Chapter 1, Problem 1.71QE

(a)

Interpretation Introduction

Interpretation:

A light-year is the distance that is travelled by light in one year.  The distance has to be calculated in miles represented by one light year, where the length of a year is 365.25 days and the light travels at a rate of $\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s}$.

Explanation of Solution

Given that light travels at a rate of $\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s}$.

The conversion factor is used to convert miters into miles.

  $\begin{array}{l}\text{1}\text{km}\text{=}{\text{10}}^{\text{3}}\text{m}\\ \text{1}\text{mi}\text{=}\text{1}\text{.609}\text{km}\\ \\ \text{Distance}\text{=}\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\frac{\text{m}}{\text{s}}\text{×}\left(\frac{\text{1}\text{km}}{{\text{10}}^{\text{3}}\text{m}}\right)\text{×}\left(\frac{\text{1}\text{mi}}{\text{1}\text{.609}\text{km}}\right)\\ \text{=}\text{1}\text{.86}\text{×}{\text{10}}^{\text{5}}\frac{\text{mi}}{\text{s}}\end{array}$

The conversion factor is used to convert seconds into light year.

  $\begin{array}{l}\text{1}\text{hr}\text{=}\text{3600}\text{s}\\ \text{24}\text{hr}\text{=}\text{1}\text{day}\\ \text{1}\text{year}\text{(light}\text{year)}\text{=}\text{365}\text{.25}\text{days}\\ \\ \text{Distance}\text{=}\text{1}\text{.86}\text{×}{\text{10}}^{\text{5}}\frac{\text{mi}}{\text{s}}\text{×}\left(\frac{\text{3600}\text{s}}{\text{1}\text{hr}}\right)\text{×}\left(\frac{\text{24}\text{hr}}{\text{1}\text{day}}\right)\text{×}\left(\frac{\text{365}\text{.25}\text{days}}{\text{1}\text{year}}\right)\\ \text{=}\text{5}\text{.87}\text{×}{\text{10}}^{\text{12}}\frac{\text{mi}}{\text{light}\text{year}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The length of a year is 365.25 days and the light travels at a rate of $\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s}$.The distance to the nearest star is 4.36 light years.  The distance has to be calculated in meters.  The result has to be expressed in scientific notation and with all the zeros.

Explanation of Solution

Given that light travels at a rate of $\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s}$.

The given distance is converted from seconds to one light year.

  $\begin{array}{l}\text{1}\text{hr}\text{=}\text{3600}\text{s}\\ \text{24}\text{hr}\text{=}\text{1}\text{day}\\ \text{1}\text{year}\text{(light}\text{year)}\text{=}\text{365}\text{.25}\text{days}\\ \\ \text{Distance}\text{=}\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\frac{\text{m}}{\text{s}}\text{×}\left(\frac{\text{3600}\text{s}}{\text{1}\text{hr}}\right)\text{×}\left(\frac{\text{24}\text{hr}}{\text{1}\text{day}}\right)\text{×}\left(\frac{\text{365}\text{.25}\text{days}}{\text{1}\text{year}}\right)\\ \text{=}\text{9}\text{.47}\text{×}{\text{10}}^{\text{15}}\frac{\text{m}}{\text{light}\text{year}}\\ \\ \text{light}\text{travels}\text{9}\text{.47}\text{×}{\text{10}}^{\text{15}}\text{m}\text{per}\text{one}\text{light}\text{year}\end{array}$

The distance travelled by light in 4.36 light years is

  $\begin{array}{l}\text{Distance}\text{=}\frac{\text{9}\text{.47}\text{×}{\text{10}}^{\text{15}}\text{m}}{\text{1}\text{light}\text{year}}\text{×}\text{4}\text{.36}\text{light}\text{years}\\ \text{=}\text{4}\text{.13}\text{×}{\text{10}}^{\text{16}}\text{m}\\ \text{=}\text{41,300,000,000,000,000}\text{m}\end{array}$