40

This is a very well known problem, but I can't find an answer in the specific case I'm looking for.

Let's consider two balls :

  • Ball 1 weighs 10 kg
  • Ball 2 weighs 1 kg
  • Balls have identical volumes (so Ball 1 is much more dense)
  • Balls have identical shapes (perfect spheres)

Let's drop them from a rather important height, on earth, WITH air. (That's the important thing, because all the proofs that I browse take place in a vaccum).

I am arguing with a colleague. He thinks that ball 1 will fall faster in air, and that the two balls will fall at the same speed in a vacuum. I think that the identical shapes and volumes make air friction identical too and that the vaccum has no importance here. Could someone tell who's right and provide a small proof?

Qmechanic
  • 201,751
FlipFlapFlop
  • 503
  • 5
    Think of throwing them next to a balloon. Or better, do the experiment with one air balloon and a water balloon. Have you ever seen an air balloon falling fast? – Davidmh Jul 23 '14 at 15:45
  • 3
    "You would think that this [ball] would fall faster than this [feather], wouldn't you?" ... "And you'd be absolutely right!" – Nick T Jul 23 '14 at 17:37
  • Yes, the force caused by the air friction will be the same. But the gravitational force will be 10 times stronger for ball 1. So ball 1 will accelerate much faster, and its speed will be much higher when the air friction and gravity forces will be in balance. – Petr Jul 24 '14 at 19:23
  • 1
    @PetrPudlák Your comment seems to suggest that a higher gravitational force for ball 1 results in more rapid acceleration, which is incorrect. The greater force is exactly balanced by greater mass, so that the gravitational component of acceleration is the same for both balls. – Aaron Novstrup Jul 24 '14 at 22:14
  • You can easily make an experiment yourself: take a metal ball and make an identical ball of Styrofoam or just paper, then drop them. You will see for yourself. – Alexander Gelbukh Jul 25 '14 at 00:26
  • And an additional consideration: even if your were right and equal friction would lead to equal speed (which is not correct), even then there's buoyancy force which counters gravitation, so the relative total force on the lighter ball is smaller and may be even zero or negative. A simple experiment: take a large enough iron ball and a helium-inflated toy balloon; watch which one will touch the ground faster. – Alexander Gelbukh Jul 25 '14 at 00:34
  • I guess the key to this question is: Does the level of air resistance factor weight as a variable? I don't know, there are people much smarter than I on here that can hopefully say. – Henry F Jul 25 '14 at 04:38
  • @AaronNovstrup True, I formulated it badly. What I meant was that only the greater gravitational force is balanced by the greater mass. The force caused by air friction is the same. So while ball 1 accelerates with $a_1=(10g-F_v)/10=g-F_v/10$, ball 2 accelerates with $a_2=(1g-F_v)/1=g-F_v$ (where $F_v$ is the air friction force at speed $v$) – Petr Jul 25 '14 at 05:26

6 Answers6

59

I am sorry to say, but your colleague is right.

Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity,

$$ m g = k v^2 $$

And thus

$$ v=\sqrt{\frac{mg}{k}}$$

So, the terminal velocity of a ball 10 times as heavy, will be approximately three times higher. In vacuum $k=0$ and there is no terminal velocity (and no friction), thus $ma=mg$ instead of $ma=mg-F$.

