2011-01-21 20:10:15 8 Comments

The common understanding is that, setting air resistance aside, all objects dropped to Earth fall at the same rate. This is often demonstrated through the thought experiment of cutting a large object in half. The halves clearly can't then fall more slowly just by being sliced in two.

However, I believe the answer is that when two objects fall together, attached or not, they do "fall" faster than an object of less mass alone does. This is because not only does the Earth accelerate the objects toward itself but the objects also accelerate the Earth toward themselves. Considering the formula: $$ F_{\text{g}} = \frac{G m_1 m_2}{d^2} $$

Given $F = ma$ thus $a = F/m$, we note that the mass of the small object doesn't seem to matter as when calculating acceleration the force is divided by the $m$ term, its mass. However, this overlooks that the force is actually applied to *both* objects, not just to the smaller one. The acceleration on the second, larger object is found by dividing $F$, in turn, by the larger object's mass. The two objects' acceleration vectors are exactly opposite, so *closing acceleration* is the sum of the two:

$$ a_{\text{closing}} = \frac{F}{m_1} + \frac{F}{m_2} $$

Since the Earth is extremely massive compared to everyday objects, the acceleration imparted on the object by the Earth will radically dominate the equation. As the Earth is $\sim 5.972 \times {10}^{24} \, \mathrm{kg} ,$ a falling object of $5.972 \, \mathrm{kg}$ (just over 13 pounds) would accelerate the Earth about $\frac{1}{{10}^{24}}$ as much, which is one part in a trillion trillion.

Of course in everyday situations, we can for all practical purposes treat objects as falling at the same rate because of this negligible difference—which our instruments probably couldn't even detect. But I'm hoping not for a discussion of practicality or what's measurable or observable, but what we think is actually happening.

Am I right or wrong?

What really clinched this for me was considering dropping a small Moon-massed object close to the Earth and a small Earth-massed object close to the Moon. This made me realize that falling isn't one object moving toward some fixed frame of reference, but that the Earth is just another object, and thus "falling" consists of *multiple objects mutually attracting in space*.

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### [SOLVED] Gravity and free fall

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## 11 comments

## @Dr. Manuel Kuehner 2018-04-17 15:26:08

Not sure if that counts as an answer but at least I use some scribbles :).

I would depict the (one dimensional) situation as this:

inertial frame of reference).The absolute acceleration of each object $\ddot{x}_1$ and $\ddot{x}_2$ can be formulated as:

## @ErikE 2018-04-17 16:11:29

Yes, that’s it. Though I thought there were adequate answers so far, I don’t begrudge your chance to explain it your way.

## @Dr. Manuel Kuehner 2018-04-17 16:14:19

I had to translate

begrudgefirst :)## @Dr. Manuel Kuehner 2018-04-17 16:29:27

@ErikE I have a picture and no square roots. Isn't that a plus? Seriously, I find the question interesting and wanted to add something to it (in my "reference system").

## @Gendergaga 2016-04-29 00:03:27

In addition to the already given answears this also might be of interest:

When hammer and feather are dropped simultaneously they arrive at the same time, when dropped independently the hammer attracts the planet more than the feather, so you are right, the total time until impact is then smaller for the hammer.

If you pick up the hammer and let it fall to the ground while the feather lies on the ground and its mass adds to the planet's mass (neglecting desity inhomogeneities) it takes the same time as when you pick up the feather and let it fall down while the hammer is on the ground and its mass adds to the planet, since m1+m2+m3=constant.

When you drop hammer and feather simultaneously, the feather will travel the longer distance in the same time, and is therefore faster than the hammer, since the planet is moving more towards the hammer than to the feather, and the feather is attracted by the largest sum of masses.

The initial distance of the point masses is 1 meter; in the first example you have 1000kg vs 100kg vs 1kg and in the second one 1000kg vs 666.̇6kg vs 500kg. As you can see the "hammer" and the "feather" arrive at the same time:

## @Tom B. 2018-02-09 20:27:19

Ha! This unambiguously explains everything! Everyone should look at this carefully.

## @JMac 2019-02-10 02:43:56

Would this still apply if the hammer and feather were dropped on opposite sides?

