Chapter 16, Problem 3P

(a)

To determine

The value of the distance.

Answer to Problem 3P

The value of the distance is $\underset{\_}{7.2\times {10}^9\text{ light years}}$.

Explanation of Solution

Write the expression for the distance.

$\begin{array}{c}R=\frac{v}{H}\\ =\frac{0.55c}{H}\end{array}$        (I)

Here, $R$ is the distance, $H$ is the Hubble’s constant, $c$ is the speed of light.

Conclusion:

Substitute, $\left(3.00\times {10}^8\text{m/s}\right)$ for $c$, $23\times {10}^{-3}\text{m/s-light year}$ for $H$ in equation (I) to find $R$.

$\begin{array}{c}R=\frac{0.55\left(3.00\times {10}^8\text{m/s}\right)}{\left(23\times {10}^{-3}\text{m/s-light year}\right)}\\ =7.2\times {10}^9\text{light year}\end{array}$

Thus, The value of the distance is $\underset{\_}{7.2\times {10}^9\text{ light years}}$.

(b)

To determine

The age of universe.

Answer to Problem 3P

The age of universe is $\underset{\_}{13\text{billion years}}$.

Explanation of Solution

Write the expression for the age of universe.

$\begin{array}{c}t=\frac{R}{v}\\ =\frac{R}{0.55c}\end{array}$        (II)

Here, $t$ is the age of universe.

Conclusion:

Substitute, $7.2\times {10}^9\text{ light years}$ for $R$ in equation (II) to find $t$.

$\begin{array}{c}t=\frac{\left(7.2\times {10}^9\text{ light years}\right)}{0.55c}\\ =\frac{\left(7.2\times {10}^{9}c\left(\text{1 year}\right)\right)}{0.55c}\\ =13\text{ billion years}\end{array}$

Thus, the age of universe is $\underset{\_}{13\text{billion years}}$.