Bernhard
  • 5,070
  • 3
  • 28
  • 41
  • 7
    Thank you for your answer. I must say I was very confident with my theory and I'm happy to have asked ;-) – FlipFlapFlop Jul 23 '14 at 12:11
  • This confuses me a bit. In a vacuum, assuming they're falling due to gravity, won't ball 1 still "fall" faster since it's more dense? – mechalynx Jul 23 '14 at 15:14
  • 1
    @ivy_lynx see Olin Lathrop's answer about falling in a vacuum – Holloway Jul 23 '14 at 15:23
  • 3
    @ivy_lynx No. $F=0$, so $ma=mg$, $m$ drops out and the acceleration is $a=g$. Which is the same for both. – Bernhard Jul 23 '14 at 16:02
  • 1
    Yeah but see my comments on Olin Lathrop's answer. Friction might be zero, but the force of gravity should depend on both objects' masses. – mechalynx Jul 23 '14 at 16:05
  • 6
    If you are talking about doing the experiment on Earth, then the Earth's gravitational field is so much larger than that of your objects, the difference can be ignored. If you're doing it in space with objects of similar mass, then yes both objects mass matters. – Jasmine Jul 23 '14 at 16:18
  • @ivy_lynx see this question (or any of the other related questions on the site) to understand why that effect is negligible. – David Z Jul 23 '14 at 21:29
  • @DavidZ I'm not arguing that the effect isn't negligible. Of course it is. And it can be ignored in this case. The answers represent what will happen accurately enough. My problem is with the explanation and the logic behind the explanation. I linked to a NASA site in Lathrop's answer that does the math. If the same logic there is applied generally we'd be forced to conclude that even if the object "falling" is a planet, it'll fall at the same rate. Technically correct, but would that really be what you'd expect to happen? What if the "falling" object is larger in mass? – mechalynx Jul 23 '14 at 22:41
  • 1
    @ivy_lynx Then you'd have to consider it in the centre of mass system, but anyhow completely irrelevant for this question. Now you are mainly confusing the OP. There is absolutely no reason to make it this complicated here. – Bernhard Jul 24 '14 at 06:11
  • Yes, this is confusing me, but at the same time I'd be really interested to understand the general case. In an empty universe, would a 10kg ball be attracted faster than a 1kg ball towards a 1000kg ball? – FlipFlapFlop Jul 24 '14 at 08:46
  • 1
    @FlipFlapFlop Maybe it is better to ask that as a separate question :) – Bernhard Jul 24 '14 at 08:59
  • @FlipFlapFlop No! The gravity force ill be proportional to the mass but the acceleration ill be inversely proportional. The mass factor cancels. Note more important to find out the "faster" one here ill be the distance between the objects (assuming all are at rest) – jean Jul 24 '14 at 11:35
  • 1
    @Bernhard If I'm confusing the OP it's only because we can arrive at a correct conclusion from many avenues. Perhaps you should be concerned about whether the OP will walk away thinking that $$a=G\frac{m}{r^2}$$ is a complete answer in all such situations. - (at)OP My argument is that the pull each object experiences is dependent on the other mass and this does not exclude the planet. So for any ball, the pull they get from the planet is the same, but the pull the planet experiences from the ball isn't. So if the ball is large enough, this matters. How large depends on our needs. – mechalynx Jul 24 '14 at 12:55
  • 1
    I think this answer should talk about the velocity during transition, and not only the maximum velocity. – Saffron Jul 24 '14 at 16:07
  • @Saffron Agreed. What if neither ball reaches terminal velocity? – Aaron Novstrup Jul 24 '14 at 21:11
  • My intuition says that the heavier ball would still fall faster in the presence of air because its greater momentum (once moving) would allow it to drive air out of its way more effectively. Is this argument correct? – Aaron Novstrup Jul 24 '14 at 21:15
  • @FlipFlapFlop also, the material of the balls matters. If they are made of a conducting material like copper, current would induce them as they would be accelerating under earth's gravity and magnetic field. the induction of current would oppose its motion due to lenz's law. Since density of the two balls is different, there would be different amount of current generation, and thus different opposition to acceleration. So this answer is wrong in that context. But true if the balls are perfectly made of wood or something... – user3459110 Jul 25 '14 at 05:37
  • @Saffron I don't agree. It is obvious that at any stage the heavy ball will fall faster (for $t>0$), due to the imbalance in forces. I don't think adding differential equations for transient terms really adds anything in the concept of air friction. Is you think it is important, feel free to add an answer where you do these derivations and show a closed form solution for the balls trajectory. Conclusions will be the same. – Bernhard Jul 25 '14 at 05:39
  • @Awal Did you estimate how large this force is compared to typical friction forced by the surrounding fluid? – Bernhard Jul 25 '14 at 05:40
  • @Bernhard nope. I didn't... am I missing something? – user3459110 Jul 25 '14 at 05:41
  • @Awal Well, I would be interested in an order of magnitude analysis. Or maybe even an analysis of the terminal velocity in absence of air friction. You might easily get into relativistic regimes. – Bernhard Jul 25 '14 at 05:44
  • @Bernhard can you come to chat please? – user3459110 Jul 25 '14 at 05:47
  • @Awal Notify me there, and I will be there when I have time – Bernhard Jul 25 '14 at 06:15
  • It would be nice to add buoyancy too. In air it will be significantly smaller than drag (friction), but it would make the answer more general. – Jan Hudec Jul 25 '14 at 15:42
21

