(a) Pressure decreases as one ascends the atmosphere. If the density of air is `rho,` What is the change in pressure dp over a differential height dh ? (b) Considering the pressure p to be proportional to the density, find the pressure p at a height h if the pressure on the sureface of the earth is `p_0.` (c ) If `p_0 = 1.03 xx 10^5 Nm^(-2), rho_0 = 1.29kg m^(-3) and g = 9.8 ms^(-2), ` at what height will the pressure drop to `(1//10)` the value at the surface of the earth ? (d) This model of the atmosphere works for relatively small distance. Identify the underlying assumption that limits the model.

Explanation

(a) Pressure of air due to height h of air is `p = h rho g` Differenting it w.r.t.h, we have `(dp)/(dh) = rho g or dp = rho g dh` Since pressure of air is decreasing upwards, so `dp = - rho g dh` (b) Let `rho_0` be the density of air on the surface of earth. As per question pressure `prop` density so `(p)/(p_0) = (rho)/(rho_0) or rho = (rho_0 p)/(rho_0)`
Pressure due to height h of air, ` p = h rho g`
`dp = -rho g dh = -(rho_0 rho)/(p_0)xx g dh`
Here -ve sign shown that pressure decreases with height `:'(dp)/(p) = -(rho_0 g)/(rho_0g)/(p_0)dh` intergrating it, within the limits as p changes from `p_0` to `p, h` changes from 0 to h. Hence
`int_(p_0)^p (dp)/(p) = -(rho_0g)/(p_0)int_0^h dh`
`log_e p- log_e p_0 = -(rho_0g)/(p_0)h or log_e (p)/(p_0) = -(rho_0g)/(p_0)h .....(i)`
`or p = p_(0)e^(-rho_0 gh//p_0)`
(c ) Here, `(p)/(p_0) = (1)/(10), h = h'`
From (i), `log_e ((1)/(10))= -(rho_0g)/(p_0)h'`
`h' = -(p_0)/(rho_0 g) log_e((1)/(10)) = -(1.03 xx 10^5)/(1.29 xx 9.8) xx(-2.303) = (1.03 xx 10^5)/(1.29 xx 9.8) xx 2.303 = 16 xx 10^3 m`
(d) The assumption that `P prop rho` is applicable only when temperature remains constant, which is true for small distance in air.