5-23 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=3.50\times10^6\ut{kg}\\ t_1&=34.0\ut{day}\\ v&=0.100c\\ c&=3.0\times10^8\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$v=v_0+at,$$ $$ \begin{aligned} a&=\frac{v-v_0}{t}\\ &=\frac{0.1c-0}{34\ut{day}}\\ &=\frac{3\times10^7\ut{m/s}}{34\ut{day}}\cdot\frac{1\ut{day}}{24\ut{h}}\cdot\frac{1\ut{h}}{3600\ut{s}}\\ &=\frac{3125}{306}\ut{m/s^2}\\ &\approx 10.21241830065359\ut{m/s^2}\\ &\approx 10.2\ut{m/s^2}\\ &\approx 1.0\times10\ut{m/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} a&=\frac{3125}{306}\ut{m/s^2}\\ \frac{a}{g}&\approx 1.041376851488897g\\ &\approx 1.04g\\ &\approx 1.0g\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \Sigma \vec F &= m\vec a,\\ &=\(3.50\times10^6\)\(\frac{3125}{306}\)\\ &=\frac{5468750000}{153}\ut{N}\\ &\approx 3.574346405228758\times10^7\ut{N}\\ &\approx 3.57\times10^7\ut{N}=35.7\ut{MN}\\ &\approx 3.6\times10^7\ut{N}=36\ut{MN}\\ \end{aligned} $$ $$\ab{d}$$ $$S=\frac{1}{2}\(v+v_0\)t,$$ $$ \begin{aligned} S_1&=\frac{1}{2}\(0.1c+0\)(34.0\ut{day})\\ &=1.7\ut{day}\cdot c\\ \end{aligned} $$ $$S = vt,$$ $$ \begin{aligned} t_2&=\frac{S_2}{v}\\ &=\frac{7.50\ut{month}\cdot c-1.7\ut{day}\cdot c}{0.1c}\\ &=\frac{223.3\ut{day}\cdot c}{0.1c}\\ &=2233\ut{day}\\ \end{aligned} $$ $$ \begin{aligned} \Ans &= t_1+t_2\\ &=(34.0+2233)\ut{day}\\ &=2267\ut{day}\\ &\approx 6.210958904109589\ut{year}\\ &\approx 6.21\ut{year}\\ \end{aligned} $$