# A balloon falling from space?

How long would it take for a store bought rubber balloon filled in space with air to fall from space just outside the atmosphere and touch ground? Would it Pop?

Added after comments: Ideally you would like to have a balloon filled with 14 PSI before it was dropped so that way it maintains its shape all the way down to the ground. But being a normal party balloon has its problems. It would only take one PSI of ambient air to fill a party balloon in space and any more than that it would pop. I didn't take in consideration on how temperature would affect the rubber. On the sunny side it would be well above the melting point of the rubber. On the freezing shadow side of the Earth side the rubber would freeze becoming brittle and break.

Temperature outside would the pressure of falling through the atmosphere pop the balloon? • Your question can't be answered unless you specify the elevation at which the balloon is released and other parameters of the balloon. Actually if it is a birthday-type balloon, I guess it won't hold much air on space (before exploding), Therefore, by the time friction becomes important, the pressure would have compressed the air inside the balloon so much that the result would probably be the same for an empty balloon. In any case, to get a figure for the fall time you will need to assume that the balloon is spherical or something like that, with a given radius and mass. Feb 6 '18 at 19:09
• @CamiloRada it would be near empty at ground level. It would depend on how much pressure it can hold in space. The quality and thickness of rubber or material would be a factor.
– Muze
Feb 6 '18 at 19:19
• It's better but still not enough to make a calculation. What I can say is that it won't pop, actually the other way around. It will shrink and on the ground it will look like an absolutely empty balloon. Because the pressure on the surface is huge. The same process would happen to a balloon taken deep underwater (so the pressure increase). Here are some experiments youtube.com/watch?v=RGTMIcAh4KM Feb 6 '18 at 19:19
– Muze
Feb 6 '18 at 19:21
• I'm not sure if I understand what you mean. At ground level, the atmospheric pressure is 14.6 psi. If you measure the pressure of a balloon or a tire, what you measure is relative pressure. That's why an empty tire shows 0 psi not 14.6 psi that is the absolute atmospheric pressure. If a balloon pops at a relative pressure of 33 psi at ground level, it will hold 33 psi also in space. However, in space it will hold less air than at ground level, because there is no external pressure, so the absolute pressure in the balloon is just 33 psi not 47.6 psi like on the surface (33+14.6=47.6). Feb 6 '18 at 20:09

You cannot leave "temperature outside", as temperature is the key factor to know if the balloon would pop or not.

Let's set up some assumptions about the problem so we can calculate something:

1. The gas inside the balloon behave according the ideal gas law: $$PV=nRT$$

2. That the balloon can hold a differential pressure of 30 mmHg (=4000 Pa) as found in this experiment.

3. That the maximum volume of a balloon is 0.015 m$$^3$$ as suggested here.
4. The balloon starts filled to its maximum size.

With this we can calculate how many moles of gas the balloon can hold when inflated in space at a given temperature.

For the temperature I'll assume it will be something between the temperature of the cosmic microwave background radiation (if it is on the shade) of ~2.7 K (-270 °C), and the temperature that a black body that is illuminated by the Sun near Earth's orbit, which would be about 278 K (about 5°C). This assumes that the balloon is small enough so that heat can be transferred efficiently by diffusion from one side to the other (the calculation of this value can be seen in the code at the end and is based on the equation here).

With this range of temperatures we can find the volume of the balloon on the surface assuming a final pressure of 1 atm (101325 Pa) and a temperature of 15 °C.

We get the following relationship: In red are the final volumes for which the balloon would explode and in blue when it wouldn't. The critical temperature in between is -262 °C (~11 K).

Therefore, the balloon would not pop as long as its initial temperature is higher than -262 °C.

This also assume that the hottest point in the trajectory is the surface. This means that we are assuming that the balloon will pass through the thermosphere fast enough to do not reach thermal equilibrium with it (it can be extremely hot). This could be a reasonable assumption given the very low density of that layer.

Of course this also assume the rubber of the balloon do not change its properties with temperature.

So, only once you specify the height at which you release the balloon and its initial temperature, then you can start calculating fall time. And that will be more difficult, because even if you assume the balloon is at free fall terminal velocity the whole time, this velocity will change along the fall line due to changes air density and balloon size. Air density and balloon size will affect air resistance and, therefore, terminal free fall speed. You will need to do an analytical or numerical integration of the fall to get the total time.

Matlab code used to generate the above figure

R=8.3145;%Ideal gas constant [J/mol K]

Lsol=3.828e26;%Solar luminosity [W]
Dsol=149597870700;%Distance earth-sun [m]
sigma=5.670373e-8;%Stefan–Boltzmann constant [W m^-2 K^-4]
alpha=0;%albedo
epsilon=1;%emissivity
% Temperature of a black body (rotating or small enough to transfer heat efficiently from the illuminated side to the dark side)
Tbb=((1/4)*((Lsol*(1-alpha))/(4*pi*epsilon*sigma*(Dsol^2))))^(1/4);

% Temperature range
T=linspace(2.7,Tbb,1000);%[K]
Vmax=0.015;% Exploding volume [m³]
Pspace=4000;%Maximum pressure hold by the balloon [Pa]
n=Pspace*Vmax./(R*T);% number of moles that the balloon would hold
Patm=101325;% Atmospheric pressure [Pa]
Tsurface=15+273.15; % temperature at the surface [K]

% Volume once back on earth surface
Vfin=n*R*Tsurface./Patm;

explodeIdx=Vfin>Vmax;
Tlim=find(explodeIdx,1,'last');

figure('Color','w');
hold on
box on
plot(Vfin(~explodeIdx)*1000,T(~explodeIdx)-273.15,'b','LineWidth',2);
plot(Vfin(explodeIdx)*1000,T(explodeIdx)-273.15,'r','LineWidth',2);
plot([0 max(Vfin)]*1000,[1 1]*T(Tlim)-273.15,':r','LineWidth',2)
ylim([0 max(T)]-273.15);

xlabel('Final volume [liters]')
ylabel('Initial temperature [°C]')

• By the way, if you have a balloon strong enough to hold 1 atm (14 psi) (something like a tire), it would be the same thing, if the air is originally very cold, you can get a lot of air inside, then when it warms up, it would explode anyway, even if started at 1 atm. The problem can be simplified if the air inside the balloon is always at the same temperature. Feb 17 '19 at 2:50