## @Gendergaga 2019-02-10 03:28:37

No, then the hammer would hit first since the planet moves in his direction and therefore away from the feather, see yukterez.net/org/1000.666.500.line.gif

## @Adrian Howard 2019-06-09 18:34:43

but isn't the feather attracted to the mass of the earth + mass of the hammer when on opposite sides?

## @Gendergaga 2019-06-09 19:53:41

It is, but that doesn't cancel out the effect that hammer and earth are accelerated stronger towards each other.

## @Guru Vishnu 2019-11-26 09:59:09

+1: If possible, could you please add another illustration which depicts the actual masses of the Earth, the hammer, and the feather? I think it will add more credit to your already excellent answer, and perhaps gives a sense of reality.

## @Gendergaga 2019-11-27 18:26:47

Then you would't see the earth move since it is so much heavier than the hammer and the feather that its displacement until impact would be less than 1 pixel on the monitor

## @Ravindra HV 2020-06-21 22:27:47

@Yukterez - Thanks for the 'yukterez.net/org/1000.666.500.line.gif' It helps confirm my understanding ! I have posted on the topic too.

## @Ron Gordon 2017-07-19 23:11:13

Let's keep it simple. Opening my textbook from 1964, "Physics, 4th ed", Hausmann & Slack, Nostrum Co., NY, 1957. When we are contemplating the laws of everyday objects in free fall near the surface of the earth (our only frame of reference), all other forces removed or neutralized, we are in the realm of Newtonian Physics.

F = G(Mm)/r^2; where G is the gravitational constant, M is the mass of the earth, m is the mass of our object, and r is the radius of the earth.

By definition, when at rest on the surface of the earth, the weight ,W, of the object is the repulsive force in the equation F = ma, and a = g, the acceleration due to gravity. Therefore, W = mg, and g = W/m. Substituting into our general gravity equation between two object, we have:

W = G(Mm)/r^2, thus W/m = GM/r^2

Therefore, W/m is a constant! Thus an object will fall with the same acceleration regardless of mass near the surface of the earth.

Had me worried there for a moment :)

## @ErikE 2017-07-19 23:13:50

You should worry some more. I discussed this. The acceleration imparted to various falling objects by the earth IS identical, but you're leaving out the part where the objects accelerate the earth toward themselves, changing the closing speed. Moreover, the question specifically says you don't get to do the normal hand-waving to ignore minuscule effects. So unfortunately, your answer doesn't add to the discussion...

## @Ron Gordon 2017-07-29 14:54:08

The point I make is one you dismiss without consideration. Why would I use a micrometer if I intend to use an axe? When would I record hourly stock market ticks if my objective is to measure daily tends? The current Core Theory has exposed a beautiful consistency up and down our various theories and has promulgated the emergence of Effective Field Theory championed by many, including Penrose and Carroll. Why tackle every problem from QM when a simpler theory that is consistent with QM is much more logical to use. Greater precision is not always the greater truth.

## @ErikE 2017-07-30 05:30:39

Why? Because that's what I was curious about. Why do you come here and tell me what kind of inquiry I am allowed or not allowed to have? You're making philosophical arguments here, not physics ones. If you want to say something different, how about asking your own question instead of mis-answering mine?

## @Ted Bunn 2011-01-21 22:28:34

The answer is yes: in principle there is such an effect. When the mass of the dropped object is small compared to the mass of the planet, the effect is very small, of course, but in principle it's there.

## @TROLLHUNTER 2011-01-21 20:57:27

Yes, a heavy object dropped from the same height will fall faster then a lighter one. This is true in the rest frame of either object. You can see this from $F=GmM/r^2=m\cdot a=m\cdot d^2r/dt^2$.

The fastest "falling" (since we are re-defining falling) object however is a photon, which has no mass.

## @ErikE 2011-01-21 21:02:03

Hmmm... where did M go in your second formula?

