4-61 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} a_{\max}&=23.0g\\ g&=9.80665\ut{m/s^2}\\ v&=0.00200c\\ c&=299792458\ut{m/s} \end{cases} $$ $$\ab{a}$$ $$a=\frac{v^2}{R},$$ $$ \begin{aligned} R&=\frac{v^2}{a}\\ &=\frac{(0.00200c)^2}{23g}\\ &=\frac{c^2}{5750000g}\\ &\approx 1.5938699604448848\times 10^9\ut{m}\\ &\approx 1.59\times 10^9\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$v=\frac{2\pi R}{T},$$ $$ \begin{aligned} \Ans&=\frac{T}{4}\\ &=\frac{2\pi R}{4v}\\ &=\frac{\pi}{2}\cdot\frac{c^2}{5750000g}\cdot\frac{1000}{2c}\\ &=\frac{\pi c}{23000 g}\\ &\approx 4175.630527795919\ut{s}\\ &\approx 69.59384212993199\ut{min}\\ &\approx 69.6\ut{min}\\ \end{aligned} $$