Certain neutron stars are believed to be rotating at about `1rev//sec` . If such a star has a radius of 20 km , the acceleration of an object on the equator of the star will be

Options

A. `20 xxx 10^(8)m//sec^(2)`
B. `8 xx 10^(5) m//sec^(2)`
C. `120 xx 10^(5) m//sec^(2)`
D. `4 xx 10^(8) m//sec^(2)`

Answer

To solve the problem of finding the acceleration of an object on the equator of a neutron star rotating at 1 revolution per second with a radius of 20 km, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Frequency of rotation (f) = 1 revolution/second
- Radius of the neutron star (r) = 20 km = 20,000 m (since 1 km = 1000 m)

2. **Convert Frequency to Angular Velocity**:
- The relationship between angular velocity (ω) and frequency (f) is given by:
\[ \omega = 2\pi f \]
- Substituting the given frequency:
\[ \omega = 2\pi \times 1 = 2\pi \, \text{radians/second} \]

3. **Use the Formula for Centripetal Acceleration**:
- The centripetal acceleration (A) at the equator is given by the formula:
\[ A = \omega^2 r \]
- Substitute the values of ω and r:
\[ A = (2\pi)^2 \times 20,000 \]

4. **Calculate ω²**:
- Calculate \( (2\pi)^2 \):
\[ (2\pi)^2 = 4\pi^2 \approx 4 \times 9.87 \approx 39.48 \]
- So, we have:
\[ A = 39.48 \times 20,000 \]

5. **Final Calculation**:
- Now, calculate the acceleration:
\[ A = 39.48 \times 20,000 = 789,600 \, \text{m/s}^2 \]
- We can express this in scientific notation:
\[ A \approx 7.896 \times 10^5 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of an object on the equator of the neutron star is approximately \( 7.896 \times 10^5 \, \text{m/s}^2 \).

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