## @arivero 2011-01-21 21:13:03

The paradox appears because the "rest frame" of the Earth is not an inertial reference frame, it is accelerating. Keep yourself in the CM reference frame and, at least for two bodies, there is no paradox. Given an Earth of mass M, a body of mass $m_i$ will fall towards the center of mass $x_\textrm{CM}=(M x_M + m_i x_i)/(M+m_i)$ with an acceleration $GM/(x_i-x_M)^2$. Note that $\ddot x_\textrm{CM}=0$

Really we have only hidden the paradox, because of course $x_\textrm{CM}$ is different for each $m_i$. But this is a first step to formulate the problem in a decent inertial frame.

The paradox resurfaces again if you want to get rid of $(x_i-x_M)$. In most applications, now that you are in a non accelerating reference system, you want to consider distances related to it, ie $x_i-X_\textrm{CM}$. The solution is to redefine the mass. As $x_i-x_\textrm{CM}= M (x_i - x_M) /(M+m_i)$, we can say that the object $i$ falls into the Mass Center with an acceleration $G{M^3 \over (M+m_ i)^2}{1 \over (x_i-x_\textrm{CM})^2}$ You could say that the actual mass of the "earth at center of mass" is this correction.

Once you are into the trick of changing the value of the mass, you can still stick to the reference frame of the earth. In this reference frame the quotient between force and acceleration is $Mm_i/M+m_i$ You can claim that this is the actual mass of the body during the calculation. This is called the

reduced mass$m_r$ of the system, and you can see that for small $m_i$, it is almost equal to $m_i$ itself. You can ever write some of the previous formulae using the reduced mass $m_r$ in combination with the original masses, for instance the above ${M^3 \over (M+m_ i)^2}= M {m_r^2\over m^2}$, but I am not sure of how useful it is. In any case, you see that you were right about the "heavier implies faster" but that it is perfectly managed.For three objects, m_1 and m_2 falling into M, the question is how to compare the case to m_1+m_2 falling into M. You separate the forces between internal, between 1 and 2, and external, against M. Look at the point $x_0= {m_1 x_1 + m_2 x_2 \over m_1+m_2}$ . This point it is not accelerated by the internal forces. And the external forces move them as $$\ddot x_0={1 \over m_1+m_2} \left(m_1 {G M \over (x_1-x_M)^2} + m_2 {G M \over (x_2-x_M)^2}\right)={F_1+F_2 \over m_1 + m_2}$$

This is becoming long... ¡I can not put all the Principia in a single answer!. So you can forget all the previous stuff, consider it is just to a mean to fix notation and get some practice, an read the

answer:If the two bodies are at the same distance $x$ of the "external" earth, they suffer the same external acceleration $g=GM/(x-x_M)^2$, and the same happens with $x_0$. If both bodies are in an approximation where $g$ can be considered constant, which was the case originally considered by Galileo (and the modern $g=9.8~{\rm m/s^2}$), then they have the same acceleration -and also the combined position $x_0$-. If they are not at the same distance nor in an approximation of constant -equal everywhere- field, then you can still save the movement of $x_0$ to work as if it were a gravitational force for some single mass $m_T$, but then the manipulation of the equations will produce in the relative positions of $x_1$ and $x_2$ some accelerations of the order of $1/(x_0-x_M)^3$. Such forces are the "

tidal forces".## @ErikE 2011-01-21 21:25:23

What is "CM reference frame" please?

## @Greg P 2011-01-21 21:28:42

Center of Mass reference frame

## @arivero 2011-01-21 21:45:05

In any case, I feel that my answer is not honest enough... but I am in hurry, sorry. Come back later tomorrow.

## @arivero 2011-01-22 03:22:05

@Emtucitor I am back, but a bit drunken. Anyway. My answer could go along the lines now of explaining the concept of reduced mass (thus resusciting the paradox) and then about the decomposition of every collective movement into CM plus local. But at the end we should go to the difference between inertial and accelerating systems, and we would do a lenghtly mathematical thing, while it seems that really you are not into maths. So my short answer is, forget gravity, the point is about if all the bodies fall freely in a similar way, and that is not Newton.

## @arivero 2011-01-22 03:23:47

@Emtucitor so Try "two new sciences", which has a very light math level (no differential calculus!), and is freely available in the internet. and the arguments therein, without invoking any accelerating frame of reference. Good night!

