Certain extremely dense neutron stars are believed to be rotating at about 1 rev/sec. If such a star has a radius of 20 km, the acceleration of an object in m//sec^(2)` on the equator of the star will be

Options

A. `8 xx 10^(5)`
B. `20 xx 10^(3)`
C. `12 xx 10^(6)`
D. `4 xx 10^(8)`

Explanation

To find 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 1: Understand the Concept of Centripetal Acceleration
Centripetal acceleration (\(a_c\)) is the acceleration that keeps an object moving in a circular path and is directed towards the center of the circle. It can be calculated using the formula:
\[ a_c = r \omega^2 \]
where \(r\) is the radius and \(\omega\) is the angular velocity in radians per second.

### Step 2: Convert the Radius from Kilometers to Meters
The radius of the star is given as 20 km. We need to convert this into meters:
\[ r = 20 \text{ km} = 20 \times 10^3 \text{ m} = 20000 \text{ m} \]

### Step 3: Calculate the Angular Velocity (\(\omega\))
The star is rotating at a frequency of 1 revolution per second (rps). To convert this to angular velocity in radians per second, we use the formula:
\[ \omega = 2\pi n \]
where \(n\) is the frequency in revolutions per second. For \(n = 1\):
\[ \omega = 2\pi \times 1 = 2\pi \text{ radians/second} \]

### Step 4: Substitute \(\omega\) into the Centripetal Acceleration Formula
Now we can substitute \(r\) and \(\omega\) into the centripetal acceleration formula:
\[ a_c = r \omega^2 = (20000) \times (2\pi)^2 \]

### Step 5: Calculate \((2\pi)^2\)
Calculating \((2\pi)^2\):
\[ (2\pi)^2 = 4\pi^2 \approx 4 \times (3.14)^2 \approx 4 \times 9.8596 \approx 39.3984 \]

### Step 6: Calculate the Centripetal Acceleration
Now substitute this value back into the equation for \(a_c\):
\[ a_c = 20000 \times 39.3984 \approx 787968 \text{ m/s}^2 \]

### Step 7: Final Result
Thus, the centripetal acceleration at the equator of the neutron star is approximately:
\[ a_c \approx 7.88 \times 10^5 \text{ m/s}^2 \]

### Conclusion
The acceleration of an object on the equator of the star is approximately \(7.88 \times 10^5 \text{ m/s}^2\).

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