Question

DATA Supernova! (a) Equation (16.30) can be written as ${\mathbf{f}}_{\mathbf{r}}\mathbf{=}{\mathbf{f}}_{\mathbf{s}}{\left(1‐\frac{v}{c}\right)}^{\frac{\mathbf{1}}{\mathbf{2}}}{\left(1+\frac{v}{c}\right)}^{\frac{\mathbf{1}}{\mathbf{2}}}$

where c is the speed of light in vacuum$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{m}\mathbf{/}\mathbf{s}$. Most objects move much slower than this (v/c is very small), so calculations made with Eq. (16.30) must be done carefully to avoid rounding errors. Use the binomial theorem to show that if v, Eq. (16.30) approximately reduces to${\mathbf{f}}_{\mathbf{r}}\mathbf{=}{\mathbf{f}}_{\mathbf{s}}\left(1-\left(\frac{v}{c}\right)\right)$. (b) The gas cloud known as the Crab Nebula can be seen with even a small telescope. It is the remnant of a supernova, a cataclysmic explosion of a star. (The explosion was seen on the earth on July 4,1054 C.E.) Its streamers glow with the characteristic red colour of heated hydrogen gas. In a laboratory on the earth, heated hydrogen produces red light with frequencyrole="math" localid="1668146026646" $\mathbf{4}\mathbf{.}\mathbf{568}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{Hz}$; the red light received from streamers in the Crab Nebula that are pointed toward the earth has frequency$\mathbf{4}\mathbf{.}\mathbf{568}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{Hz}$Estimate the speed with which the outer edges of the Crab Nebula are expanding. Assume that the speed of the centre of the nebula relative to the earth is negligible. (c) Assuming that the expansion speed of the Crab Nebula has been constant since the supernova that produced it, estimate the diameter of the Crab Nebula. Give your answer in meters and in light-years. (d) The angular diameter of the Crab Nebula as seen from the earth is about 5 arc-minutes$\mathbf{(}\mathbf{1}\mathbf{arc}\mathbf{‐}\mathbf{min}\frac{\mathbf{1}}{\mathbf{60}}\mathbf{degree}\mathbf{)}$Estimate the distance (in light-years) to the Crab Nebula, and estimate the year in which the supernova actually, took place.

Short answer

a) The expression of the frequency of the sound measured by the receiver when v c is${\mathrm{f}}_{\mathrm{s}}=\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)$ b) The speed of gas cloud Crab Nebula is$1.18\times {10}^8\mathrm{m}/\mathrm{sc}$ c) The diameter of the supernova in terms of meters is $7.5\times {10}^{16}\mathrm{m}$and the diameter of the supernova in light year is 7.6ly d) The distance of the Crab Nebula from the earth is $10.45\times {10}^3\mathrm{ly}$, and the explosion of the supernova actually took place before$10.45\times {10}^3\mathrm{ly}$

Step-by-step solution

Calculate the frequency of the sound

The speed of the light in vacuum is 3.0 x 108 m/s.

Formula to calculate the frequency of the sound measured by the receiver according to the Doppler effect is${\mathrm{f}}_{\mathrm{r}}={\mathrm{f}}_{\mathrm{s}}{\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)}^{\frac{1}{2}}{\left(1+\frac{\mathrm{v}}{\mathrm{c}}\right)}^{-\frac{1}{2}}$ where, is the frequency of the light measured by the receiver.is the frequency of the light emitted by the source, v is the speed of the light in the medium, c is the speed of the light in vacuum.

According to binomial theorem, if then

Applying binomial theorem in the equation (l) whento find,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{r}}={\mathrm{f}}_{\mathrm{s}}{\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)}^{\frac{1}{2}}{\left(1+\frac{\mathrm{v}}{\mathrm{c}}\right)}^{-\frac{1}{2}} \\ ={\mathrm{f}}_{\mathrm{s}}\left(1-\frac{1}{2}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)\right)\left(1+\frac{1}{2}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)\right) \\ \begin{array}{rl}& ={\mathrm{f}}_{\mathrm{s}}{\left(1-\frac{1}{2}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)\right)}^2\end{array} \\ ={\mathrm{f}}_{\mathrm{s}}\left(1-\frac{1}{2}\left(2\right)\left(\frac{\mathrm{v}}{\mathrm{c}}\right)\right) \\ ={\mathrm{f}}_{\mathrm{s}}\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right) \end{gathered}$$

