91影视

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${\mathrm{f}}_{\mathrm{r}}={\mathrm{f}}_{\mathrm{s}}{\left(1鈭�\frac{\mathrm{v}}{\mathrm{c}}\right)}^{\frac{1}{2}}{\left(1+\frac{\mathrm{v}}{\mathrm{c}}\right)}^{鈭�\frac{1}{2}}$ 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,

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

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

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

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

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

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

The supernova is occurred 960 years ago from the year 2015 and the speed of the gas of the supernova is$-1.18脳{10}^6\mathrm{m}/\mathrm{s}$ 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.

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

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

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

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

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

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

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

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