# Which ball touches the ground first?

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?

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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
"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 Pudlák Jul 24 '14 at 19:23
@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 0:26

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 $ma=mg-F$.

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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
@ivy_lynx see Olin Lathrop's answer about falling in a vacuum –  Trengot Jul 23 '14 at 15:23
@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
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. –  ivy_lynx Jul 23 '14 at 16:05
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

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 accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate 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 decelleration 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 accelleration, and ball 2 experiences 6.8 N / 1 kg = 6.8 m/s2 downward accelleration. This means ball 1 will continue on falling faster than ball 1.

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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. –  ivy_lynx 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. –  ivy_lynx Jul 23 '14 at 15:39
@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
@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. –  Jefromi Jul 23 '14 at 16:18
@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. –  ivy_lynx Jul 23 '14 at 17:59

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.

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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
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

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.

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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 8:42
@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 8:48

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.

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