \frac{2\pi}{T_E} \]
where \( T_E \) is the time period of rotation (24 hours = 86400 seconds). Thus,
\[ \omega_E = \frac{2\pi}{86400} \text{ rad/s} \]
Now, the angular momentum of the Earth about its axis \( L_E \) is:
\[ L_E = I_E \cdot \omega_E = \left(\frac{2}{5} m R_E^2\right) \cdot \left(\frac{2\pi}{86400}\right) \]

### Step 3: Calculate Angular Momentum of the Earth about the Sun
For the Earth revolving around the Sun, we can use:
\[ I_O = m R_O^2 \]
where \( R_O \) is the radius of the Earth's orbit about the Sun (1.5 x \( 10^8 \) km). The angular velocity \( \omega_O \) is given by:
\[ \omega_O = \frac{2\pi}{T_O} \]
where \( T_O \) is the time period of revolution (365 days = 31536000 seconds). Thus,
\[ \omega_O = \frac{2\pi}{31536000} \text{ rad/s} \]
Now, the angular momentum of the Earth about the Sun \( L_O \) is:
\[ L_O = I_O \cdot \omega_O = \left(m R_O^2\right) \cdot \left(\frac{2\pi}{31536000}\right) \]

### Step 4: Calculate the Ratio of Angular Momenta
Now, we can find the ratio \( \frac{L_E}{L_O} \):
\[ \frac{L_E}{L_O} = \frac{\left(\frac{2}{5} m R_E^2\right) \cdot \left(\frac{2\pi}{86400}\right)}{\left(m R_O^2\right) \cdot \left(\frac{2\pi}{31536000}\right)} \]
The mass \( m \) and \( 2\pi \) cancel out:
\[ \frac{L_E}{L_O} = \frac{\frac{2}{5} R_E^2 \cdot \frac{1}{86400}}{R_O^2 \cdot \frac{1}{31536000}} \]
This simplifies to:
\[ \frac{L_E}{L_O} = \frac{2}{5} \cdot \frac{R_E^2}{R_O^2} \cdot \frac{31536000}{86400} \]

### Step 5: Substitute Values
Substituting \( R_E = 6400 \text{ km} \) and \( R_O = 1.5 \times 10^8 \text{ km} \):
\[ \frac{L_E}{L_O} = \frac{2}{5} \cdot \frac{(6400)^2}{(1.5 \times 10^8)^2} \cdot \frac{31536000}{86400} \]

### Step 6: Calculate the Numerical Value
Calculating each part:
1. \( (6400)^2 = 40960000 \)
2. \( (1.5 \times 10^8)^2 = 2.25 \times 10^{16} \)
3. \( \frac{31536000}{86400} = 365 \)

Now substituting these values:
\[ \frac{L_E}{L_O} = \frac{2}{5} \cdot \frac{40960000}{2.25 \times 10^{16}} \cdot 365 \]

Calculating this gives:
\[ \frac{L_E}{L_O} \approx 2.66 \times 10^{-7} \]

### Final Answer
The ratio of the angular momentum of the Earth about its axis due to its spinning motion to that about the Sun due to its orbital motion is approximately:
\[ \frac{L_E}{L_O} \approx 2.66 \times 10^{-7} \]