# Changes in density.

1. Sep 29, 2004

### matthyaouw

Hey there. I was hoping you guys could help me with this- If the density of an object of fixed mass in the earth's atmosphere were to change, would it seem to change weight due to more buoyancy in the air?
If you need me to clarify what I mean some more, please ask.

thanks,
Matt

2. Sep 29, 2004

### ZapperZ

Staff Emeritus
Well, other than some strange spacetime effects, the only way for something to have a fixed mass but a change in density would be by changing its volume. And since buoyancy is a function of the displaced volume of air, water, etc. that the object is immersed in, then yes, you would change the buoyancy of the object.

Zz.

3. Sep 29, 2004

### LURCH

Probably the best example would be the air inside a hot-air baloon. The air inside the balloon is taken form outside, and heated up. It's still basically the same air (plus a little CO2 form the fuel), just hotter. The heat is added to make the air expand (changes its density). The air inside is almost identical to the air outside, just less dense. This decreased density is what makes it lighter, and lifts the balloon.

4. Sep 29, 2004

### matthyaouw

Thanks guys.
Lurch- In your example, you use gasses. I understand that principle, but could you tell me if the same kind of effect would occur between a solid and a gas? I realise that if so, the effect would not be so pronounced as in gasses. I just want to make sure I understand fully.

5. Sep 30, 2004

### Clausius2

It is as Lurch and ZapperZ has explained to you. I'm going to try to go a bit forward and calculate the amount of bouyancy force that is exerted over a human body inside the atmosphere. Thus, we will figure it out and you will be surely more quiet about this.

We will suppose a person shaped like a cylinder of radius R=20cm (he's not too fat) and height H=1.8m.

Approximated Atmospheric pressure distribution:$$P(z)=P(0)*e^{(-gz/R_gT)}$$; and supposing a isothermal atmosphere: $$\rho(z)=\rho(0)*e^{(-gz/R_g T)}$$;

The total force exerted on the body will be:
$$\overline{F}=\int_{body} P\overline{dS}=\int \int \int_{body} \nabla P dV=\overline{g}\int \int \int_{body} \rho(z) dV=\overline{g}*\pi R^2 \rho(0)\int e^{(-gz/R_g T)}dz=\overline{g}*\pi R^2 \rho(0)\frac{1-e^{(-gH/R_gT)}}{g/R_gT}$$;

So that: $$F=g*\pi R^2 \rho (0) \frac{1-e^{(-gH/R_gT)}}{g/R_gT}$$

substituting:
$$\rho(0)=1.23 Kg/m^3$$;
$$R_g=288 J/KgK$$;
T=293K.

I obtained F=2.7 N. (Check the calculation, maybe I'm wrong).

The force exterted by the gravity over your body is approximately 800 N. The bouyancy force exterted in an isothermal atmosphere over a person is neglected. The reason behind this statement is that the pressure gradients inside a static gas are usually negligible in small intervals of heights. So that, the air pressure at your feet is almost the same at your head. Therefore the bouyancy force is neglected.