What constant force, tangential to the equator should be applied to the earth to stop its rotation in one day

Options

A. `1.3xx10^(22)` N
B. `8.26xx10^(28)` N
C. `1.3xx10^(23)` N
D. None of these

Explanation

To solve the problem of determining the constant force tangential to the equator that should be applied to the Earth to stop its rotation in one day, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Initial Angular Velocity (\( \omega_0 \))**:
The Earth completes one rotation in 24 hours. To find the angular velocity in radians per second:
\[ \omega_0 = \frac{2\pi \text{ radians}}{24 \text{ hours}} = \frac{2\pi}{24 \times 3600} \text{ rad/s} \]
\[ \omega_0 = \frac{2\pi}{86400} \text{ rad/s} \approx 7.272 \times 10^{-5} \text{ rad/s} \]

2. **Calculate the Time (\( t \))**:
The time to stop the rotation is given as 1 day, which is:
\[ t = 24 \text{ hours} = 86400 \text{ seconds} \]

3. **Determine Angular Acceleration (\( \alpha \))**:
Since the final angular velocity (\( \omega_f \)) is 0 (the Earth stops rotating), we can use the formula:
\[ \alpha = \frac{\omega_f - \omega_0}{t} = \frac{0 - \omega_0}{t} = -\frac{\omega_0}{t} \]
Substituting the values:
\[ \alpha = -\frac{7.272 \times 10^{-5}}{86400} \approx -8.4 \times 10^{-10} \text{ rad/s}^2 \]

4. **Calculate the Moment of Inertia (\( I \))**:
The Earth can be approximated as a solid sphere, and its moment of inertia is given by:
\[ I = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the Earth (\( m \approx 6 \times 10^{24} \) kg) and \( r \) is the radius of the Earth (\( r \approx 6400 \) km = \( 6.4 \times 10^6 \) m).

5. **Substituting Values into the Moment of Inertia**:
\[ I = \frac{2}{5} \times (6 \times 10^{24}) \times (6.4 \times 10^6)^2 \]

6. **Calculate the Torque (\( \tau \))**:
The torque is related to angular acceleration by:
\[ \tau = I \alpha \]

7. **Relate Torque to Force**:
The torque can also be expressed as:
\[ \tau = F \cdot r \]
where \( F \) is the tangential force. Setting the two expressions for torque equal gives:
\[ F \cdot r = I \alpha \]
Thus, we can solve for \( F \):
\[ F = \frac{I \alpha}{r} \]

8. **Substituting Values**:
Substitute \( I \) and \( \alpha \) into the equation to find \( F \):
\[ F = \frac{\left(\frac{2}{5} \times (6 \times 10^{24}) \times (6.4 \times 10^6)^2\right) \times (-8.4 \times 10^{-10})}{6.4 \times 10^6} \]

9. **Calculate the Final Result**:
After performing the calculations, we find:
\[ F \approx 1.3 \times 10^{22} \text{ N} \]

### Final Answer:
The constant force tangential to the equator that should be applied to the Earth to stop its rotation in one day is approximately \( 1.3 \times 10^{22} \) Newtons.

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