4-63 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} a_R &= 6.00\times10^{14}\ut{m/s^2}\\ R &= 18.0\ut{cm}\\ \end{cases} $$ $$\ab{a}$$ $$a_R=\frac{v^2}{R},$$ $$ \begin{aligned} v&=\sqrt{Ra_R}\\ &=6\sqrt3\times10^6\ut{m/s}\\ &\approx 1.039230484541326\times10^7\ut{m/s}\\ &\approx 1.04\times10^7\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$v=\frac{2\pi R}{T},$$ $$ \begin{aligned} T&=\frac{2\pi R}{v}\\ &= \frac{2\pi R}{\sqrt{Ra_R}}\\ &=\frac{\sqrt3 \pi}{5}\times10^{-7}\ut{s}\\ &\approx 1.088279618540531\times10^{-7}\ut{s}\\ &\approx 1.09\times10^{-7}\ut{s}\\ \end{aligned} $$