1-22 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 1\ut{AU}=&92.9\times10^6\ut{mi}\\ 1\ut{ly}=&\cfrac{186000\ut{mi}}{1\ut{sec}}\cdot1\ut{year} \end{cases} $$ $$\ab{a}$$ $$L=r\theta$$ $$ \begin{aligned} 1\ut{AU}=&1\ut{pc}\cdot1\ut{''}\cdot\frac{1\ut{\degree}}{3600\ut{''}}\cdot\frac{\pi\ut{rad}}{180\ut{\degree}}\\ =&\frac{\pi}{648000}\ut{pc}\\ \approx&4.84813681109536\times10^{-6}\ut{pc}\\ \approx&4.85\times10^{-6}\ut{pc}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans=&1\ut{AU}\\ =&1\ut{AU}\times\frac{92.9\times10^6\ut{mi}}{1\ut{AU}}\times\frac{1\ut{ly}\cdot1\ut{sec}}{186000\ut{mi}\cdot1\ut{year}}\\ &\times\frac{1\ut{h}}{3600\ut{sec}}\times\frac{1\ut{day}}{24\ut{h}}\times\frac{1\ut{year}}{365.25\ut{day}}\\ =&\frac{92.9\cdot10^6}{186000\cdot3600\cdot24\cdot365.25}\ut{ly}\\ =&\frac{929}{58697136}\ut{ly}\\ \approx&0.000015827007300662847\ut{ly}\\ \approx&1.58\times10^{-5}\ut{ly} \end{aligned} $$