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum.

In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The acceleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accelerate the same, which is the accelleration due to gravity for the local conditions (about 9.8 m/s2 on the surface of the earth).

However, in air there is the additional upwards force due to friction with the air. That force is a function of the speed and shape of the falling object. If both balls were falling at the same speed, both would have the same upwards force on them due to air resistance. This force is not proportional to the mass of the object, so causes a higher deceleration on the object with less mass.

For example, the 10 kg ball is pulled downwards due to gravity with 98 N force, whereas the 1 kg ball is only pulled downwards with 9.8 N. Let's say they are falling at the same speed thru air and that each is experiencing 3 N upwards force due to the air. Ball 1 is now being pulled down by a total of 95 N, and ball 2 by 6.8 N. That means ball 1 experiences 95 N / 10 kg = 9.5 m/s2 downward acceleration, and ball 2 experiences 6.8 N / 1 kg = 6.8 m/s2 downward acceleration. This means ball 1 will continue on falling faster than ball 2.

Saaru Lindestøkke
  • 189
Olin Lathrop
  • 12,916
  • 1
  • 31
  • 47
  • Sorry, but while I understand that for all practical purposes the balls will fall at the same rate, I don't see how this can be the general case. I'm assuming your explanation is pretty much the same as this. However, the force of gravity is dependent on both masses, correct? At least according to newtonian gravity. It seems that if one replaced F with m g then replace g with the Newtonian expression, g would be larger for ball 1, thus it will accelerate faster. – mechalynx Jul 23 '14 at 15:33
  • correction (since I can't edit the above comment) It isn't g that will be different but F (sorry for the silly mistake). F will be different between the balls and larger for ball 1, so I think it should accelerate faster. – mechalynx Jul 23 '14 at 15:39
  • 1
    @ivy_lynx You are absolute right that the total gravitational force will be larger. However, the acceleration will be the same, as force = mass times acceleration. So the effect of a larger mass on the larger force is exactly cancelled in the acceleration term. – Bernhard Jul 23 '14 at 16:10
  • 1
    @ivy_lynx More specifically, you said "replace g with the Newtonian expression". If you think about what the Newtonian expression actually is, remembering that g is what you're calling the acceleration due to gravity, $F = m g = G M m / r^2$, thus $g = G M / r^2$, and it depends only on the radius and mass of the earth, not on the mass of the object you're dropping. – Cascabel Jul 23 '14 at 16:18
  • @Jefromi I corrected that (see the next comment) - sorry, my math is really rusty. – mechalynx Jul 23 '14 at 16:20
  • @ivy_lynx Your next comment still says that the force is different so the acceleration will be different. Have another look at my comment: I explicitly explained how the different force results in the same acceleration. (Feel free to change $g$ to $a$ if it makes it more obvious.) – Cascabel Jul 23 '14 at 17:02
  • @ivy_lynx Consider a train, and a basketball. You see them both accelerating at you, at the same rate. Which one has a higher force being applied to it? Same thing when the 10kg and 1kg ball are accelerating down toward you. You know the heavier ball has more force acting on it. That's simply because it takes more force to accelerate the bigger ball. – Cruncher Jul 23 '14 at 17:31
  • @Cruncher their acceleration however is due to a force produced regardless of mass. It's a release of energy from a fuel. Gravity is directly proportional to mass. – mechalynx Jul 23 '14 at 17:34
