Question

What is the density of neutron star ?

• A. $6.87\times {10}^{6}kg{m}^{-3}$
• B. $5.46\times {10}^{12}kg{m}^{-3}$
• C. $2.42\times {10}^{17}kg{m}^{-3}$
• D. none

Answer 1

Answer: (A)

$6.87\times {10}^{6}kg{m}^{-3}$

We have $\omega =\frac{2\pi }{T}$ and $K=\frac{1}{2}I{\omega }^2$
or
$K=\frac{2{\pi }^{2}I}{T^2}$
or
$\frac{dK}{dt}=-\frac{4{\pi }^{2}I}{T^2}\frac{dT}{dt}$.
This is the rate of energy loss.
Now moment of inertia
$I=\frac{P{T}^3}{4\pi }\frac{1}{dT/dt}=\frac{(5\times {10}^{31})(0.0331{)}^3}{4{\pi }^2}\frac{1}{4.22\times {10}^{-13}}=1.09\times {10}^{38}kg{m}^2$
Now radius of neutron star
$R=\sqrt{\frac{5I}{2M}}=\sqrt{\frac{5(1.09\times {10}^{38})}{2\left(1.4\right)(1.99\times {10}^{30})}}=9.9\times {10}^{3}m$
Now velocity is given as
$v=\frac{2\pi R}{T}=\frac{2\pi (9.9\times {10}^3)}{0.0331}=1.9\times {10}^{6}m{s}^{-1}$
Now the density is given as
$\rho =\frac{M}{V}=\frac{M}{4\pi /3R^3}=6.9\times {10}^{17}kg{m}^{-3}$