Thus, the expression of the frequency of the sound measured by the receiver in the case of v $֏$s is $={\mathrm{f}}_{\mathrm{s}}\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)$

Calculate the speed of gas cloud Crab Nebula

The frequency of the red light emitted by the Nebula is$4.568\times {10}^{14}\mathrm{Hz}$, the frequency of the light received in earth is$4.568\times {10}^{14}\mathrm{Hz}$, the speed of the light in vacuum is$3.0\times {10}^8\mathrm{m}/\mathrm{s}$

Formula to calculate the frequency of the sound measured by the receiver forfrom the solution of part

$$\begin{gathered} {\mathrm{f}}_{\mathrm{r}}={\mathrm{f}}_{\mathrm{s}}\left(1-\left(\frac{\mathrm{v}}{\mathrm{c}}\right)\right) \\ 4.568\times {10}^{14}\mathrm{Hz}=4.568\times {10}^{14}\mathrm{Hz}\left(1-\frac{\mathrm{v}}{3.0\times {10}^8\mathrm{m}/\mathrm{s}}\right) \\ \frac{\mathrm{v}}{3.0\times {10}^8\mathrm{m}/\mathrm{s}}=1-\frac{4.568\times {10}^{14}\mathrm{Hz}}{4.568\times {10}^{14}\mathrm{Hz}} \\ \mathrm{V}=1.18\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of gas cloud Crab Nebula is $1.18\times {10}^8\mathrm{m}/\mathrm{s}$

STEP 3 Calculate the diameter of the supernova

The supernova is occurred 960 years ago from the year 2015 and the speed of the gas of the supernova is$-1.18\times {10}^6\mathrm{m}/\mathrm{s}$ Formula to calculate the diameter of the supernova is, D = yv where,D is the diameter of the supernova, v is the number of years that the supernova were occurred, v is the speed of the supernova gas.

$$\begin{gathered} \mathrm{D}=960\times 1.18\times {10}^6\mathrm{m}/\mathrm{s} \\ =3.52\times {10}^{16}\mathrm{m}/\mathrm{s} \end{gathered}$$

Convert the diameter of the supernova from meter to light years is, ${\mathrm{D}}_{\mathrm{y}}={\mathrm{D}}_{\mathrm{m}}\times \left(\frac{1\mathrm{ly}}{9.46\times {10}^{15}\mathrm{m}}\right)$

Substitute $3.52\times {10}^{16}\mathrm{m}/\mathrm{s}$ for in the above equation to find D_ly

$$\begin{gathered} {\mathrm{D}}_{\mathrm{ly}}=3.52\times {10}^{16}\mathrm{m}/\mathrm{s}\times \left(\frac{1\mathrm{ly}}{9.46\times {10}^{15}\mathrm{m}}\right) \\ =7.6\mathrm{ly} \end{gathered}$$

Therefore, the diameter of the supernova in terms of meters is $7.5\times {10}^{16}\mathrm{m}$and the diameter of the supernova in light year is 7.6ly

Calculate the distance of the Nebula from the earth surface

Formula is$\mathrm{s}=\left(\frac{2}{2\mathrm{x}}\right){\left({\mathrm{D}}_{\mathrm{ij}}\right)}_{\mathrm{N}}\left(\frac{1}{\mathrm{a}}\right)$ were,s is the distance of the Nebula from the earth.

${\left({\mathrm{D}}_{\mathrm{ly}}\right)}_{\mathrm{N}}$diameter of the Nebula, a is the arc-minutes. Substitute the values in the above equation

$$\begin{gathered} \mathrm{s}=\left(\frac{2}{2\mathrm{x}}\right){\left({\mathrm{D}}_{\mathrm{ij}}\right)}_{\mathrm{N}}\left(\frac{1}{\mathrm{a}}\right) \\ \mathrm{s}=\left(\frac{2}{2\mathrm{x}}\right)(7.6\mathrm{ly})\left(\frac{1}{5\mathrm{arc}-\mathrm{m}\mathrm{inutes}\times \frac{1\mathrm{deg}}{360\times 60-\mathrm{arcmin}}}\right) \\ =10.45\times {10}^3\mathrm{ly} \end{gathered}$$

Therefore, the distance of the Crab Nebula from the earth is $10.45\times {10}^3\mathrm{ly}$, and the explosion of the supernova actually took place before$10.45\times {10}^3\mathrm{ly}$