## @ErikE 2011-01-22 10:15:00

I'm actually quite into math and did get a 5 in the calculus BC AP exam. So hit me with your best shot... If I get too bogged down in understanding I'll stop asking questions.

## @Fattie 2014-06-08 09:58:33

what "paradox" is under discussion here? there is no "paradox" whatsoever.

## @David Z 2011-01-21 21:47:29

Using your definition of "falling," heavier objects do fall faster, and here's one way to justify it: consider the situation in the frame of reference of the center of mass of the two-body system (CM of the Earth and whatever you're dropping on it, for example). Each object exerts a force on the other of

$$F = \frac{G m_1 m_2}{r^2}$$

where $r = x_2 - x_1$ (assuming $x_2 > x_1$) is the separation distance. So for object 1, you have

$$\frac{G m_1 m_2}{r^2} = m_1\ddot{x}_1$$

and for object 2,

$$\frac{G m_1 m_2}{r^2} = -m_2\ddot{x}_2$$

Since object 2 is to the right, it gets pulled to the left, in the negative direction. Canceling common factors and adding these up, you get

$$\frac{G(m_1 + m_2)}{r^2} = -\ddot{r}$$

So it's clear that when the total mass is larger, the magnitude of the acceleration is larger, meaning that it will take less time for the objects to come together. If you want to see this mathematically, multiply both sides of the equation by $\dot{r}\mathrm{d}t$ to get

$$\frac{G(m_1 + m_2)}{r^2}\mathrm{d}r = -\dot{r}\mathrm{d}\dot{r}$$

and integrate,

$$G(m_1 + m_2)\left(\frac{1}{r} - \frac{1}{r_i}\right) = \frac{\dot{r}^2 - \dot{r}_i^2}{2}$$

Assuming $\dot{r}_i = 0$ (the objects start from relative rest), you can rearrange this to

$$\sqrt{2G(m_1 + m_2)}\ \mathrm{d}t = -\sqrt{\frac{r_i r}{r_i - r}}\mathrm{d}r$$

where I've chosen the negative square root because $\dot{r} < 0$, and integrate it again to find

$$t = \frac{1}{\sqrt{2G(m_1 + m_2)}}\biggl(\sqrt{r_i r_f(r_i - r_f)} + r_i^{3/2}\cos^{-1}\sqrt{\frac{r_f}{r_i}}\biggr)$$

where $r_f$ is the final center-to-center separation distance. Notice that $t$ is inversely proportional to the total mass, so larger mass translates into a lower collision time.

In the case of something like the Earth and a bowling ball, one of the masses is much larger, $m_1 \gg m_2$. So you can approximate the mass dependence of $t$ using a Taylor series,

$$\frac{1}{\sqrt{2G(m_1 + m_2)}} = \frac{1}{\sqrt{2Gm_1}}\biggl(1 - \frac{1}{2}\frac{m_2}{m_1} + \cdots\biggr)$$

The leading term is completely independent of $m_2$ (mass of the bowling ball or whatever), and this is why we can say, to a leading order approximation, that all objects fall at the same rate on the Earth's surface. For typical objects that might be dropped, the first correction term has a magnitude of a few kilograms divided by the mass of the Earth, which works out to $10^{-24}$. So the inaccuracy introduced by ignoring the motion of the Earth is roughly one part in a trillion trillion, far beyond the sensitivity of any measuring device that exists (or can even be imagined) today.

## @Ravindra HV 2013-10-30 07:52:44

I agree. My understanding is the same as well.

Assuming that the earth, mars and moon are of the same size - If the earth and mars where to be suspended in space (mars falling on earth), they would come into contact faster than - if earth and moon were to be suspended in space (moon falling on earth) owing to the fact that mars would cause the earth to accelerate towards it more than moon. This is provided that the distance between the two objects are the same initially. The earth would attract both at the same rate for any given distance.

I have also posted here regarding what would fall first in case of three objects being involved, asking if my understanding is correct. It is the classic apple-feather experiment revisited. I hope it clarifies @KeithThompson 's question above.

