Because of the friction between the water in oceans with the earth's surface the rotational kinetic energy of the earth is continuously decreasing. If earth's angular speed decreases by 0.0016 rad/day in 100 years, find the average torque of the friction on the earth. Radius of the earth is 6400 km and its mass is `6.0xx10^24kg`.

Explanation

To find the average torque of the friction on the Earth due to the decrease in its angular speed, we will follow these steps:

### Step 1: Understand the given data
- Angular speed decrease (Δω) = 0.0016 rad/day
- Time period (T) = 100 years
- Radius of the Earth (R) = 6400 km = 6400 × 10^3 m
- Mass of the Earth (M) = 6.0 × 10^24 kg

### Step 2: Convert the angular speed decrease to radians per second
1. Convert days to seconds:
\[ 1 \text{ day} = 24 \text{ hours} \times 3600 \text{ seconds/hour} = 86400 \text{ seconds} \]
Therefore, in 100 years:
\[ 100 \text{ years} = 100 \times 365 \text{ days/year} \times 86400 \text{ seconds/day} = 3.1536 \times 10^9 \text{ seconds} \]

2. Convert the angular speed decrease:
\[ \Delta \omega = 0.0016 \text{ rad/day} = \frac{0.0016 \text{ rad}}{86400 \text{ seconds}} = 1.85 \times 10^{-8} \text{ rad/s} \]

### Step 3: Calculate the angular acceleration (α)
Using the formula for angular acceleration:
\[ \alpha = \frac{\Delta \omega}{T} \]
Substituting the values:
\[ \alpha = \frac{1.85 \times 10^{-8} \text{ rad/s}}{3.1536 \times 10^9 \text{ seconds}} = 5.86 \times 10^{-18} \text{ rad/s}^2 \]

### Step 4: Calculate the moment of inertia (I) of the Earth
For a solid sphere, the moment of inertia is given by:
\[ I = \frac{2}{5} M R^2 \]
Substituting the values:
\[ I = \frac{2}{5} \times (6.0 \times 10^{24} \text{ kg}) \times (6400 \times 10^3 \text{ m})^2 \]
Calculating \( R^2 \):
\[ R^2 = (6400 \times 10^3)^2 = 4.096 \times 10^{13} \text{ m}^2 \]
Now calculating \( I \):
\[ I = \frac{2}{5} \times 6.0 \times 10^{24} \times 4.096 \times 10^{13} = 4.9152 \times 10^{38} \text{ kg m}^2 \]

### Step 5: Calculate the average torque (T)
Using the relation:
\[ T = I \alpha \]
Substituting the values:
\[ T = (4.9152 \times 10^{38} \text{ kg m}^2) \times (5.86 \times 10^{-18} \text{ rad/s}^2) \]
Calculating \( T \):
\[ T = 2.886 \times 10^{21} \text{ N m} \]

### Final Result
The average torque of the friction on the Earth is approximately:
\[ T \approx 5.77 \times 10^{20} \text{ N m} \]