# Why are non-equally heavy objects falling at the same speed on the moon? [duplicate]

This is no duplicate. I have own questions to it, that I want to have explained.

I recently read about this thing Galileo discovered with severel experiments. Everything, with any mass, falls at the same speed, on the moon.

Ok maybe... explanations always mention the missing air resistance and that two falling objects with different mass are attracted with the same gravity... I'm no expert. All the calculations seem right in any way, but I have a logical problem while thinking it through.

Assumption: I have two cuboids, one made of solid steal, the other one is of hollow plastic. These two objects have the exact same dimensions. You can only differ them by lifting.

Now, I drop them both from the same hight on a flat surface. I assume, that the heavy one hits the ground first. Is that the case, because of density? If yes: If I resize/inflate the heavy cubiod to a height (same bottom dimensions as the other, that its air resistance during falling stays the same) where it has the same density as the lighter object (weight stays also the same), will they fall at the same speed now?

Would they (same size and shape, different weight) hit the surface at the same time, if I dropped them in a vacuum chamber?

## marked as duplicate by sammy gerbil, stafusa, Jon Custer, Kyle Kanos, Qmechanic♦Jun 29 '18 at 12:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 4 Answers

So this is one of those things that may not seem completely obvious at first. I once took two plastic soda bottles, that I had laying around, and illustrated this to my mother. I filled one up maybe one third of the way with water, and the other one the whole way with water, so that one was 3 times heavier than the other. And then I dropped them side-by-side. And they both hit the ground at the same time, to the naked eye. She laughed and said “I just saw you prove it and I still don’t believe you! Why doesn’t the heavier bottle hit the ground first?”

Had I dropped them off a bridge, maybe they could have gotten closer to their terminal velocities, so the heavier one could have pulled ahead. That is our experience with flowers and feathers after all, they fall slower. But just in dropping bottles to the floor, they don't get enough time to accelerate to that speed, so you don't see the effect of air resistance. Feel free to replicate this! One is a third the density of the other one, but they both hit the ground at the same time.

Now you can even go a step further, and understand why it has to be this way. If you think about dropping two balls side by side, with like a hairs-breadth distance between them, they will take a certain time $T$ to hit the ground. Now if I were to say that something with twice the mass were to fall, say, twice as fast, then I would be making a concrete statement about what happens when you glue these two things together, I would be saying that gluing them together makes them fall faster. I’d be saying that unglued, they fall in time $T$, but glued together they fall in time $T / 2$. That seems implausible. But if gravity only cares about mass, and does not care about shape, then two balls glued together need to fall as fast as one ball that is twice as heavy.

Now of course gravity might not know what shape something is, but wind resistance we know does. This is the principle behind a parachute: it has the same mass before and after you pull the ripcord, but before it is the shape of a backpack and after it is the shape of a big balloon or sail. This is not quite density, but it's sort of related. You are not changing the density of paper if you crumple up a flat sheet into a tiny ball, but you are changing its shape, and it will take a shorter time to fall to the ground. But density can certainly help guide your intuitions. Because if the geometry is the same, and it's just a matter of the same mass and a different size of this geometry, then one can define the density and observe that the more dense thing will fall faster: A loosely crumpled paper ball can be regarded as less dense than a tightly crumpled paper ball even though the paper itself has the same density. (In this sense I am “cheating” when I compare the ball to the plain sheet because those have different geometries and it's harder to mentally include some air with the plain sheet the way you can when talking about crumpled balls, averaging out the paper’s mass over the ball’s spherical volume.) And the tightly crumpled ball hits the ground first.

But on a space rock that does not have an atmosphere, like the Moon, it is much easier to neglect air drag, and so everything falls independent of geometry: then it is easy to see that whether two balls hit the ground faster or slower should not depend on whether they are glued together.

When lacking any external dampening (such as air resistance), any two objects will fall at the same rate and hit the ground at the same time when released from the same height - no matter by how much their masses differ.

The 'easy' explanation for this is like you mentioned: gravity will accelerate both objects at the same rate, regardless of their mass.

The 'real' and more complex explanation is that gravity is not really a force, but rather an apparent or pseudo force that manifests because the objects are residing in an accelerating frame of reference, much like how you feel a pseudo force pushing you backward in an accelerating car. The accelerating frame of reference is the region of space-time which is curved due to the planet's (presumably Earth:) great mass - this is explained by the theory of general relativity.

It's not because of density, it's because of air resistance, or drag, which depends, approximately, on cross-sectional area.

So when you make your heavier mass longer so that it has the same density as the lighter one, it still has the same cross-sectional area (as you've carefully arranged that). So to a good approximation the drag on it will be the same as it was previously, and it will still fall faster than the light object. (In fact it might fall a little more slowly than it did because there is some drag associated with the air flowing along its sides, but this is a complicated question.)

Finally, yes, if you drop them in a vacuum chamber they will fall at the same speed.

## Some details

If we ignore air resistance then the force on something falling under gravity is $F_G = mg$, where $g$ is the acceleration due to gravity, and $m$ is the mass of the object. And Newton tells us that $F = ma$, where $a$ is the acceleration of the object. We can put these together to get the equation of motion:

$$ma = mg$$

And now look: $m$ is on both sides so we can cancel it to get $a = g$. And this is independent of the mass of the object, so we know that all objects fall with the same acceleration.

But this is ignoring air resistance.

Air resistance is in general complicated to model because it depends on a lot of factors. But a simple-minded expression for it which is approximately correct in many cases is:

$$F_D = - cAv^2$$

Where

• $F_D$ is the drag force;
• $A$ is the cross-sectional area of the thing falling in the direction of motion;
• $v$ is the speed it is falling at;
• $c$ is a bundle of other factors which takes account of the density of the air, the shape of the body and so on.

So now we can get a much more complicated equation of motion for something falling with air resistance

$$ma = mg - cAv^2$$

Or, since $a = dv/dt$:

$$m\frac{dv}{dt} = mg - cAv^2$$

And if we divide through by $m$:

$$\frac{dv}{dt} = g - \frac{A}{m}cv^2$$

I'm not going to solve this, but you can see two things:

• $m$ does not vanish any more;
• the thing that matters is this $A/m$ term.

What this means is that the thing that controls the influence of air resistance on falling bodies (to a reasonable approximation) is the ratio of cross-sectional area, $A$ to mass, $m$. And in your example this ratio remained unchanged as you simply made the object longer while keeping $A$ the same.

The force due to gravitational attraction experienced by an object is proportional to the mass of the object. This results in objects of differing masses accelerating at the same rate. On Earth, the air affects experimental results due to friction and turbulence. On the moon a feather and iron bar accelerate towards the surface at the same rate due to the (almost complete) lack of atmosphere. Changing the density of the objects as you describe in your question does not affect the mass, so the dense and less-dense objects will experience the same force and accelerate at the same rate in the absence of air.