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Expert-verified Found in: Page 1364 ### Fundamentals Of Physics

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 ${\mathbit{z}}{\mathbf{=}}\frac{\mathbf{\Delta }\mathbf{\lambda }}{{\mathbf{\lambda }}_{\mathbf{0}}}$(a)Show that, in terms of,${\mathbit{z}}$ the recessional speed parameter${\mathbit{\beta }}{\mathbf{=}}\frac{\mathbf{v}}{\mathbf{c}}$ is given by ${\mathbit{\beta }}{\mathbf{=}}\frac{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{z}}{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{+}\mathbf{2}}$.(b) A quasar in 1987 has ${\mathbit{z}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.43}}$ . 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, $r=1.28×{10}^{10}\text{ly}$width="113" style="max-width: none; vertical-align: -4px;" $r=1.28×{10}^{10}\text{ly}$

See the step by step solution

## Step 1:Explain given information

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

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

## Step 2: Show the recessional speed parameter in terms of .z

(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,

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

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

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

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

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

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

## Step 3: Calculate the speed parameter of the quasar.

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.$\beta =0.934$

## Step 4:Find the distance to the quasar, assuming that Hubble’s law is valid to these distances.

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,

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

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

$r=1.28×{10}^{10}\text{ly}$

Therefore, The distance to the quasaris $r=1.28×{10}^{10}\text{ly}$$r=1.28×{10}^{10}\text{ly}$ ### Want to see more solutions like these? 