Answer to the Question 10/98

FALLING CLOUDS

The question was:

Clouds are made of water droplets, and water is about 800 times denser than air. Why don't clouds fall like a rock to the ground?


(11/98) The problem has been solved with varying degrees of accuracy and completness, by Oren Melinger (e-mail melinger@internet-zahav.net), Kerry MacArthur (e-mail MacArthu@firsttrust.com), Dmitriji Podolsky from Ukraine, and William Chess (e-mail chess@kennett.net). Short solution of the problem is presented below.


The answer: The clouds do fall, but they fall very slowly...

The solution:

A falling drop feels a frictional force from the air's viscosity that opposes gravity (the drop has to push its way through the air), the drop reaches a constant terminal velocity, and this terminal velocity is smaller the smaller the radius of a fluid particle or ice crystal. It turns out that this terminal velocity is quite small for drops the size of a few millimeters and so, although the drops in a cloud fall, they fall so slowly that only careful measurements would reveal this fact.

More quantitatively we can say the following: The weight of a drop of air of radius R is 4{pi}R3{rho}g/3 where {rho} is the density of the water (=103 kg/m3, and g (=9.8 m/sec2) free-fall acceleration. If the drop is falling through air with velocity v, the viscosity of the air n (=1.75*10-5 N*sec/m2) will cause friction (Stokes law) 6{pi}nRv. [Here we neglect the density of the air and assume that the viscosity of the drop is much larger than viscosity of the air; we also asume that the drop is approximately spherical.] The terminal velocity of the falling drop is obtained by equating the weight and the friction force, and is given by

v=2{rho}gR2/9n .

Thus 1 micron drop will have velocity of 0.13 mm/sec, or 11 m/day. Larger, 10 micron drops will still fall slowly, - 1.1km/day. Such fall rates can be neglected, especially since the motion of the air itself can be faster than that.

Drops significantly smaller than 1 micron are not visible and will not be percieved as clouds, while 0.1 mm drops will fall with velocity of about 1 m/sec, i.e. it will rain. Larger drops will fall with even larger velocities; however, the air-friction starts increasing faster than than v, i.e. the velocity will not increase as fast with increasing drop size. At such large speeds the weight of the drop is balanced by the drag force (1/2){rho}airC{pi}R2v2, where {rho}air is the density of the air (=1.2kg/m3) and C=1.2 is the drag coefficient for sphere. According to this equation 1 mm drop will fall with velocity of 4.3 m/sec and 10 mm drop will fall with velocity of 13.6 m/sec. However, drops larger than 5 mm are usually broken into smaller drops that fall more slowly.

Comment 1: This problem is related to the Milliken oil-drop experiment in which the absolute magnitude of the electric charge was measured for the first time.

Comment 2: There is a nice book that discusses the question of falling clouds and other meteorological phenomena: W. E. Knowles Middleton, A History of the Theories of Rain and Other Forms of Precipitation, Franklin Watts, Inc. (NY), 1965.


We thank Henry S. Greenside for his help in formulation of this solution.
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