Respuesta :
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
[tex]h = \dfrac{P}{\rho\ g}[/tex]
[tex]h = \dfrac{101300}{1.3\times 9.8}[/tex]
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
[tex]\rho_x = \dfrac{\rho_{sl}}{h}\times x[/tex]
now, Pressure at depth x
[tex]dP = \rho_x g dx[/tex]
[tex]dP = \dfrac{\rho_{sl}}{h}\times x g dx[/tex]
integrating both side
[tex]P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx[/tex]
[tex] P =\dfrac{\rho_{sl}\times g h}{2}[/tex]
now,
[tex]h=\dfrac{2P}{\rho_{sl}\times g}[/tex]
[tex]h=\dfrac{2\times 101300}{1.3\times 9.8}[/tex]
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
