# Could you use a balloon to show the drag from the gas/atmosphere of something orbitting near the ISS?

If a balloon were floating next to the International Space Station (ISS), how big, light, and/or dense would it need to be such that the gas/atmosphere at that distance from the earth's surface would cause it to visibly accelerate relative to the ISS? I'll say provisionally that the acceleration would need to be at least $$10^{-3}\text{ m }\text{s}^{-2}$$.

I looked all over just now, trying to find out what the density of the atmosphere is at $$400\text{ km}$$ but all I could find were figures for the lower atmosphere, and for the thermosphere, which is what the ISS orbits inside, log graphs that I can't decipher and equations that I am not smart enough to be able to use.

https://space.stackexchange.com/a/38130/40252 has a graph (is it reliable?) which if accurate implies that the density at four hundred kilometers is about one trillionth of a kilogram per cubic meter. This figure is conveniently a round number, and is close to one trillionth of the density of air at sea level (Wikipedia says: At $$101.325$$ kPa (abs) and $$15$$°C, air has a density of approximately $$1.225$$ kg m$$^{-3}$$.)

• This is essentially a problem of air drag which depends on velocity, air density, size of object etc. It would be helpful , if you can put in some more research effort to include some numbers on the density of air outside the ISS Apr 19, 2021 at 5:50
• @silverrahul That's a good idea. I looked all over just now, but all I could find were figures for the lower atmosphere, and for the thermosphere, which is what the ISS orbits inside, log graphs that I couldn't decipher and equations that I am not smart enough to be able to use. Apr 19, 2021 at 7:57
• @silverrahul I posted it as a question at Space Exploration Beta. Apr 19, 2021 at 8:19

I used some approximate numbers to do an approximate calculation.

Assuming that:

speed of ISS = 8000 m s$$^{-1}$$

Density of air outside ISS = $$10^{-12}$$ kg m$$^{-3}$$

Drag coefficient = 1

The drag force on the balloon is a half * density * speed$$^2$$ * drag coefficient * cross-sectional area

So, I put in the numbers into that formula. A is in square meters and m is kilograms.

1/2 * $$10^{-12}$$ kg m$$^{-3}$$ * (8000 m s$$^{-1}$$)^2 * 1 * A = m * $$10^{-3}\text{ m}\ \text{s}^{-2}$$

According to the result, to get the required acceleration, the balloon has to have a mass/area ratio of around 0.032 kg m$$^{-2}$$. That's thirty-two grams per square meter of cross-sectional area.

So, you could use a balloon to show the drag if the balloon had a ratio of mass/cross sectional area of .032 in SI units AND provided that the balloon was built of a material strong enough. There might be an issue with feasibility as any material strong enough to survive the pressure differential may not be light enough to have this kind of mass/area.

• Is that a typo where you multiply speed by velocity squared? Apr 19, 2021 at 9:28
• yes, i corrected it . It will be density multiplied by velocity squared Apr 19, 2021 at 9:30
• Is this correct:..................................................................................................................?: The drag force on the balloon is 0.5 * density * velocity^2 * drag coefficient * area Apr 19, 2021 at 9:42
• Yes, that is correct Apr 19, 2021 at 10:04
• Do you mean cross sectional area? Apr 19, 2021 at 11:16