Re: How much vacuum pressure is needed in a tube to remove air?

For this answer I am going to assume dropping two balls, each having a diameter of 40 mm. One will be made of steel, which has a density of about 7.8 g/cm³, and the other a ping pong ball which has a mass of 2.7 g (See, for instance, this rules page.). A ball of radius 20 mm (2 cm) has a volume of 33.5 cm³, so the steel ball has a mass of about 261 g. If a 40-mm-diameter steel ball is not available, any steel ball will do, or any dense ball, because the effect of air resistance is nearly negligible for any dense ball of reasonable size over the small distance I am considering (2 m), as you will see below.

In addition, I found some information on the Internet which gives the density of the atmosphere at various altitudes, from which I made the following table:

There are some previous answers in our archives dealing with terminal velocity and related subjects, which you can find using the Google side of our search engine to search on "terminal velocity" (including the quotation marks). Using the equations at this terminal-velocity calculator and knowing that the usual coefficient of drag for a sphere is taken to be C_d = 0.5, I used the program Excel to do some computations for me. I found that, in the usual air density of the atmosphere at sea level, the steel ball would reach a terminal velocity of about 81.3 m/s (Which is verified by the online calculator!) if it had sufficient distance to fall freely, and will fall 2 meters (starting from rest) in about 0.639 seconds. (I chose 2 meters as being a reasonable height to build a vacuum tube to do these experiments.) However, the ping pong ball would reach a terminal velocity of only about 8.27 m/s and will fall 2 meters (starting from rest) in about 0.670 seconds. You could probably detect the difference in the fall of those two objects, especially since at 0.639 seconds the ping pong ball has only fallen 1.835 m, putting it 0.165 m (165 mm) behind the steel ball.

Using the air density equivalent to 20000 meters altitude (roughly 12 miles) the steel ball would reach a terminal velocity of about 307 m/s if it had sufficient distance to fall, and will again fall 2 meters in about 0.639 seconds. The steel ball doesn't act much differently across only 2 meters of falling distance. The ping pong ball, though, at an air density equilvalent to 20000 meters altitude would reach a terminal velocity of about 30.9 meters per second and will fall 2 meters in about 0.641 seconds. That is faster than at sea-level air density, and almost as quick a 2-meter fall as the steel ball. At 0.639 seconds the ping pong ball is only about 0.013 m (13 mm) behind the steel ball.

To obtain a vacuum equivalent to the air density of 20000 meters requires a vacuum pump that can provide 0.088/1.23 atmospheres, which is about 0.0715 atmospheres, and is, in usual vacuum-speak, about 27.8 inches of mercury vacuum, or about 54 Torr. That is a pretty good vacuum (A perfect vacuum would give 1 atmosphere, or about 29.92 inches mercury, of vacuum.). I did a quick search online of typical laboratory vacuum pumps and find that, for about $600, a pump can be found that can provide an appropriate vacuum.
Welch
Cole Parmer

Lesser vacuum pumps could be used. But if the vacuum is half as good as that given above, the steel ball will fall 2 meters in about 0.639 seconds, still, and the ping pong ball will fall 2 meters in about 0.653 seconds. That may still be good enough to demonstrate the effect of vacuum. At 0.639 seconds the ping pong ball is about 0.082 m (82 mm) behind the steel ball, so it is discernibly later, but by only about half the distance as using sea-level-density air.

If you are interested in the Excel files I used, send me email requesting the files and I will email them to you.

John Link, MadSci Physicist