According to Newton's third law of motion that states that every action has an equal and opposite reaction.

So, if the Earth exerts a gravitational pull on us (people) then even we should exert a force equal and opposite (in terms of direction) on the Earth.

It is intuitive to think that this force is really small to get the Earth to move. But, if we take a look at the second law of motion that states that F = ma we see that however small the force, there will be some amount of acceleration. Therefore, even though we exert a very small gravitational force on the Earth it should be enough to get the Earth to move even though the acceleration is a very small amount.

But what I said clearly does not happen. So there must be some flaw with my reasoning. What is that flaw?

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    $\begingroup$ What do you mean that it clearly does not happen? If the acceleration is exceptionally tiny (because the mass is exceptionally large), then maybe you just don't notice? $\endgroup$
    – Steeven
    Commented Jan 15, 2019 at 8:39
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    $\begingroup$ You say "this thing should happen, but it clearly does not". You should get out of that habit as a scientist. Better: "this thing should happen, I will compute the expected effect and either design an experiment that confirms or denies the expectation, or explain why such an experiment is infeasible given my budget". $\endgroup$ Commented Jan 15, 2019 at 15:24
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    $\begingroup$ @EricLippert From now I will try to be a better scientists by not making assumptions that are not supported by hard evidence $\endgroup$
    – whae
    Commented Jan 15, 2019 at 15:39
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    $\begingroup$ "or explain why such an experiment is infeasible given my budget" I've heard funding of science is being cut all the time, but I hope we can still afford to jump! $\endgroup$
    – JiK
    Commented Jan 15, 2019 at 15:46
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    $\begingroup$ That's great. While you are thinking about this problem: unlike you, the moon is hanging over the earth, and unlike you, the moon does have enough mass to seriously pull on something the mass of the earth. Is the earth observed to accelerate towards the moon? $\endgroup$ Commented Jan 15, 2019 at 15:46

8 Answers 8


The acceleration that your gravitational pull causes in the Earth is tiny, tiny, tiny because the Earth's mass is enormous. If your mass is, say, $70\;\mathrm{kg}$, then you cause an acceleration of $a\approx 1.1\times 10^{-22}\;\mathrm{m/s^2}$.

A tiny, tiny, tiny acceleration does not necessarily mean a tiny, tiny, tiny speed, since, as you mention in comments, the velocity accumulates. True. It doesn't necessarily mean that - but in this case it does. The speed gained after 1 year at this acceleration is only $v\approx 3.6×10^{-15}\;\mathrm{m/s}$. And after a lifetime of 100 years it is still only around $v\approx 3.6×10^{-13}\;\mathrm{m/s}$.

If all 7.6 billion people on the planet suddenly lifted off of Earth and stayed hanging in the air on the same side of the planet for 100 years, the planet would reach no more than $v\approx 2.8\times 10^{-3}\;\mathrm{m/s}$; that is around 3 millimeters-per-second in this obscure scenario of 100 years and billions of people's masses.

Now, with all that being said, note that I had to assume that all those people are not just standing on the ground - they must be levitating above the ground.

Because, while levitating (i.e. during free-fall), they only exert the gravitational force $F_g$:

$$\sum F=ma\quad\Leftrightarrow\quad F_g=ma,$$

causing a net acceleration according to Newton's 2nd law. If they were standing on the ground, on the other hand, they apart from their gravitational force also exert a downwards pushing force equal to their weight, $w$:

$$\sum F=ma\quad\Leftrightarrow\quad F_g-w=ma.$$

Then there are two forces exerted on the planet, pushing in opposite directions. And in fact, the weight exactly equals the gravitational force (because those two correspond directly to an action-reaction pair via Newton's 3rd law). So the pressing force exerted on the planet cancels out the gravitational pull in the planet. Then the above formula results in zero acceleration. The forces cancel out and nothing accelerates any further.

In general, no system can ever accelerate solely via it's own internal forces. If we consider the Earth-and-people as one system, then their gravitational forces on each other are internal. Each part of the system may move individually - the Earth can move towards the people and the free-falling people can move towards the Earth. But the system as a whole - defined by the centre-of-mass of the system - will not move anywhere.

So, the Earth can move a tiny, tiny, tiny bit towards you while you move a much larger distance towards the Earth during your free-fall so the combined centre-of-mass is still stationary. But when standing on the ground, nothing can move because that would require you to break through the ground and move inwards into the Earth. If the Earth was moving but you weren't, then the centre of mass would be moving (accelerating) and that is simply impossible both because of the momentum conservation law as well as the energy conservation law. The system would be gaining kinetic energy without any external energy input; creating free energy out of thin air is just not possible. So this never happens.

