4-34 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} R&=5.00\times10^6\ut{m}\\ T&=22.5\ut{h}=81000\ut{s} \end{cases} $$ $$\ab{a}$$ $$v=\frac{2\pi R}{T},$$ $$ \begin{aligned} a_R&=\frac{v^2}{R}\\ &=\frac{1}{R}\(\frac{2\pi R}{T}\)^2\\ &=\frac{4\pi^2 R}{T^2}\\ &=\frac{4\pi^2 (5.00\times10^6\ut{m})}{(81000\ut{s})^2}\\ &=\frac{20 \pi ^2}{6561}\ut{m/s^2}\\ &\approx 0.03008567109004529\ut{m/s^2}\\ &\approx 0.0301\ut{m/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$a_R=\frac{4\pi^2 R}{T^2},$$ $$ \begin{aligned} T&=\sqrt{\frac{4\pi^2 R}{a_R}}\\ &=\sqrt{\frac{4\pi^2 (5.00\times10^6)}{9.8}}\\ &=\frac{10000 \pi }{7}\ut{s}\\ &=\frac{500 \pi }{21}\ut{min}\\ &\approx 74.79982508547127\ut{min}\\ &\approx 74.8\ut{min}\\ \end{aligned} $$