Question

A neutron star has a density equal to that of nuclear matter (2.8×10^17 kg m^-3). Assume the star to be spherical, find the radius of the neutron star whose mass is 4.0×10^10 kg (twice the mass of the sun) .

Answer 1

given,
density of neutron star,
$p = 2.8 \times {10}^{17} kg \: {m}^{ - 3}$
mass of neutron star,
$m = 4.0 \times {10}^{10} kg$
density(p) is given by,

$p = \frac{m}{v}$
$p = \frac{m}{ \frac{4}{3}\pi \: {r}^{3} } \: \: \: \: \: \: \:$
because,
$(v = \frac{4}{3} \pi \: {r}^{3} )$
${r}^{3 } = \frac{3m}{4\pi \: p}$
${r}^{3 } = \frac{3 \times 4.0 \times {10}^{10} }{4 \times 3.14 \times 2.8 \times \ {10}^{17} }$
after calculation,
${r}^{3} = 0.341 \times {10}^{ - 7}$
$r = 0.32 \times {10}^{ - 2} m$
hope this help........:)

Answer 2

Answer:

Radius, $r=3.24\times 10^{-6}\ km$

Explanation:

It is given that,

Density of nuclear matter, $\rho=2.8\times 10^{17}\ kg/m^3$

Mass of the neutron star, $m=4\times 10^{10}\ kg$

We know that mass per unit volume is called the density of the star.

$\rho=\dfrac{m}{V}$

$V=\dfrac{m}{\rho}$

It is assumed that star to be spherical. So,

$\dfrac{4}{3}\pi r^3=\dfrac{m}{\rho}$

$r^3=\dfrac{3m}{4\pi\rho}$

$r^3=\dfrac{3\times 4\times 10^{10}}{4\times 3.14\times 2.8\times 10^{17}}$

$r^3=0.00324\ m$

$r=3.24\times 10^{-6}\ km$

Hence, the radius of the neutron star is $r=3.24\times 10^{-6}\ km$