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    $\begingroup$ Alright so if I am standing on the Earth's surface then I get squeezed to the Earth's surface and exert a pushing force on the Earth's surface that cancels out the pulling gravitational force that I exert on the Earth. If I somehow manage to levitate then the velocity will take a really really really long time to build up for us to notice. $\endgroup$
    – whae
    Commented Jan 15, 2019 at 9:40
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    $\begingroup$ @AdityaBharadwaj Exactly. $\endgroup$
    – Steeven
    Commented Jan 15, 2019 at 9:43
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    $\begingroup$ @Aditya note also that, unless you're at the North or South poles, if you were levitating over a constant spot of the ground, then the acceleration you'd cause on the Earth would be rotating once a day, and the bulk of the contributions would cancel, as the velocity accumulated during the day would point opposite to that accumulated at night. Only the North-South component, proportional to the sine of your latitude, would survive. $\endgroup$ Commented Jan 15, 2019 at 10:31
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    $\begingroup$ Reminds me on what-if.xkcd.com/8 $\endgroup$
    – Holger
    Commented Jan 15, 2019 at 10:57
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    $\begingroup$ @AdityaBharadwaj Also consider that if you jump up from the surface of the planet you are first pushing Earth away from you. As you fall back to earth you return to each other and end up right back where you started. To effect any real delta-V you have to have some mass (or at least momentum) leaving the system. So long as we remain at the surface we can't leave the system and if we can't leave the system then we can't effect an outside acceleration on it. A spaceship to the moon, though, would effect a net acceleration as it leaves the earth completely and less of it returns than what left. $\endgroup$
    – J...
    Commented Jan 15, 2019 at 12:16

Short answer, it does. It's just too small for you to notice. So, why isn't it noticeable?

The earth doesn't noticeably move towards you despite your gravitational pull because you're resting on it, the same way you don't move towards the center of the earth if you have stable footing on its surface.

If you're not standing on it (e.g. skydiving, or in orbit) then the earth does start to move towards you (or, if you jump up, move in the opposite direction of your body). But since the earth's acceleration imparted by you is so very small, although it's calculable, it's basically immeasurable. As @Steeven calculated, a 70Kg person would impart an acceleration of $a\approx 1.1\times 10^{-22}\;\mathrm{m/s^2}$ which isn't noticeable with our human perception.

Adding to the very small levels of acceleration, one more reason you don't notice earth's movement is that many, many things happen on earth. It's optimistic/unrealistic to model the earth as a perfectly rigid body, but let's do it for the sake of this example: there's so much stuff happening all around earth, each imparting a small acceleration on the whole body (cranes lifting things, mines being excavated with explosives, landslides and avalanches, earthquakes etc.) that even if we were able to measure the tiny acceleration that you exert on earth it would be near impossible to isolate the impact that you have from all the noise created by everything else.

  • $\begingroup$ what if everyone on earth somehow accumulate on one region on the Earth's surface and jump together.During their jump , would the net force of all the people be enough to cause a noticeable acceleration of the Earth? $\endgroup$
    – whae
    Commented Jan 15, 2019 at 15:03
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    $\begingroup$ @AdityaBharadwaj there's a Vsauce video on youtube answering the very question! youtube.com/watch?v=jHbyQ_AQP8c if you don't know the chanel I encourage you to check it out. $\endgroup$
    – Aubreal
    Commented Jan 15, 2019 at 15:05
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    $\begingroup$ @AdityaBharadwaj there's 7.53B people on earth, with average mass of ~65.5Kg, meaning the total mass of all humans is 4.93x10^11Kg. Earth's mass is 5.97x10^24, several orders of magnitude mroe than us. If we assume everyone jumps 40cm up at the same time, the earth would move down 3.3x10^-14m. As we fall down, the earth would also move towards us and the total displacement (after all that) will be 0. But yes, the earth would move a tiny bit, but not a significant amount due to just how much heavier earth is than humans. $\endgroup$
    – Aubreal
    Commented Jan 15, 2019 at 15:13
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    $\begingroup$ @AdityaBharadwaj There's also an XKCD on it: what-if.xkcd.com/8 Short answer: almost everyone dies of starvation. $\endgroup$ Commented Jan 15, 2019 at 18:29

From Newton's third law, we know one thing: every action has an equal and opposite reaction. This means that the force we act on earth is equal to the force that the earth acts on us.

