When an object is experiencing free fall, it has a constant acceleration and hence an increasing velocity (neglecting friction). Thus its momentum is increasing. But according to law of conservation of momentum, shouldn't there be a corresponding decrease in momentum somewhere else ? Where is it ?

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    $\begingroup$ Consider this in the light of Newton's 3rd law of motion. $\endgroup$
    – DanDan面
    Aug 25, 2019 at 6:48
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    $\begingroup$ Hint: momentum is conserved for closed systems. So you need to consider the change in momentum of the earth due to the force exerted on it by the object as well as the force the ball exerts on the Earth. A quick google will give you a specific numerical example. $\endgroup$ Aug 25, 2019 at 7:09
  • $\begingroup$ A hammer does fall faster on the moon than a feather: Because it attracts the moon more which then moves towards it. $\endgroup$ Aug 26, 2019 at 9:45
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    $\begingroup$ I'm bothered by all the "clever" indirect answers. Short answer is: The Earth itself accelerates (ever so slightly) toward the object at the same time. The opposite-direction velocity of the Earth will be exactly proportional to the ratio of the object's mass to Earth's mass, thus exactly balancing ("conserving") momentum. $\endgroup$
    – Jeff Y
    Aug 26, 2019 at 15:30
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    $\begingroup$ @JeffY Don't be bothered. The clever indirect answers are not answers at all, but comments. They are hints, which is a standard way that we encourage the OP to rethink the problem. If they were intended to be answers, they would be posted as "Answers". By the way, your comment would be better posted as an answer (although a short answer without detail is arguably appropriate as a comment. $\endgroup$
    – garyp
    Aug 26, 2019 at 16:41

5 Answers 5


Linear Momentum is conserved only in systems with net external force equal to zero. For a body falling on Earth, it experiences Earth's gravitational force so its linear Momentum increases. But if you include Earth in your system then definitely, momentum is conserved, as an equal amount of momentum of Earth is increased in upward direction. But individually for both it's not conserved, there is an external force of gravity on each.

  • $\begingroup$ If you consider only particular states (rather than continuous evaluation) you can generalize further by loosening the restriction to "zero net external impulse". But at that point we are deep into the weeds. $\endgroup$ Aug 26, 2019 at 18:51
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    $\begingroup$ Momentum is always conserved! It might not always be constant in the system, but conservation laws must consider a current or flow/flux. In the case of momentum, the current is the impulse. $\vec{p}_{after}=\vec{p}_{before}+\int \vec{F}_{ext}~dt$. People get confused when they fail to distinguish between constancy and conservation. Noether's theorem tells us momentum is always conserved, along with energy, angular momentum, and charge: en.wikipedia.org/wiki/Noether%27s_theorem $\endgroup$
    – Bill N
    Aug 26, 2019 at 19:59

Linear momentum of a system remains conserved unless an external force acts on it. Since during free fall, a gravitational force acts on the body, it's momentum will not remain conserved. However, if we change the reference in such a manner that the gravitational force becomes an internal force of the system, i.e. regard both the body and Earth together as a system, and consider this system to be isolated in the universe, with no other body present near the system, we can now apply the law of conservation of linear momentum as there are no external forces acting on the system now.


Sciencisco's is the best, but I thought I would add one thought: the external potential $V = mgy$ does not exhibit translational symmetry in the $y$ direction. Noether's theorem says that each symmetry gives a conservation law. Furthermore, if you don't have a symmetry, then you don't have the associated conservation law. Translational symmetry gives us conservation of momentum. Because this potential is not translationally invariant in the $y$ direction, momentum is not conserved in that direction.

  • $\begingroup$ This is the right answer IMO. The fundamental law is Noether's theorem, not conservation of linear momentum, and here the Lagrangian has only two directions of translational symmetry. $\endgroup$
    – user2617
    Aug 26, 2019 at 19:07

Momentum is not conserved for you alone, because an external force acts on you. But if you consider both the earth and you. Then since $F_{ext}= 0 $, net momentum change is definitely zero.

Let's say you are starting from rest. Now let force by earth on you be $F$. So $a=\frac{F}{m_{you}}$ and your velocity after time $t$ is $v= \frac{Ft}{m_{you}}$ Now your momentum is $m_{you}v= Ft$ .

Similarly $F$ on earth by you is $-F$. {Negative as the direction is opposite}.So $A_{earth}=-\frac{F}{M_{earth}}$ and earth's velocity after time $t$ is $V= -\frac{Ft}{M_{earth}}$ Now earth's momentum is -$M_{earth}V= -Ft$. Thus net $\Delta P = Ft-Ft=0$.


Neither energy nor momentum is individually conserved during free-fall.

During free-fall the ratio of energy to momentum changes - full stop.

Kinetic Energy is Momentum in disguise.

The idea of GPE being converted to KE is born out of ignorance and double accounting (somehow thinking KE is more than just a different representation of momentum).

When a mass is in free-fall

$dE/ds = d\rho/dt$

There is nothing more needed, apart from some scaling.

Did Einstein's famous mass-energy equation teach you guys nothing about the relationships between energy, momentum, time, distance and light speed?


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