Chapter 24, Problem 26Q

To determine

(a)

The speed at which the blazar is moving away from the Earth.

Answer to Problem 26Q

The speed at which the blazar is moving away from the Earth is $1.80\times {10}^5\text{km}/\text{s}$.

Explanation of Solution

Given:

The red shift of the given blazar is, $z=1.00$.

Formula used:

The expression for the ratio of the speed of the object moving away from the earth and speed of the light is given by,

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

Here, $v$ is velocity of the object moving away, $c$ is the speed of the light and $z$ is the amount of the red shift.

Calculation:

The speed of the blazar in moving away from the Earth is calculated as,

$\begin{array}{c}\frac{v}{c}=\frac{{\left(z+1\right)}^2-1}{{\left(z+1\right)}^2+1}\\ \frac{v}{c}=\frac{{\left(1+1\right)}^2-1}{{\left(1+1\right)}^2+1}\\ \frac{v}{c}=0.6\\ \frac{v}{3\times {10}^5\text{}\text{km}/\text{s}}=0.6\end{array}$

Solve further,

$\begin{array}{c}\frac{v}{3\times {10}^5\text{km}/\text{s}}=0.6\\ v=1.8\times {10}^5\text{km}/\text{s}\end{array}$

Conclusion:

The speed at which the blazar is moving away from the earth is $1.80\times {10}^5\text{km}/\text{s}$.

To determine

(b)

The time for which the fluctuation lasted for an astronomers within the blazar’s host galaxy.

Answer to Problem 26Q

The time for which the fluctuation lasted for an astronomer within the blazar’s host galaxy is $134\text{h}$.

Explanation of Solution

Given:

The time span for which the fluctuation was seen on the earth is $\Delta t=168\text{}\text{h}$.

Formula used:

The expression for the time dilation as per theory of relativity is given by,

$\Delta t=\frac{\Delta t_0}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Here, $\Delta t$ is the time of the observer and $\Delta t_0$ is the estimated time of the receiver.

Calculation:

The time for which the fluctuation lasted for an astronomer within the blazar’s host galaxy is calculated as,

$\begin{array}{c}\Delta t=\frac{\Delta t_0}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ 168\text{h}=\frac{\Delta t_0}{\sqrt{1-{\left(\frac{1.8\times {10}^5\text{km}/\text{s}}{3\times {10}^5\text{km}/\text{s}}\right)}^2}}\\ \Delta t_0=168\text{h}\times \sqrt{1-{\left(\frac{1.8\times {10}^5\text{km}/\text{s}}{3\times {10}^5\text{km}/\text{s}}\right)}^2}\\ \Delta t_0=134\text{h}\end{array}$

Conclusion:

Therefore, the time for which the fluctuation lasted for an astronomer within the blazar’s host galaxy is $134\text{h}$.

To determine

(c)

The maximum size of the region for emitting energy by the blazar.

Answer to Problem 26Q

The maximum size of the region for emitting energy by the blazar is $995.3\text{au}$.

Explanation of Solution

Given:

The time for which the fluctuation lasted for an astronomer within the blazar’s host galaxy is $134\text{h}$.

Formula used:

The size of the blazar is given by,

$\text{size of the blazar}=\left(t\right)\left(c\right)$

Here, $t$ is the time for which the fluctuation lasted.

Calculation:

The maximum size of energy emitting region of the blazar is calculated as,

$\begin{array}{c}\text{size of the blazar}=\left(t\right)\left(c\right)\\ =\left(138\text{h}\right)\left(3\times {10}^8\text{m}/\text{s}\right)\\ =\left(138\text{h}\right)\left(3\times {10}^8\text{m}/\text{s}\times \frac{{10}^{-3}\text{km}}{1\text{m}}\times \frac{3600\text{s}}{1\text{h}}\right)\end{array}$

Solve further,

$\begin{array}{c}\text{size of the blazar}=\left(1.49\times {10}^{11}\text{km}\times \frac{6.68\times {10}^{-9}\text{au}}{1\text{km}}\right)\\ =995.32\text{au}\end{array}$

Conclusion:

The maximum size of the region for emitting energy by the blazar is $995.3\text{au}$.