Chapter 1, Problem 69P

(a)

To determine

To write: The number of parsec in $1\text{AU}$ .

Answer to Problem 69P

  $1\text{AU}$ is equal to $4.848\times {10}^{-6}\text{pc}$ .

Explanation of Solution

Given:

The arc length is equal to $1\text{AU}$ .

The angle of arc covered is $1\text{s}$ .

Formula used:

Write the expression for arc.

  $a=r\times \theta$

Rearrange the above expression in terms of $r$ .

  $r=\frac{a}{\theta }$   .............. (1)

Here, r is the radius, $a$ is the arc and $\theta$ is the angle covered.

Calculation:

Substitute $1\text{AU}$ for $a$ and $1\text{s}$ for $\theta$ in equation (1).

  $\begin{array}{c}r=\frac{1\text{AU}}{1\text{s}\left(\frac{1°}{3600\text{s}}\right)\left(\frac{\left(\frac{\pi }{180}\right)}{1°}\right)}\\ =206264.81\text{AU}\end{array}$

The radius of circle is $206264.81\text{AU}$ that is equal to 1pc.

Calculate the number of parsec in $1\text{AU}$ .

  $\begin{array}{c}d=1\text{AU}\left(\frac{1\text{pc}}{206264.81\text{AU}}\right)\\ =4.8481\times {10}^{-6}\text{pc}\end{array}$

Conclusion:

Thus, $1\text{AU}$ is equal to $4.8481\times {10}^{-6}\text{pc}$ .

(b)

To determine

To write: The number of meter in $1\text{pc}$ .

Answer to Problem 69P

  $1\text{pc}$ is equal to $3.086\times {10}^{16}\text{m}$ .

Explanation of Solution

Given:

  $1\text{AU}$ is equal to $1.496\times {10}^{11}\text{m}$ .

Formula used:

Write the conversion for parsec into astronomical units as follows:

  $1\text{ Pa}=206264.81\text{AU}$

Calculation:

Calculate the number of meters in $1\text{pc}$ .

  $\begin{array}{c}d=1\text{pc}\left(\frac{206264.81\text{AU}}{1\text{pc}}\right)\left(\frac{1.496\times {10}^{11}\text{m}}{1\text{AU}}\right)\\ =3.086\times {10}^{16}\text{m}\end{array}$

Conclusion:

Thus, $1\text{pc}$ is equal to $3.086\times {10}^{16}\text{m}$ .

(c)

To determine

To write: The number of meters in a light-year.

Answer to Problem 69P

A light-year is equal to $9.46\times {10}^{15}\text{m}$ .

Explanation of Solution

Given:

  $1\text{AU}$ is equal to $1.496\times {10}^{11}\text{m}$ .

Formula used:

Write the expression for distance covered.

  $d=v\times t$   .............. (2)

Here, d is the distance covered, v is the speed of object and t is the time taken.

Calculation:

Substitute $3\times {10}^8\text{m}/\text{s}$ for v and $1\text{y}$ for t in equation (2).

  $\begin{array}{c}d=\left(3\times {10}^8\text{m}/\text{s}\right)\left(1\text{y}\right)\left(\frac{365\text{days}}{1\text{y}}\right)\left(\frac{86400\text{s}}{1\text{day}}\right)\\ =9.46\times {10}^{15}\text{m}\end{array}$

Conclusion:

Thus, a light-year is equal to $9.46\times {10}^{15}\text{m}$ .

(d)

To determine

To write: The number of AU in a light-year.

Answer to Problem 69P

A light-year is equal to $6.324\times {10}^4\text{AU}$ .

Explanation of Solution

Given:

  $1\text{AU}$ is equal to $1.496\times {10}^{11}\text{m}$ .

Formula used:

Write the conversion for light years into meter as follows:

  $1\text{ ly}=9.46\times {10}^{15}\text{m}$

Calculation:

Calculate the number of AU in a light-year.

  $\begin{array}{c}d=1\text{ly}\left(\frac{9.46\times {10}^{15}\text{m}}{1\text{ly}}\right)\left(\frac{1\text{AU}}{1.496\times {10}^{11}\text{m}}\right)\\ =6.324\times {10}^4\text{AU}\end{array}$

Conclusion:

Thus, a light-year is equal to $6.324\times {10}^4\text{AU}$ .

(e)

To determine

To write: The number of light-year in a parsec.

Answer to Problem 69P

  $1\text{pc}$ is equal to $3.262\text{ ly}$ .

Explanation of Solution

Given:

  $1\text{AU}$ is equal to $1.496\times {10}^{11}\text{m}$ .

Formula used:

Write the conversion for parsec into meter as follows:

  $1\text{ pc}=3.086\times {10}^{16}\text{m}$

Calculation:

Calculate the number of light-year in a parsec.

  $\begin{array}{c}d=1\text{pc}\left(\frac{3.086\times {10}^{16}\text{m}}{1\text{pc}}\right)\left(\frac{1\text{ly}}{9.46\times {10}^{15}\text{m}}\right)\\ =3.262\text{ ly}\end{array}$

Conclusion:

Thus, $1\text{pc}$ is equal to $3.262\text{ ly}$ .