자연계열 1번

(1) 변화비율에 대한 관계식 $\displaystyle \frac1{P(h)}\frac{dP}{dh}=f(h)$의 양변을 $\displaystyle 0$부터 $\displaystyle h$까지 정적분하면$\displaystyle \int_0^h\frac1{P(h)}\frac{dp}{dh}\,dh=\ln P(h)-\ln P(0)=\int_0^h f(h)\,dh$와 같은 관계식을 얻을 수 있다.$\displaystyle \ln P(h)=\ln P_0+\int_0^h f(h)\,dh$이므로, $\displaystyle P(h)=P_0e^{\int_0^h f(h)\,dh}$.

(2) $\displaystyle P(h)=P_0e^{\int_0^h k\,dh}=P_0e^{kh}$이다. $\displaystyle P(5680)=P_0e^{k\cdot5680}=\frac12P_0$이므로, $\displaystyle k=\frac1{5680}\ln\frac12=-\frac{\ln2}{5680}$이다.따라서 $\displaystyle P(h)=P_0e^{\frac h{5680}\ln\frac12}=P_0\left(\frac12\right)^{h/5680}$이고, $\displaystyle P(8520)=P_0\left(\frac12\right)^{8520/5680}=P_0\left(\frac12\right)^{3/2}=P_0\frac1{2\sqrt2}$.

(3) $\displaystyle T\left(\frac12\right)=\frac{373}{1-a\log0.5}=\frac{373}{1.05}$이므로, $\displaystyle a=\frac{0.05}{\log2}$ ($\displaystyle =\frac{0.05}{0.3}=\frac16$)이다.$\displaystyle T(P_H)=\frac{373}{1.0125}=\frac{373}{1-a\log P_H}$이므로, $\displaystyle \log P_H=-\frac{0.0125}a=-\frac{0.0125}{0.05}\log2$($\displaystyle =-\frac{0.3}4$)이고, $\displaystyle P_H=\left(\frac12\right)^{1/4}$이다.$\displaystyle P(H)=\left(\frac12\right)^{H/5680}=\left(\frac12\right)^{1/4}$이므로 $\displaystyle H=\frac{5680}4=1420\,(m)$.