  • @Jefromi I got the same, but I feel like this explanation is inadequate. I made a mathjax rendering of why I think it'll still have more acceleration. I feel that my explanation demonstrates that the effect is both negligible but also present. I can't reconcile balls accelerating the same, with this kind of explanation, while if we had planets we'd clearly use the Newtonian formula, which would also clearly show that the acceleration would be dependent on both masses. (btw, in the mathjax, I got neglibility backwards, another mistake :P). – mechalynx Jul 23 '14 at 17:36
  • @ivy_lynx Perhaps, step by step will help. How much force is being applied on the 1kg ball? How much force is being applied on the 10kg ball? (Answer: ~10N and ~100N respectively. Do you agree?) – Cruncher Jul 23 '14 at 17:38
  • @Cruncher I understand that already and I did the math that demonstrates that thought process. However, I feel the math that ends up with $$g = \frac{m_1}{r^2}$$ in this case contains the assumption that only the ball is accelerating. This is not in agreement with Newtonian gravity (or any other gravity for that matter). The only way this could be true is if the planet was held in place by something, preventing it from accelerating. In other words, it's a partial answer. – mechalynx Jul 23 '14 at 17:41
  • @ivy_lynx Well of course the earth is accelerating as well. It wasn't clear that that's what you were talking about. It also doesn't play into effect if you drop the two balls beside eachother which is how the problem traditionally goes. This does not actually change the instantaneous acceleration of the ball. Since you have to be in an inertial reference frame. – Cruncher Jul 23 '14 at 17:48
  • @Cruncher I know it wasn't clear, since I struggle a bit with this (not much experience with physics since college). My disagreement has to do with the conclusion that all objects, regardless of mass will fall with the same speed. While obviously when you have two tiny balls it won't matter (lol "tiny balls"), if you generalize the though and the math, you'll get the obviously false assumption that a planet would also fall at the same rate towards another. The problem is the frame of reference. Falling at the same rate according to who? For two planets the mistake would be obvious. – mechalynx Jul 23 '14 at 17:51
  • (Neglibility fixed: http://imgur.com/iWeRL3k) – mechalynx Jul 23 '14 at 17:53
  • @ivy_lynx Both balls are still accelerating to the planet at the same rate. You state: "The problem is the frame of reference". Absolutely, I agree. However you need an inertial frame of reference(http://en.wikipedia.org/wiki/Inertial_frame_of_reference) to do the math properly. If you're on earth, you would see the bigger ball accelerating faster. But you wouldn't be on an inertial frame of reference. – Cruncher Jul 23 '14 at 17:56
  • 1
    @Cruncher I'm not talking about inertial frames, just perspective. There is no frame of reference that you can observe from, that will give you the exact same time to reach the surface, if you were observing the phenomenon. If you're the ball, the planet is accelerating faster towards you. If you're the planet, you're accelerating faster towards the ball. If you're some dude in space, you'll see them collide faster than if the ball had less mass. – mechalynx Jul 23 '14 at 17:59
  • Please continue this discussion in [chat]. – Manishearth Jul 25 '14 at 10:08
12

Other answers & comments cover the difference in acceleration due to drag, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider.

The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, the acceleration resulting from this force will differ based on the mass of the ball.

This is most easily illustrated by considering one as a lead ball and one as a helium balloon - obviously the helium balloon doesn't fall, because it is lighter than the air it displaced. The upward buoyancy force is greater than the downward gravitational force.

In a heavier fluid, like water, this effect is even more pronounced.

