91Ó°ÊÓ

1. The period of rotation of the pulsar, ${\mathrm{T}}_0=0.033\text{ s}$.
2. The rate of increase of the period,$1.26×{10}^{−5}\text{ s}/\text{yr}$ .

The pulsar rotates about its axis. So, we can use the rotational kinematic equations to determine the time of decay and time of its origin.

$\mathbf{Î\pm }\mathbf{=}\mathbf{}\frac{\mathbf{»åÓ\neg }}{\mathbf{dt}}$

$\mathbf{Ó\neg }\mathbf{=}{\mathbf{Ó\neg }}_{\mathbf{0}}\mathbf{+}\mathbf{Î\pm ³Ù}$

The period of rotation is increasing at the rate

$\frac{\text{d}\mathrm{T}}{\text{dt}}=1.26×{10}^{−5}\text{ s}/\text{yr}$

Let’s rewrite it as

$\begin{array}{c}\frac{\text{d}\mathrm{T}}{\text{dt}}=1.26×{10}^{−5}\text{ s}/\text{yr}\\ =1.26×{10}^{−5}\text{ s}/\text{yr}×\frac{1\text{ y°ù}}{365\text{ d²¹²â²õ}×24\text{ h°ù²õ}×3600\text{ s}}\\ =3.99×{10}^{−13}\end{array}$

It is known that,

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{\mathrm{T}}$

The angular acceleration can be calculated using the definition as

$\begin{array}{c}\mathrm{Î\pm }=\frac{\mathrm{»åÓ\neg }}{\mathrm{dt}}\\ =\frac{\mathrm{»åÓ\neg }}{\mathrm{dT}}.\frac{\mathrm{dT}}{\mathrm{dt}}\\ =\frac{−2\mathrm{Ï€}}{{\mathrm{T}}^2}.\frac{\mathrm{dT}}{\mathrm{dt}}\\ =\frac{−2×3.14}{{\left(0.033\text{ s}\right)}^2}×3.99×{10}^{−13}\\ =−2.30×{10}^{−9}\text{ r²¹»å}/{\text{s}}^{\text{2}}\end{array}$

The pulsar’s angular acceleration is $−2.30×{10}^{−9}\text{ r²¹»å}/{\text{s}}^{\text{2}}$

We will calculate initial angular velocity of the pulsar as

$\begin{array}{c}\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{\mathrm{T}}\\ {\mathrm{Ó\neg }}_0=\frac{2\mathrm{Ï€}}{{\mathrm{T}}_0}\\ =\frac{2\mathrm{Ï€}}{0.033\text{ s}}\\ {\mathrm{Ó\neg }}_0=1.90×{10}^2\text{ r²¹»å}/\text{s}\end{array}$

We want to find the time at which the pulsar stops rotating, i.e. $\mathrm{Ó\neg }=0\text{ r²¹»å}/\text{s}$

So we use the kinematical equation as

$\begin{array}{c}\mathrm{t}=\frac{\mathrm{Ó\neg }−{\mathrm{Ó\neg }}_0}{\mathrm{Î\pm }}\\ =\frac{0−1.90×{10}^2\text{ r²¹»å}/\text{s}}{–2.30×{10}^{−9}\text{ }{\text{rad}/\text{s}}^2}\\ =8.27×{10}^{10}\text{ s}\\ =8.27×{10}^{10}\text{ s}×\frac{1\text{ y°ù}}{3.15×{10}^7\text{ s}}\end{array}$

$\begin{array}{c}\mathrm{t}=2625.39\text{ y°ù²õ}\\ ≈2.6×{10}^3\text{ y°ù²õ}\end{array}$

The time when the pulsar will stop rotating is $2.6×{10}^3\text{ y°ù²õ}$.

The present life of pulsar,

$\begin{array}{c}\mathrm{t}=2018\text{ y°ù²õ}−1054\text{ y°ù²õ}\\ =964\text{ y°ù²õ}\end{array}$

Then period at the time of birth can be calculated as

$\mathrm{T}=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}$

And

$\begin{array}{c}\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }\mathrm{t}\\ =\frac{2\mathrm{Ï€}}{{\mathrm{Ó\neg }}_0+\mathrm{Î\pm }\mathrm{t}}\\ =\frac{2\mathrm{Ï€}}{1.90×{10}^2\text{ r²¹»å}/\text{s}−(−2.30×{10}^{−9}\text{ }{\text{rad}/\text{s}}^2×964\text{ y°ù²õ}×3.15×{10}^7\text{ s})}\\ \mathrm{Ó\neg }=0.024\text{ s}\end{array}$

The initial time period at which the pulsar originated is $0.024\text{ s}$.