Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Buoyant Vacuums?
Name: Steve T.
Status: other
Age: 20s
Location: N/A
Country: N/A
Date: Thursday, April 25, 2002


Question:
is it possible to take advantage of the fact that a vacuum is lighter than helium by creating an envelope that was engineered to be very light while also strong enough to resist acute internal strain? could purging a container like this cause it to float?


Replies:
Good question. I made an estimate with some excitement, but the answer to your question is no (if by floating, you mean in the atmosphere). Steel vessels can be easily constructed that float in water, as is obvious from any visit to any harbor. They take advantage of the fact that water is some 833 times as dense as air is near the earth's surface (1.3 kg/m^3).

To make an estimate of the feasibility of the idea, three equations are needed:

B = 4/3 pi r^3 dair,  M = 4 pi r^2 t dsteel,   
                                               
   F       =    A     S
pi r^2 patm = 2 pi r t S

Here B is the mass of air displaced by a steel sphere of radius r in air of density dair. pi = 3.14.

M is the mass of the sphere given by the area of the sphere times its wall thickness times the density of steel, dsteel.

The third equation estimates the thickness of the wall required to withstand the force of atmospheric pressure (patm) which must be withstood by the steel around a diameter. A is the area of this steel (circumference times thickness) and S is the maximum compressive stress that can be withstood by the steel.

Putting these equations together, I get:
B/M = (2 S dair)/(3 patm dsteel) = 0.56

I used dair = 1.3 kg/m^3, patm = 1.0 x 10^5 N/m^2,

dsteel = 7800 kg/m^3, and S = 5 x 10^8 N/m^2.

Notice that the buoyant force is only a little more than half the weight of the steel sphere. As the equation shows, this can be improved by using a stronger material (S), and/or a less dense material (dsteel). dair and patm obviously cannot be changed.

For a 10m diameter steel sphere, B = 680 kg and M = 1215 kg.

Best, Dick Plano...



Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory