Question
(a) It is known that density ρ of air decreases with height y as ρ0e-y/y0
Where ρ0 = 1.25 kg m-3
is the density at sea level, and y0 is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of gremains constant.
Solution
Volume of the balloon, V = 1425 m3
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s2
y0 = 8000 m
ρHe = 0.18 kg m-3
ρ0 = 1.25 kg m-3
Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as,
ρ = ρ0e-y/y0
ρ / ρ0 = e-y/y0 ...(i)
This density variation is called the law of atmospherics.
From equation (i), we can say that the rate of decrease of density with height is directly proportional to
.
That is,
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s2
y0 = 8000 m
ρHe = 0.18 kg m-3
ρ0 = 1.25 kg m-3
Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as,
ρ = ρ0e-y/y0
ρ / ρ0 = e-y/y0 ...(i)
This density variation is called the law of atmospherics.
From equation (i), we can say that the rate of decrease of density with height is directly proportional to
.That is,
