91影视

Given in the question,

The power released by the Nebula is equal to the energy lost by the Neutron star,$P=5脳{10}^3\text{鈥塛}$

\Time period,$T=0.331\text{鈥夆赌塻}$

Increase in period per second,$\frac{dT}{dt}=4.22脳{10}^{鈭�13}\text{鈥塻}$

Mass of neutron star$m=2.786脳{10}^{30}\text{鈥夆赌块驳}$

Rotational Motion

If the motion of an object is around a circular path, in a fixed orbit, then that motion is known as rotational motion.

Rotational energy

Rotational kinetic energy is also known as angular kinetic energy, this iskinetic energy due to the rotation of any object or body and is part of the total kinetic energy of the system.

The formula use are given below:

$P=\frac{dK}{dt}$

Power can be defined as the change in kinetic energy per time

Where P is power, K is kinetic energy and t is time

Angular speed

$蝇=\frac{2蟺}{T}$

Where $蝇$is angular speed and T is time period.

Linear speed

$v=\frac{2蟺R}{T}$

Where v is linear speed R is radius and T is time period

Rotational kinetic energy

$K_r=\frac{1}{2}I{蝇}^2$

Where$K_r$is rotational kinetic energy,$l$is angular momentum and$蝇$is angular velocity.

We know that the rotational kinetic energy of neutrons can be given as

$K=\frac{1}{2}I{蝇}^2$

Where I is the moment of inertia of neutron and $蝇$is angular velocity

We know

$蝇=\frac{2蟺}{T}$

So

$K=\frac{1}{2}I{\left(\frac{2蟺}{T}\right)}^2$

To find the power differentiating the energy with respect to time

$\begin{array}{l}\frac{dK}{dt}=\frac{d}{dt}\left[\frac{1}{2}I{\left(\frac{2蟺}{T}\right)}^2\right]\\ \frac{dK}{dt}=\frac{鈭�4蟺I}{T^3}\frac{dT}{dt}\end{array}$

We know

$P=\frac{dK}{dT}$

So, power loss by neutron

$\begin{array}{c}P=\frac{4{蟺}^{2}I}{T^3}\frac{dT}{dt}\\ I=\frac{P{T}^3}{4{蟺}^2\frac{dT}{dt}}\\ I=\frac{\left(5脳{10}^3\text{鈥塛}\right){\left(0.331\text{鈥夆赌塻}\right)}^3}{4{蟺}^2\left(4.22脳{10}^{鈭�13}\text{鈥塻}\right)}\\ I=1.09脳{10}^{38}{\text{鈥塳驳.尘}}^{\text{2}}\end{array}$

Hence the moment of inertia of the neutron is$1.09脳{10}^{38}{\text{鈥塳驳.尘}}^{\text{2}}$

The moment of inertia of a solid sphere can be given as

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

Where m is mass and r is the radius of the sphere.

Since we are modeling the neutron as a solid sphere, we can apply this formula to find its radius.

$\begin{array}{c}1.09脳{10}^{38}{\text{鈥塳驳.尘}}^{\text{2}}=\frac{2}{5}\left(2.786脳{10}^{30}\text{鈥夆赌块驳}\right)r^2\\ r=9889.92\text{鈥夆�尘}\\ r=9.89\text{鈥夆赌塳尘}\end{array}$

The radius of a neutron star is $9.89km$

The linear speed of neutron can be given by using the formula

$v=\frac{2蟺R}{T}$

Where v is linear speed R is radius and T is time period

$\begin{array}{l}v=\frac{2蟺R}{T}\\ v=\frac{2蟺\left(9889.92\text{鈥夆�尘}\right)}{\left(0.0331\text{鈥夆赌塻}\right)}\\ v=1.88脳{10}^6\text{鈥尘}/\text{s}\end{array}$

Hence the linear speed of a point on the equator of the neutron star is

$1.88脳{10}^6\text{鈥尘}/\text{s}$

This speed is less than the speed of light which is

$c=3脳{10}^8\text{鈥尘}/\text{s}$

We know that

$\text{density=}\frac{\text{mass}}{\text{volume}}$

$蚁=\frac{m}{v}$

Since we assume that the neutron star is a solid sphere, the volume of the neutron star can be given by

$v=\frac{4}{3}蟺r^3$

Where r is the radius of the neutron star.

Therefore

$\begin{array}{l}蚁=\frac{m}{v=\frac{4}{3}蟺r^3}\\ 蚁=\frac{\left(2.786脳{10}^{30}\text{鈥夆赌块驳}\right)}{\frac{4}{3}蟺{\left(9889.92\text{鈥夆�尘}\right)}^3}\\ 蚁=6.8756脳{10}^{17}\text{鈥夆赌块驳}/{\text{m}}^3\end{array}$

Hence the density of neutron stars is $6.8756脳{10}^{17}\text{鈥夆赌块驳}/{\text{m}}^3$

The density of rock is ${蚁}_r=3000\text{鈥夆赌块驳}/{\text{m}}^3$

The density of an atomic nucleus${蚁}_n={10}^{17}\text{鈥夆赌块驳}/{\text{m}}^3$

So, we can conclude that the density of neutron stars is much higher than the density of the ordinary rock.

Now we can see that the density of a neutron star is of the same order as the density of an atomic nucleus, hence a neutron star is essentially a larger atomic nucleus.