# Don't heavier objects actually fall faster because they exert their own gravity?

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 of which 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:

$F_g = G m_1 m_2/d^2$

We can see that the force of gravity is dependent on BOTH the masses, not just that of the more massive object.

Of course in everyday situations, we can for all practical purposes treat objects as falling at the same speed. 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 Earth. 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 falling is multiple objects mutually attracting in space.

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Now that's one piece of trivia to blast unsuspecting nerds :D – SF. Dec 8 '12 at 21:17

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The classic experiment, as performed by Galileo and David Scott, involves dropping two objects simultaneously. In that case, the pull of the larger mass also affects the falling time of the smaller mass, as the Earth falls up and meets both objects slightly sooner. With <handwave>appropriate assumptions</handwave>, do they still hit simultaneously, or is there some tiny third-order effect I'm not allowing for? – Keith Thompson Dec 8 '12 at 21:45
@Keith I haven't been considering the case of dropping both dissimilar-mass objects simultaneously--the question is considering the two objects separately. The accepted answer says that the difference in the times between two objects could be on the order of 1 part in a trillion trillion. So there won't be much to measure, and according to human perception & instrumentation we would always call it "simultaneous." Dropping the items together would take even less time. But one way to eliminate the complexity of the number of items is just to consider each atom individually. – ErikE Dec 8 '12 at 21:59
I removed the CW status on your question, but keep in mind that this is something we try to do quite sparingly. Just try to be careful with your edits on future questions :-) – David Zaslavsky♦ Dec 9 '12 at 2:02
– ErikE Dec 9 '12 at 3:09
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## 9 Answers

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.

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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 $x_{CM}=(M x_M + m_i x_i)/(M+m_i)$ with an acceleration $GM/(x_i-x_M)^2$. Note that $\ddot x_{CM}=0$

Really we have only hidden the paradox, because of course $x_{CM}$ 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 $x_i-X_{CM}$. The solution is to redefine the mass. As $x_i-x_{CM}= M (x_i - x_M) /(M+m_i)$, we can say that the object $i$ falls into the Mass Center with an acceleration $G{M^3 \over (M+m_ i)^2}{1 \over (x_i-x_{CM})^2}$ 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 $$\ddot x_0={1 \over m_1+m_2} (m_1 {G M \over (x_1-x_M)^2} + m_2 {G M \over (x_2-x_M)^2})={F_1+F_2 \over m_1 + m_2}$$

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 $g=9.8m/s^2$), 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".

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What is "CM reference frame" please? – ErikE Jan 21 '11 at 21:25
Center of Mass reference frame – Greg P Jan 21 '11 at 21:28
In any case, I feel that my answer is not honest enough... but I am in hurry, sorry. Come back later tomorrow. – arivero Jan 21 '11 at 21:45
@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 Jan 22 '11 at 3:22
@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! – arivero Jan 22 '11 at 3:23
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The new version of the question is much clearer than before, and I removed my down vote accordingly.

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.

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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 $F=GmM/r^2=m*a=m*d^2r/dt^2$.

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

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Hmmm... where did M go in your second formula? – ErikE Jan 21 '11 at 21:02

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.

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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 Feb 19 '11 at 1:19
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. – ErikE Feb 19 '11 at 1:21
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. – HAL Feb 19 '11 at 4:41
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? – ErikE Feb 19 '11 at 4:53
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. – HAL Feb 19 '11 at 5:00
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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.

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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. – ErikE Jan 21 '11 at 20:52

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@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. – Jerry Schirmer Jan 21 '11 at 22:12
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? – ErikE Jan 21 '11 at 22:34
@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. – user346 Jan 22 '11 at 3:45
@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. – ErikE Jan 22 '11 at 10:23
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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 sum of 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 m1 = m2, or m1 >> m2.

When a body is picked up to a certain height and then dropped, the time to fall to the Earth does not depend 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 external body is brought to a certain height above the Earth and then dropped, the free fall time does depend 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.

"Most bodies fall at the same rate on earth, relative to the earth, because the earth's mass M is extremely large compared with the mass m of most falling bodies. The body and the earth each fall toward their common center of mass, which for most cases is approximately the same as relative to the earth. In principle, the results of a free fall experiment depend on whether falling masses originate on earth, are extraterrestrial, are sequential or concurrent, or are simultaneous for coincident or separated bodies, etc. When falling bodies originate from the earth, all bodies fall at the same rate relative to the earth because the sum m + M remains constant.
-- ArXiv:Deterrents to a Theory of Quantum Gravity

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.)

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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. – ErikE Dec 8 '12 at 18:12
@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). – Nick Dec 8 '12 at 20:45
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 when the 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, they both accelerate towards each other. The mass of #2 affects the time until impact. – ErikE Dec 8 '12 at 21:09

I always thought the reason a penny falls faster than a feather is air resistance. There is real physics involved, but it's aerodynamics, not tides or gravity.

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It seems you didn't read carefully. Please see section "Guidelines", point 5. I'd appreciate an answer that is more on topic and responds to the points I actually raised in my question. – ErikE Jan 21 '11 at 20:54
Now that I shortened my answer, I should clarify that in my original post I asked answerers to completely ignore wind resistance. – ErikE Jan 21 '11 at 22:35
F=ma F = GMm/r^2 a = F/m = GM/r^2 Hence to first approximation  a  one kilogram weight falls down at exactly the same speed as a two kilogram weight. However, the earth "falls up" twice as fast for the two kilogram weight,  and would "fall up” one hundred times as fast for a one hundred kilogram weight.  Of course if the one kilogram effect is of order one part in 10^24 as suggested byanother poster,  the one hundred kilogram effect would still be only one part in 10^22 – Jim Graber Jan 27 '11 at 11:31

With gravity and wind resistance as factors, you may already know that an object dropped from a point of rest goes 9 meters per second per second. This differs in a vacuum environment such as the moon.

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The $9\,\mathrm{m}\,\mathrm{s}^2$ acceleration is due to gravity alone, not air resistance. Moreover, the point of the question is that the fact that this acceleration is constant is only approximation because the Earth will also move a bit due to gravity from the falling object. – Ondřej Černotík Feb 4 at 19:19
I said "setting air resistance aside"... – ErikE Feb 4 at 23:14

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