## @Ravindra HV 2013-10-30 07:56:48

PS: I agree with @Nick in that there are two cases - one where the total mass in the system is the same and the other where it is not. The above understanding holds good only if the total mass in the system varies.

## @ErikE 2013-10-30 21:06:49

And the total mass in the system varies when you have two separate cases: dropping a small mass and dropping a large mass.

## @Fattie 2014-06-08 10:00:18

Yes, of course, obviously the total mass of the system varies in the two experiments. Experiment 1: you have the earth and mars 100,000 km apart. How long until they are 10,000 km apart? Experiment 1: you have the earth and moon 100,000 km apart. How long until they are 10,000 km apart? Simple.

## @Fattie 2014-06-08 10:05:02

Is there a chance some genius could simply calculate, the two cases: Mars->Earth 100k->10k how many minutes and Moon->Earth 100k->10k how many minutes. If so, you rock.

## @Nick 2012-12-08 15:03:02

The free fall time of two point masses is $ t = \frac{\pi}{2} \sqrt{ \frac{r^3}{2 G(m1+m2)}} $.

The free fall time is dependent on the

sumof the two masses. For a given total mass, the free fall time is independent of the ratio of the two masses. The free fall time is the same whether m_{1}= m_{2}, or m_{1}>> m_{2}.When a body is

to a certain height and then dropped, the time to fall to the Earth doespicked upnotdepend on the mass of the object. If you lift a ping-pong ball and then drop it, it will take the same time to fall to the Earth as a bowling ball. Splitting the Earth into two masses does not change the sum of those masses, or the free fall time.However, when an

is brought to a certain height above the Earth and then dropped, the free fall timeexternal bodydoesdepend on the mass of the external body. Because the sum of the Earth and the external body obviously does depend on the mass of the external body.Assumptions:

The Earth is isolated (there is no moon, sun, etc).

The Earth is non-rotating.

The Earth has no atmosphere. (A hot air balloon would fall upwards because it is less dense than the atmosphere it displaces.)

## @ErikE 2012-12-08 18:12:36

This makes no sense at all. There is no sensible distinction between local and external masses. Please read the accepted answer which provides the proper proof with formulas.

## @Nick 2012-12-08 20:45:05

@ErikE There are two scenarios. In the first, the total mass of the system is constant (you split the Earth into two pieces). In the second, the total mass of the system increases (you introduce new mass).

## @ErikE 2012-12-08 21:09:18

It makes no difference. It's a human reference frame to imagine the Earth to be still. But that is illogical. The Earth is not fixed in space. It moves due to acceleration imparted on it by other objects! Stop thinking about

whenthe second object is introduced. Calculate everything after that. The scenario is: two objects in an otherwise empty universe, one very massive, one of unknown mass, are in airless free fall, held apart by a force. When the force separating them is removed, theybothaccelerate towards each other. The mass of #2 affects the time until impact.## @ErikE 2015-05-22 23:18:21

I take it back, at least a little, due to reading another answer of yours that I either read more carefully or that explained better. I'm open to the possibility that serially picking up a light object and a heavy object would encounter the effect you're mentioning (where the heavy object would not fall faster because the "Earth" has had its mass reduced by more in that case), but I'm not sure at this point.

## @Timaeus 2015-08-20 19:55:40

@ErikE DavidZ's accepted answer has the same formula in a more general form just put in $r_f=0$ and you get the formula here.

## @HAL 2011-02-18 22:50:41

A simple explanation is that it takes more FORCE to ACCELERATE an object of greater MASS. A=F/M....or with a constant FORCE, the ACCELERATION is inversely proportional to the MASS. The is a result of inertia.

A larger mass dropped (on a massive objects) requires a GREATER force to accelerate; additionally a GREATER mass exerts a GREATER gravitaional force.

Setting newtons second law to the gravitional force law cancles mass out and therefore renders MASS not related to the ACCELERATION of two gravitationally related objects.

