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Q40P

Expert-verifiedFound in: Page 1364

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Because the apparent recessional speeds of galaxies and quasars at great distances are close to the speed of light, the relativistic Doppler shift formula (Eq.37-31) must be used. The shift is reported as fractional red shift ${\mathit{z}}{\mathbf{=}}\frac{\mathbf{\Delta}\mathbf{\lambda}}{{\mathbf{\lambda}}_{\mathbf{0}}}$**

**(a)Show that, in terms ****of,${\mathit{z}}$**** the recessional speed ****parameter${\mathit{\beta}}{\mathbf{=}}\frac{\mathbf{v}}{\mathbf{c}}$ is**** given by**

** **** ${\mathit{\beta}}{\mathbf{=}}\frac{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{z}}{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{+}\mathbf{2}}$.**

**(b) A quasar in 1987 has ${\mathit{z}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.43}}$ . Calculate its speed parameter.**

**(c) Find the distance to the quasar, assuming that Hubble’s law is valid to these distances.**

(a) It can be shown that,the recessional speed parameter$\beta =\frac{v}{c}$ can been expressed in terms of $z$is given by .$\beta =\frac{{z}^{2}+2z}{{z}^{2}+2z+2}$

(b) Speed parameter, $\beta =0.934$

(c) The distance to the quasar, $r=1.28\times {10}^{10}\text{ly}$width="113" style="max-width: none; vertical-align: -4px;" $r=1.28\times {10}^{10}\text{ly}$

Consider the relativistic Doppler shift formula (Eq.37-31) as follows,

${\mathit{\lambda}}{\mathbf{=}}{{\mathit{\lambda}}}_{{\mathbf{0}}}\sqrt{\frac{\mathbf{1}\mathbf{+}\mathbf{\beta}}{\mathbf{1}\mathbf{-}\mathbf{\beta}}}$ …… (1)

(a)

Consider the Equation (1),

$\lambda ={\lambda}_{0}\sqrt{\frac{1+\beta}{1-\beta}}$ .

From, $f=\frac{c}{\lambda}$, The Equation (1) can be written as,

${\lambda}_{0}=({\lambda}_{0}+\Delta \lambda )\sqrt{\frac{1-\beta}{1+\beta}}$ ,

Divide both the sides of the equation by ${\lambda}_{0}$ ,

$1=(1+z)\sqrt{\frac{1-\beta}{1+\beta}}$ , where $z=\frac{\Delta \lambda}{{\lambda}_{0}}$

Solve the above equation for $\beta $ as follows,

$\beta =\frac{{(1+z)}^{2}-1}{{(1+z)}^{2}+1}$

$P=\frac{{z}^{2}+2z}{{z}^{2}+2+2}$…… (2)

Therefore, From the Equation (2) It has been shown that the recessional speed parameter$\beta =\frac{v}{c}$ can been expressed in terms of .data-custom-editor="chemistry" $\mathrm{z}$

b)

Consider the given value of quasar$z=4.43$ and substitute the value of $z$ in the Equation (3) from the solution of (a) as follows,

$P=\frac{{z}^{2}+2z}{{z}^{2}+2+2}$

$\begin{array}{l}\beta =\frac{{\left(4.43\right)}^{2}+2\left(4.43\right)}{{\left(4.43\right)}^{2}+2\left(4.43\right)+2}\\ \beta =0.934\end{array}$

Therefore, the speed parameter of the quasar is.$\beta =0.934$

c)

Consider the Hubble’s law,

$v=Hr$

The equation can be rewritten as,

$r=\frac{v}{H}$,

Substitute $\beta c$ to $v$ into the equation as follows,

$r=\frac{\beta c}{H}$

$r=\frac{\left(0.934\right)(3.0\times {10}^{8}\text{m}/\text{s})}{0.0218\text{m}/\text{s}\cdot \text{ly}}$

$r=1.28\times {10}^{10}\text{ly}$

Therefore, The distance to the quasaris $r=1.28\times {10}^{10}\text{ly}$$r=1.28\times {10}^{10}\text{ly}$

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