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I came across a statement online that said that an average cloud has a weight of 1 million pounds. My main question is, does a cloud actually have weight on Earth or was the term "weight" a colloquial expression meaning "mass"?

From my understanding, weight is a force defined as W = mass × gravity, or essentially F = ma. Because the cloud is not falling, its acceleration is 0 -- that is, the acceleration of -9.8m/s2 exerted by Earth's gravity must be counteracted by another equal and opposite force in order for the cloud to be stationary in mid-air, hence the "a" portion of F=ma is 0, so F, and therefore weight, must be 0. Is this thinking correct?

However, when the state changes from gas to liquid -- that is, when it starts raining - the water molecules in the liquid state obviously have weight. So my secondary question is, does a molecule's state change affect its weight? For gases which are "lighter than air", is their weight negative? (I suspect yes because of the force equation).

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  • $\begingroup$ Note that the clouds that you "see" are all liquid water - water vapour itself is transparent. However the cloud droplets are so small that they are suspended in the updraft. $\endgroup$ – milancurcic Jul 10 '14 at 17:49
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    $\begingroup$ The major flaw in your premise is confusing gravity for net acceleration. F=ma=0 because net acceleration is 0, but gravity is still roughly -9.81 m s$^{-2}$ $\hat{\mathbf{k}}$. To verify this, go stand on a scale and stand still. Your net acceleration is 0, but you still have weight. $g \ne a_{net}$ except in inviscid free fall. $\endgroup$ – casey Jul 11 '14 at 14:18
  • $\begingroup$ Link to the article in question: mentalfloss.com/article/49786/how-much-does-cloud-weigh Also, this is based on a cumulus cloud approximately 1km cubed. $\endgroup$ – Richard Dec 9 '14 at 15:01
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There are different senses of weight used within branches of physics and engineering. In the strictest physical sense "weight", is given by W = mg. i.e. weight is the force on an object due to gravity alone. Clouds are not "weightless" in that sense, since they are not far enough away from the Earth to escape its gravitational pull.

The reason that clouds stay up in the sky is, as you say, because they are acted on by an equal and opposite force, namely the surrounding buoyancy of the air. Rain drops are formed by water molecules condensing around some sort of nucleus (a speck of dust, for example). This is triggered by the cooling of the cloud, or lifting to lower pressures, which reduces the moisture-carrying capacity of that particular "parcel" of air.

While the arrangement of water molecules in the cloud make it light enough to be buoyant in the air, the arrangement of those same molecules in rain drops are not – so they begin to fall.

By analogy, I would think of the fact that a ship is buoyant, despite its huge mass, but if you 'condensed' the ship into a lump of steel with the same mass, it would obviously sink. It's weight remains the same however, no mass has been removed, and the value of g has not changed. So to answer your second question, no, phase changes do not affect weight. (Although strictly speaking, in quantum physics it can do – but the amount is negligible here)

Gases which are "lighter than air" will eventually settle in a portion of the atmosphere where the surrounding gases are the same density. They are not weightless, nor do they have a negative weight. The molecules are still experiencing a force due to the Earth's gravity.

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    $\begingroup$ You may actually get a better answer over at Physics Stack Exchange. $\endgroup$ – decvalts Jul 10 '14 at 10:20
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    $\begingroup$ If you wanted to weigh a cloud on Earth (a very small cloud): Seal the cloud in a strong container. Place the container on a set of scales in a vacuum chamber. Subtract the weight of the empty container and you now have your cloud weight :) $\endgroup$ – decvalts Jul 10 '14 at 14:34
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    $\begingroup$ Ahh, I see the part i was missing is the fact that W is defined by the effects of gravity alone - I assumed W = mg is the same as F = ma, but the "a" is the sum of all forces, whereas g is just one force, namely gravity. In other words, I assumed that F= W = ma = mg, but a is not g (even if g is a), if that makes sense. $\endgroup$ – Mirkules Jul 10 '14 at 20:06
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No, the weight of the cloud can't be zero; it has to be about the same as an equivalent volume of air. If a parcel of air-plus-visible-matter is lighter (less dense) than the air around it, it will rise; if it's heavier (more dense), it will fall.

Clouds are visible because of water droplets or crystals (or smoke particles or whatever) suspended in the air. Air holding suspended liquid or solid matter will be denser than "empty air", all other things being equal -- but if the air in the cloud is warmer or more humid than surrounding air (since water vapor is less dense than air), it's easy to achieve more or less neutral buoyancy.

So, yes, clouds are big and heavy -- but they easily float in the atmosphere, which is even bigger and heavier!

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