A star 2.5 times the mass of the sun and collapsed to size of 12 km rotates with a speed of 1.5 rev per second ( extremely compact stars of this kind are known as neutron stars. Certain observed stellar objects called pulsars are believed to belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? ( Mass of the sun `= 2 xx 10^(30) kg ` )

Explanation

To determine whether an object placed on the equator of a neutron star will remain stuck to its surface due to gravity, we need to compare the gravitational acceleration acting on the object with the centripetal acceleration due to the star's rotation.

### Step 1: Calculate the gravitational acceleration (g) at the surface of the neutron star.

The formula for gravitational acceleration is given by:

\[ g = \frac{G \cdot M}{R^2} \]

Where:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) (gravitational constant)
- \( M = 2.5 \times M_{\text{sun}} = 2.5 \times 2 \times 10^{30} \, \text{kg} = 5 \times 10^{30} \, \text{kg} \) (mass of the neutron star)
- \( R = 12 \, \text{km} = 12 \times 10^3 \, \text{m} \) (radius of the neutron star)

Substituting the values into the formula:

\[ g = \frac{(6.67 \times 10^{-11}) \cdot (5 \times 10^{30})}{(12 \times 10^3)^2} \]

Calculating the denominator:

\[ (12 \times 10^3)^2 = 144 \times 10^6 = 1.44 \times 10^8 \, \text{m}^2 \]

Now substituting back into the equation for \( g \):

\[ g = \frac{(6.67 \times 10^{-11}) \cdot (5 \times 10^{30})}{1.44 \times 10^8} \]

Calculating \( g \):

\[ g \approx \frac{3.335 \times 10^{20}}{1.44 \times 10^8} \approx 2.32 \times 10^{12} \, \text{m/s}^2 \]

### Step 2: Calculate the centripetal acceleration (a_c) at the surface of the neutron star.

The formula for centripetal acceleration is given by:

\[ a_c = \omega^2 \cdot R \]

Where:
- \( \omega = 2 \pi n \) (angular velocity)
- \( n = 1.5 \, \text{rev/s} \)

First, convert revolutions per second to radians per second:

\[ \omega = 2 \pi \cdot 1.5 = 3 \pi \, \text{rad/s} \]

Now calculating \( a_c \):

\[ a_c = (3 \pi)^2 \cdot (12 \times 10^3) \]

Calculating \( (3 \pi)^2 \):

\[ (3 \pi)^2 \approx 9 \cdot 9.87 \approx 88.9 \]

Now substituting back into the equation for \( a_c \):

\[ a_c \approx 88.9 \cdot (12 \times 10^3) \approx 1.0668 \times 10^6 \, \text{m/s}^2 \]

### Step 3: Compare gravitational acceleration and centripetal acceleration.

Now we compare \( g \) and \( a_c \):

- \( g \approx 2.32 \times 10^{12} \, \text{m/s}^2 \)
- \( a_c \approx 1.0668 \times 10^6 \, \text{m/s}^2 \)

Since \( g \) is much greater than \( a_c \):

\[ g \gg a_c \]

### Conclusion

An object placed on the equator of the neutron star will remain stuck to its surface due to the overwhelming gravitational force compared to the centripetal force acting outward due to the star's rotation.

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