Question

A star with the mass (M=2.0×1030kg) and size (R=3.5×108m) of our sun rotates once every 30.0 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.200 s . By treating the neutron star as a solid sphere, deduce its radius.

Answer 1

Answer:

$r=97.22x10^{3} m$

Explanation:

Using the angular formulas can determine the radius using both values neutron star and the the knowing star so

$L=I*w$

$L_{1}=I_{1}*w_{1}=L_{2}=I_{2}*w_{2}$

$I_{1}*w_{1}=I_{2}*w_{2}$

I=Inertia of the star

w=angular velocity

$I=\frac{2*m*r^{2}}{5}$

$w=\frac{2\pi}{t}$

Notice the angular velocity determinate by the time and the Inertia have the radius value so

$\frac{2}{5}*m*r_{sn}^{2}*\frac{2\pi }{t_{1}}=\frac{2}{5}*m*r_{s}^{2}*\frac{2\pi }{t_{2}}$

$r_{sn}^{2}*\frac{1}{t_{1}}=r_{s}^{2}*\frac{1}{t_{2}}$

$r_{sn}^{2}=r_{s}^{2}*\frac{t_{1}}{t_{2}}$

$t_{1}=0.2s\\t_{2}=30day*\frac{24hr}{1day}*\frac{60minute}{1hr}*\frac{60seg}{1minute}=2.592x10^{6}s$

$r_{sn}=3.5x10^{8}m*\sqrt{\frac{0.2s}{2.592x^{6}s}}$

$r_{sn}=97.22x10^{3} m$