Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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 $z = \frac{Δ λ}{λ_{0}}$
(a)Show that, in terms of, $z$ the recessional speed parameter $β = \frac{v}{c}$ is given by
$β = \frac{z^{2} + 2 z}{z^{2} + 2 z + 2}$ .
(b) A quasar in 1987 has $z = 4 .43$ . 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 $β = \frac{v}{c}$ can been expressed in terms of $z$ is given by . $β = \frac{z^{2} + 2 z}{z^{2} + 2 z + 2}$

(b) Speed parameter, $β = 0.934$

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:Explain given information

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

$λ = λ_{0} \sqrt{\frac{1 + β}{1 - β}}$ …… (1)

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

(a)

Consider the Equation (1),

$λ = λ_{0} \sqrt{\frac{1 + β}{1 - β}}$ .

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

$λ_{0} = (λ_{0} + Δ λ) \sqrt{\frac{1 - β}{1 + β}}$ ,

Divide both the sides of the equation by $λ_{0}$ ,

$1 = (1 + z) \sqrt{\frac{1 - β}{1 + β}}$ , where $z = \frac{Δ λ}{λ_{0}}$

Solve the above equation for $β$ as follows,

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

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

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

Step 3: Calculate the speed parameter of the quasar.

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,

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

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

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

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

Consider the Hubble’s law,

$v = H r$

The equation can be rewritten as,

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

Substitute $β c$ to $v$ into the equation as follows,

$r = \frac{β c}{H}$

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

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

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1:Explain given information

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

Step 3: Calculate the speed parameter of the quasar.

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1:Explain given information

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

Step 3: Calculate the speed parameter of the quasar.

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

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