paulw1128
  • 221
  • 1
    The buoyancy force is proportional to the volume of the object. Since both balls have the same volume, they experience the same buoyancy force. A lead ball and a helium baloon of the same volume experience the same buoyance. The difference is that the lead ball experiences more gravity, much more than the buoyancy force. For the helium balloon, the gravity force is less than the buoyance force. – Olin Lathrop Jul 23 '14 at 18:47
  • Yes - this is what I tried to explain in my second paragraph - the force is the same but the acceleration is dependent on the mass. – paulw1128 Jul 23 '14 at 19:56
  • @OlinLathrop Just as the force due to air resistance is the same for both balls (at least when they're traveling at the same velocity)? So it seems that a complete explanation for the heavier ball falling faster in air must include both direct air resistance and buoyancy. As a side comment, the word "friction" seems out-of-place here. – Aaron Novstrup Jul 24 '14 at 21:31
  • 1
    It would be nice to quantify this answer. Is the effect of air resistance really always the largest effect? I would guess it's not: consider a spherical balloon full of air and a spherical balloon full of lead starting at rest. At t=0, wouldn't air resistance exert 0 force, while buoyancy exerts some non-zero force? – Aaron Novstrup Jul 24 '14 at 21:58
  • You're right Aaron. If I get a chance over the weekend I'll work up the proper analysis. – paulw1128 Jul 25 '14 at 20:40
2

I am not satisfied with the way @Bernhard answered, since it just shows the maximum velocity, thus only answering partially the question.

The air resistance can be written as : $$ R = \frac{1}{2}\,C_x\, \rho\, S\, v^2 $$ Note : The mass of the object is not in this equation. This is very important.

Applying Newton's law to one of the object gives at any moment of the fall: $$ a = g - \frac{1}{2m}\,C_x\, \rho\, S\, v^2 $$

As you can see the acceleration is function of the mass of the object $m$. A heavier object will accelerate more than a lighter one, therefore, will go faster during the whole fall. Both objects will at one point reach the maximum velocity that is explained well in @Bernhard answer.

So at any point of the fall, your heavier object will be faster than the lighter one.

Saffron
  • 151
  • Your answer is not complete, as $a$ also depends on the velocity. If the heavier object accelerates quicker, the velocity is higher, and thus acceleration lower. I think if you want to claim that, you will need a further analysis. – Bernhard Jul 25 '14 at 08:42
  • @Bernhard Let's assume velocity is relevant, and decelerate the object more than the mass accelerates it. At one point, the two object will have the same velocity, and the heavier object will accelerate more. There is no need for a big analysis to show that velocity is not relevant for what we want to prove. Should I put that in the answer though ? – Saffron Jul 25 '14 at 08:48
1

Since air creates a force that is approximately proportional to the square of the velocity, the acceleration for each sphere is $a_r = kv^2/m (where \text{ } k = \frac{1}{2} C_x\rho\ S) $ The net acceleration on each sphere is $ a_n = g - a_r$. As the velocity increases, the $a_r $ increases until the net acceleration $a_n $ becomes zero $(a_r = g)$, and thus each sphere reaches its terminal velocity. $$Given: m_1 = 10kgr, \text{ } \text{ } m_2 =1kgr, \text{ } k = 0.01, \text{ }g = 9.8m/s$$ $$ For \text{ } m_2, ( v_2 = m_2g/k)^{1/2} = (1x9.8/.01)^{1/2} = 31.3 m/s$$ $$For \text{ } m_1, (v_1 = m_1g/k)^{1/2} = (10x9.8/.01)^{1/2} = 98.99m/s$$

After using an iterative method, I determined that the 1kgr mass $(m_2)$ reaches the terminal velocity in about 10 seconds and the 10kgr mass $m_1$ in about 33 seconds. Although the spheres reach their terminal velocity at different times, the larger mass reaches a higher velocity because the lighter mass reaches its terminal velocity sooner and does not increase after that. The heavier mass, takes longer to reach its terminal velocity, and thus it becomes larger. So, the heavier mass will reach the ground sooner.

Guill
  • 2,493
0

This problem can be easily solved by the formula “F=ma”. You must be familiar with the reason why it would fall at the same rate in vacuum. But if we talk about the free fall in atmosphere, as you said there will be friction off course, and as the objects have the same shape, it’ll be same.

As the force of friction is the same on the two bodies, the one with the larger mass will have a smaller (negative) acceleration , and the one with the smaller mass have a bigger (negative) acceleration. So the ball with smaller mass will be slowed down by a great extent ( than the ball with larger mass).

ALWAYS remember, F=ma. Force depends ONLY on the mass and NOT on the density!

PS - I don’t know why are others making the problem so complicated with those formulas!

Aaryan Dewan
  • 1,780