Question

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 10^14 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0×10^5 km (comparable to our sun); its final radius is 18km .

Answer 1

The angular speed of the neutron star is 3130.5 rad/s.

Step-by-step explanation:

Given that,

Initial radius
`r_(1)=7*10^(5)\ km`

Final radius
`r_(2)=18 km`

Density of a neutron
`\rho= 10^(14)`

Equal masses of two stars
`m_(1)=m_(2)`

Suppose, If the original star rotated once in 35 days, find the angular speed of the neutron star

Time period of original star T = 35 days = 3024000 s

We need to calculate the initial angular speed of original star

Using formula of angular star

`\omega=(2\pi)/(T)`

Put the value into the formula

`\omega_(1)=(2\pi)/(3024000)`

`\omega_(1)=0.00207*10^(-3)\ rad/s`

Let the initial moment of inertia of the star is

`I_(1)=m_(1)r_(1)^2`

Final moment of inertia of the star is

`I_(2)=m_(2)r_(2)^2`

From the conservation of angular momentum

`I_(1)\omega_(1)=I_(2)\omega_(2)`

`\omega_(2)=(I_(1)\omega_(1))/(I_(2))`

`\omega_(2)=(m_(1)r_(1)^2\omega_(1))/(m_(2)r_(2)^2)`

Put the value into the formula

`\omega_(2)=((7.0*10^(5))^2*0.00207*10^(-3))/(18^2)`

`\omega_(2)=3130.5\ rad/s`

Hence, The angular speed of the neutron star is 3130.5 rad/s.