Question

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation of $\mathbf{T}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.033}\mathbf{\text{\hspace{0.17em}s}}\mathbf{}$ that is increasing at the rate of $\mathbf{1.26}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{\text{\hspace{0.17em}s}}\mathbf{/}\mathbf{\text{y}}$ .

(a) What is the pulsar’s angular acceleration $\mathbf{\text{α}}$?

(b) If $\mathbf{\text{α}}$is constant, how many years from now will the pulsar stop rotating?

(c) The pulsar originated in a supernova explosion seen in the year $\mathbf{1054}$ .Assuming constant a, find the initial T.

Short answer

1. The pulsar’s angular acceleration is $-2.3\times {10}^{-9}\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$.
2. The time when the pulsar will stop rotating is$2.6\times {10}^3\text{\hspace{0.17em}yrs}$.
3. The initial time period at which the pulsar originated is $0.024\text{\hspace{0.17em}s}$.

Step-by-step solution

Listing the given quantities

1. The period of rotation of the pulsar, ${\mathrm{T}}_0=0.033\text{\hspace{0.17em}s}$.
2. The rate of increase of the period,$1.26\times {10}^{-5}\text{\hspace{0.17em}s}/\text{yr}$ .

Understanding the kinematic equations

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.

$\mathbf{\alpha }\mathbf{=}\mathbf{}\frac{\mathbf{d\omega }}{\mathbf{dt}}$

$\mathbf{\omega }\mathbf{=}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{+}\mathbf{\alpha t}$

(a) Calculation of angular acceleration of pulsar

The period of rotation is increasing at the rate

$\frac{\text{d}\mathrm{T}}{\text{dt}}=1.26\times {10}^{-5}\text{\hspace{0.17em}s}/\text{yr}$

Let’s rewrite it as

$\begin{array}{c}\frac{\text{d}\mathrm{T}}{\text{dt}}=1.26\times {10}^{-5}\text{\hspace{0.17em}s}/\text{yr}\\ =1.26\times {10}^{-5}\text{\hspace{0.17em}s}/\text{yr}\times \frac{1\text{\hspace{0.17em}yr}}{365\text{\hspace{0.17em}days}\times 24\text{\hspace{0.17em}hrs}\times 3600\text{\hspace{0.17em}s}}\\ =3.99\times {10}^{-13}\end{array}$

It is known that,

$\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}$

The angular acceleration can be calculated using the definition as

$\begin{array}{c}\mathrm{\alpha }=\frac{\mathrm{d\omega }}{\mathrm{dt}}\\ =\frac{\mathrm{d\omega }}{\mathrm{dT}}.\frac{\mathrm{dT}}{\mathrm{dt}}\\ =\frac{-2\mathrm{\pi }}{{\mathrm{T}}^2}.\frac{\mathrm{dT}}{\mathrm{dt}}\\ =\frac{-2\times 3.14}{{\left(0.033\text{\hspace{0.17em}s}\right)}^2}\times 3.99\times {10}^{-13}\\ =-2.30\times {10}^{-9}\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\end{array}$

The pulsar’s angular acceleration is $-2.30\times {10}^{-9}\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$

(b) Calculation of time for which pulsar stop rotating.

We will calculate initial angular velocity of the pulsar as

$\begin{array}{c}\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}\\ {\mathrm{\omega }}_0=\frac{2\mathrm{\pi }}{{\mathrm{T}}_0}\\ =\frac{2\mathrm{\pi }}{0.033\text{\hspace{0.17em}s}}\\ {\mathrm{\omega }}_0=1.90\times {10}^2\text{\hspace{0.17em}rad}/\text{s}\end{array}$

We want to find the time at which the pulsar stops rotating, i.e. $\mathrm{\omega }=0\text{\hspace{0.17em}rad}/\text{s}$

So we use the kinematical equation as

$\begin{array}{c}\mathrm{t}=\frac{\mathrm{\omega }-{\mathrm{\omega }}_0}{\mathrm{\alpha }}\\ =\frac{0-1.90\times {10}^2\text{\hspace{0.17em}rad}/\text{s}}{–2.30\times {10}^{-9}{\text{rad}/\text{s}}^2}\\ =8.27\times {10}^{10}\text{\hspace{0.17em}s}\\ =8.27\times {10}^{10}\text{\hspace{0.17em}s}\times \frac{1\text{\hspace{0.17em}yr}}{3.15\times {10}^7\text{\hspace{0.17em}s}}\end{array}$

$\begin{array}{c}\mathrm{t}=2625.39\text{\hspace{0.17em}yrs}\\ \approx 2.6\times {10}^3\text{\hspace{0.17em}yrs}\end{array}$

The time when the pulsar will stop rotating is $2.6\times {10}^3\text{\hspace{0.17em}yrs}$.

(c) Calculation of initial time period at which pulsar originated

The present life of pulsar,

$\begin{array}{c}\mathrm{t}=2018\text{\hspace{0.17em}yrs}-1054\text{\hspace{0.17em}yrs}\\ =964\text{\hspace{0.17em}yrs}\end{array}$

Then period at the time of birth can be calculated as

$\mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}$

And

$\begin{array}{c}\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha }\mathrm{t}\\ =\frac{2\mathrm{\pi }}{{\mathrm{\omega }}_0+\mathrm{\alpha }\mathrm{t}}\\ =\frac{2\mathrm{\pi }}{1.90\times {10}^2\text{\hspace{0.17em}rad}/\text{s}-(-2.30\times {10}^{-9}{\text{rad}/\text{s}}^2\times 964\text{\hspace{0.17em}yrs}\times 3.15\times {10}^7\text{\hspace{0.17em}s})}\\ \mathrm{\omega }=0.024\text{\hspace{0.17em}s}\end{array}$

The initial time period at which the pulsar originated is $0.024\text{\hspace{0.17em}s}$.