## @ErikE 2011-02-19 01:19:18

Halston, you performed a presto-change-o in the middle of your answer. You started out with accelerating one mass, but suddenly we're accelerating two. My point is that even with just dropping "one mass" there are really two masses involved, two separate accelerations, each imparted on one object

by the other. You can't just cancel one out. The Earth imparts 9.80665 ms^2 of acceleration. But each object imparts its own acceleration on the Earth. Falling is not just acceleration of a single mass but the two toward each other. I think you have missed the point.## @ErikE 2011-02-19 01:21:11

Adding a third mass to the equation (the "second" object) changes things enough that the closing acceleration between the Earth and the objects will not be the same as either mass alone.

## @HAL 2011-02-19 04:41:46

The same will apply to your "two mass". There is a force exerted on object one, there is a force exerted on object two. In fact, it is an action-reaction force. The book exerts a gravitational force on the earth, the earth exerts a gravitational force on the book. The book seems to be accelerating only because its mass (and inertia) is insignificant in comparison to the earth. Now, if the book had a mass of equal magnitude to the earth, the equal masses would demonstrate equal (in magnitude) and opposite acceleration.

## @ErikE 2011-02-19 04:53:25

What do you mean, "seems to accelerate?" it does, in fact. I doubt I can explain it at this point. Do you really think that if you had a moon-massed object the size of a basketball that it would only close with the Earth at 9.8 m/s/s?

## @HAL 2011-02-19 05:00:37

the acceleration of the book differs from the acceleration of the earth towards the book. seemingly just simply notes the availability of observation. I dont know if you are saying that the force law of gravity changes with a given mass. If you are invalidating Newton's law then perhaps.

## @ErikE 2011-05-11 18:30:56

This i about BOTH the object accelerating and moving independently. No invalidating of Newton's law (chuckle).

## @Jerry Schirmer 2011-01-21 20:36:54

Drop a 5 lb iron barbell and a 25 lb iron barbell simultaneously. The'll hit the ground simultaneously. The only possible caveat is the recoil effect Ted Bunn brings up above.

## @ErikE 2011-01-21 20:52:14

It seems you didn't read carefully. Please see section "Practically Speaking". I'd appreciate an answer that is more on topic and responds to the points I made in my question.

## @Jerry Schirmer 2011-01-21 22:12:58

@Emtucifor: At this point, why not fac tor in the gravitational radiation of the bodies? Or the one or two electron ionization of the falling masses, or a million other effects? If the effect you're talking about is completely unmeasurable, is it an effect at all? And I did, in fact, mention the recoil effect, which is going to be the leading order correction to the equivalence principle, anyway.

## @ErikE 2011-01-21 22:34:16

Now that I shortened my answer, I should clarify that in my post I made it clear that I know normal observation would seem to indicate all objects fall at the same rate, but that wasn't what I was interested in. Now to Jerry's last comment: Ted made no mention of any recoil effect. Could you tell me more about that? Also, I'm quite interested in any other factors that could affect falling. Would you care to elaborate on gravitational radiation, electron ionization, and so on?

## @user346 2011-01-22 03:45:59

@Emtucifor I get the feeling that you're seeking replies more for attention than any other reason. You mention you have a 20 year old high school education. That is fine. But one needs to recognize what their limitations are or they just end up sounding silly.

## @ErikE 2011-01-22 10:23:49

@space Your feeling is misplaced. And I'll thank you kindly to not discuss my supposed mental limitations. It's all very well for you to claim I am misunderstanding, but please give me the benefit of the doubt and post an answer to help disabuse me of my mistaken notions. Otherwise you're just here to insult and try to feel good about yourself. I dare you: set me (and us all) straight with superior understanding, in a proper answer. Please.

## @Joe Fitzsimons 2011-01-22 18:05:25

@Jerry: This isn't true in the earth's atmosphere, which is why people find it so counter intuitive in the case of a vacuum (a caveat which you should probably add).

## @Fattie 2014-06-08 10:03:04

This answer / comments is utterly bizarre from a moderator with 15,000 points. Erik, I would encourage you to simply not respond further.

## @Jerry Schirmer 2014-06-08 15:44:13

@JoeBlow: your comment is the first activity on this thread in three years. The original question was edited with further commentary. Read my other answers if you think I don't know physics.