Question

During most of its lifetime, s star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen fuel is consumed by the nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These pulsing stars were discovered in the 1960's and are called pulsars.a) A star with the mass (M=2.0x10^30 Kg) and size (R=7.0x10^8 m) of our sun rotates once every 30 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.10 s. By treating the neutron star as a solid sphere, deduce its radius.b) What is the speed of a point on the equator of the neutron star? Your answer will be somewhat too large because a star cannot be accurately modeled as a solid sphere. Even so, you will be able to show that a star, whose mass is 10^6 larger than the earth's, can be compressed by gravitational forces to a size smaller than a typical state in the United States!

Answer 1

Answer:

$r_f=137493m$

$v=8638940m/s$

Explanation:

During this process the mass $M=2\times10^{30}Kg$ will be considered constant. We start from a radius $r_i=7\times10^8m$ and a period $T_i=30\ days=(30)(24)(60)(60)s=2592000s$. The final period is $T_f=0.1s$.

Angular momentum L is conserved in this process. We can use the formula $L=I\omega$, where I is the momentum of inertia (which for a solid sphere is $I=\frac{2mr^2}{5}$) and $\omega=\frac{2\pi }{T}$ is the angular velocity, so we can write the star's angular momentum as:

$L=I\omega=\frac{2mr^2}{5}\frac{2\pi }{T}=\frac{4\pi mr^2 }{5T}$

Since $L_f=L_i$ we have:

$\frac{4\pi mr_f^2 }{5T_f}=\frac{4\pi mr_i^2 }{5T_i}$

Which can be simplified as:

$\frac{r_f^2 }{T_f}=\frac{r_i^2 }{T_i}$

Which means:

$r_f=\sqrt{\frac{r_i^2 T_f}{T_i}}=r_i \sqrt{\frac{T_f}{T_i}}$

Which for our values is:

$r_f=r_i \sqrt{\frac{T_f}{T_i}}=(7\times10^8m) \sqrt{\frac{0.1s}{2592000s}}=137493m$

And we calculate the speed of a point on the equator by dividing the final circumference over the final period:

$v=\frac{C_f}{T_f}=\frac{2\pi r_f}{T_f}=\frac{2\pi (137493m)}{(0.1s)}=8638940m/s$