Question

95 [SSM As mentioned in the discussion of the law of atmospheres, the fractional decrease in atmospheric pressure is proportional to the change in altitude. Expressed as a differ- ential equation we have dP/P -C/dh, where C is a con- stant. (a) Show that P(h) -Poe-Ch is a solution of the differential equation. (b) Show that if Ah << ho where ho - 1 /C, then Ph + ΔΑ) ~P(h)(1-ΔΙ/ho). (c) Given that the pres- sure at h-5.5 km is half that at sea level, find the constant C

Answer 1

A small typo was in the question which has been corrected and solved.

All the final answers have been boxed.

All the used formulas are mentioned.
A small typo was in the question which has been corrected and solved. All the final answers have been boxed. All the used formulas are mentioned. dP/P = -c/dh dh (Given) P(h) = P_o e^-ch (Given) differentiate w.r.t. p h dp/dh = P_o e^-ch times (- c) dp/p = p_o e^-ch (- c). dh dp/p = -c dh. So p(h) = p_o e^-ch p(h + delta h) = p_o e^-c(h + delta h) = e^-ch times e^-c delta h p_o = p_o e^-ch times e^-c delta h = p(h) e^-c delta h since delta h c < 1, we can expand e^-c delta h By Tayler expension. P(h + delta h) asymptoticallyequalto P(h) [1 + (- c delta h)] = c = 1/h_o P(h + delta h) = P (h) [1 - delta h/h_o] \r\n h = 5.5 km. P(h) = 1/2 P(0) h = 0 for sec level. C = ? P_o e^-c times 5.5 times 10^3 = (P_o times 1/2)