Archimedes' Principle — Question 9

Question 9

A sealed rigid balloon reaches neutral lift through air density

A sealed rigid balloon has outer volume $V=1.0\,\mathrm{m^3}$ and total mass, including gas and envelope, $M=0.80\,\mathrm{kg}$. The surrounding atmosphere has density $\rho_a(z)=1.20e^{-z/H}\,\mathrm{kg/m^3}$, with $H=8000\,\mathrm{m}$. Treat gravity as constant and neglect drag for the static force calculation.

1. Find the altitude at which buoyancy equals weight.

2. Is the vertical force restoring near that altitude for this rigid balloon? Explain why a freely expanding balloon cannot automatically use the same conclusion.

Question 9 - Solution

A sealed rigid balloon reaches neutral lift through air density

(a) Derivation and calculation

$$B(z)=\rho_a(z)gV=1.20gVe^{-z/H},\quad W=Mg.$$$$B(z_*)=W\Rightarrow e^{-z_*/H}=\frac{M}{1.20V}.$$$$\boxed{z_*=H\ln\frac{1.20V}{M}=8000\ln1.50=3243.72\,\mathrm m}.$$

(b) Result and interpretation

For fixed $V$ and $M$,$$F_z=g[V\rho_a(z)-M],\quad\left.\frac{dF_z}{dz}\right|_*=-\frac{gV\rho_a(z_*)}{H}=\boxed{-Mg/H<0}.$$$$F_z(z_*+\delta z)\simeq-\frac{Mg}{H}\delta z.$$For an expanding balloon,$$\frac{dF_z}{dz}=g\left(V\frac{d\rho_a}{dz}+\rho_a\frac{dV}{dz}\right),$$$$\text{so the additional volume term can change stability}.$$

Physical check

Total mass includes both gas and envelope; excluding the gas would overestimate lift.