6-65 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=1\ut{kg}\\ R&=6.40\times10^6\ut{m}\\ v&=465\ut{m/s}\\ F_0&=9.8\ut{N}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Sigma F_R &= \frac{mv^2}{R}\\ &=\frac{8649}{256000}\ut{N}\\ &=0.03378515625\ut{N}\\ &\approx 33.8\ut{mN}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Sigma F_R&= mg-F_N\\ F_N&=mg-\Sigma F_R\\ &=\frac{2500151}{256000}\ut{N}\\ &=9.76621484375\ut{N}\\ &\approx 9.77\ut{N}\\ \end{aligned} $$