Chapter 23, Problem 30Q

(a)

To determine

The recessional velocity, if it is given that the galaxy RD1 has a redshift value of $z=5.34$, in $\text{km/s}$. Also, determine it as a fraction of the speed of light.

Answer to Problem 30Q

Solution:

$v=2.85\times {10}^5\text{km/s}$ or $v=0.951c$.

Explanation of Solution

Given data:

The value of the redshift (z) is $5.34$.

Formula used:

The expression for the the recessional velocity is written as,

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

Here, $v$ is the recessional velocity and $z$ is the redshift value.

Explanation:

Recall the expression for the recessional velocity.

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

Substitute $5.34$ for $z$.

$\begin{array}{c}\frac{v}{c}=\frac{{\left(5.34+1\right)}^2-1}{{\left(5.34+1\right)}^2+1}\\ =\frac{{\left(6.34\right)}^2-1}{{\left(6.34\right)}^2+1}\\ =0.951\end{array}$

Further, solve the above expression for $v$.

$v=0.951c$

Substitute $3\times {10}^5\text{km/s}$ for $c$.

$\begin{array}{c}v=0.951\left(3\times {10}^5\text{km/s}\right)\\ =2.85\times {10}^5\text{km/s}\end{array}$

Conclusion:

Thus, the recessional velocity of the galaxy is $2.85\times {10}^5\text{km/s}$ or $0.951c$.

(b)

To determine

The recessional velocity, if it is given that galaxy RD1 has a redshift value of $z=5.34$, by using the low-speed formula. Also, check whether the result obtained by this formula gives a small or large error.

Answer to Problem 30Q

Solution:

$1.6\times {10}^6\text{km/s}$, it is a large value, which is more than the speed of light. So, this is not a possible value.

Explanation of Solution

Given data:

The value of redshift (z) is $5.34$.

Formula used:

The expression for low speed formula is written as,

$\frac{v}{c}=z$

Here, $z$ is the redshift value, $v$ is the recessional velocity, and $c$ is the speed of light.

Explanation:

Recall the expression for low speed formula.

$\frac{v}{c}=z$

Substitute $5.34$ for $z$ and $3\times {10}^5\text{km/s}$ for $c$.

$\frac{v}{3\times {10}^5}=5.34$

Solve the above expression for $v$.

$\begin{array}{c}v=5.34\left(3\times {10}^5\text{km/s}\right)\\ =1.6\times {10}^6\text{km/s}\end{array}$

The recessional velocity from low-speed formula comes out to be $1.6\times {10}^6\text{km/s}$. It is impossible to have such a high speed.

Conclusion:

Thus, the recessional velocity as calculated from a low-speed formula comes out to be very large, which is not possible.

(c)

To determine

The distance to galaxy RD1 from Earth in MPc, as well as in light-years, with the help of Hubble’s law, if the Hubble constant is $73\text{km/s/MPc}$.

Answer to Problem 30Q

Solution:

$3904\text{MPc}$ or $12730\text{Mly}$

Explanation of Solution

Given data:

The value of the Hubble’s constant is $73\text{km/s/MPc}$.

Formula used:

According to Hubble's law,

$v=H_{o}d$

Here, $v$ is the recessional velocity, $H_o$ is the Hubble’s constant and $d$ is the distance.

Explanation:

Recall the expression for Hubble’s law and rearrange it for $d$.

$\begin{array}{l}v=H_{o}d\\ d=\frac{v}{H_o}\end{array}$

Substitute $2.85\times {10}^5\text{km/s}$ for $v$ and $73\text{km/s/MPc}$ for $H_o$.

$\begin{array}{c}d=\frac{2.85\times {10}^5\text{km/s}}{73\text{km/s/MPc}}\\ =3904\text{MPc}\end{array}$

The conversion factor of megaparsecs into light year is $1\text{MPc}=3.26\times {10}^6\text{light-years}$.

Thus,

$\begin{array}{c}3904\text{MPc}=3904\text{MPc}\left(\frac{\text{3}\text{.26}\times {\text{10}}^6\text{ light-years}}{1\text{ MPc}}\right)\\ =\text{12730}\times {10}^6\text{light-years}\end{array}$

Conclusion:

The distance to galaxy RD1 is $3904\text{MPc}$ or $\text{12730}\times {10}^6\text{light-years}$.