This means $$ f = ma \rlap{~~~~ \left( \text{by person} \right)} $$

The average mass of a person is $70 \, \mathrm{kg}$ and acceleration due to gravity is nearly $10 \, \mathrm{m}/\mathrm{s}^2 .$

So the force we apply on earth is nearly $700 \, \mathrm{N} .$

Now $$ F = MA \rlap{~~~~\left( \text{by Earth} \right)} $$ The approximate mass of Earth is $6 \times {10}^{24} \, \mathrm{kg} ,$ but the force remains $700 \, \mathrm{N} .$

Now $A = F/M$ $$ A = \frac{700}{6 \times {10}^{24}} ~~ \Rightarrow ~~ \sim 116 \times {10}^{-24} \, \frac{\mathrm{m}}{\mathrm{s}^2} \,.$$

That acceleration applied by one person is so minuscule that it does not need to be considered.

The acceleration of Earth is in negative powers of $24 .$ So we need more that just billions of people to accelerate the earth upwards.


Why doesn't the Earth accelerate towards us?

It does, but as noted elsewhere the acceleration is negligibly small.

Consider a person of 98 kilograms who jumps high enough to remain in the air for one second (Michael Jordan, for example). At an acceleration of $9.8\;\mathrm{m/s^2}$, that's a height of $1.225\;\mathrm{m}$, consistent with reports that Michael Jordan had a four-foot vertical leap (four feet is $121.92\;\mathrm{cm}$).

Multiply the jumper's displacement by the ratio of masses to get the displacement of the earth (of course, this is an approximation: more precisely, the total displacement is the sum of two quantities that are inversely proportional to the masses, but since the masses are so different in magnitude this approximation yields the same result):

$$1.2\;\mathrm{m}\times{\frac{9.8\times 10^1\;\mathrm{kg}}{6.0\times 10^{24}\;\mathrm{kg}}}\approx2.0\times10^{-23}\;\mathrm{m}$$

By contrast, the charge radius of a proton appears to be at least $8.4\times10^{-16}\;\mathrm{m}$, which means that the displacement of the earth during a Michael Jordan jump shot is less than one 42 millionth of the charge radius of a proton.


Why doesn't the Earth accelerate towards us?

It does accelerate towards us, as it also accelerates towards the moon. The flaw in you reasoning in the first instance is that we are standing upon the earth so that acceleration is cancels out; and in the second, the earth and moon are rotating around each other, which although a form of acceleration, is not linearly directed and so doesn't look like the usual, everyday notion of what accelerate means.


I will go against the grain and play devils advocate since I did not see this mentioned in any of the answers that I read. If the Earth is moving towards us per the law then that means its moving towards all people, animals, and anything with mass. Since it is a sphere calculations have to include cancelling out the pull in x direction by the same linear pull from the converse of a similarly massed object on the other side of the Earth. As this would then mean that the Earth is expanding at a rate consistent with the gravitational pull of the objects with mass in every direction all the time. Thus, the Earth is not moving toward "us" because "us" was used as a singularity (me) rather than us as an inclusive to all objects exerting or absorbing force. Not to mention that the Earth also rotates as a sphere with an object proven to exert gravitational force, that is the moon and it's effect on tides, that would pull "us" away from the Earth on the side of the Earth that is closest to the moon. Expanding on that idea would be cosmic forces exerting a cancelling out effect by other astral bodies, the Sun for example.


Since he/she asked why we couldn't observe his results, I understand that he/she thought his model was realistic.

On one side, modeling the earth as a rigid body with one person jumping on them is inacurate. We should take in count many other things. So, the acceleration of what you'd like to call strictly earth (not the earth+humans+animals+natural phenomena system) you would compute would be, averaged in a period of time, equal to $0$ (I would think that mass (humans, animals, air, etc) above earth's surface is distributed homogenly).

That should answer your question.

On the other hand, the center of mass of the system will have an acceleration due to external forces (gravitational forces exerced by the Sun, Moon and other planets). Internal forces such as gravitational interaction between humans and earth wouldn't change the systems center of mass. And, since the earth is considerably more massive than us and we can't be too far from the surface, the center of mass of the system would be the same as the earth's center of mass. (This is equivalent to the small acerleration argument).


Do any of the posted answers take account of the planet's inertia?

I'm certainly NOT about to be the first poster to say that the Earth does accelerate towards us. Because plenty of the previous posters have said exactly that, while mentioning that in practice the effect would be too small to notice or to measure.

I am going to complain about the answer that the Moon can accelerate toward the Earth, because it is fairly widely known that the Moon is in orbit, and as frame-dragging effects caused by the Earth's gravitational field accelerate any orbiting object, that effect causes the Moon to move away from the Earth at a continuously accelerating rate, which I have a memory from childhood of being a rate of approx one inch per century (or metric equivalent).

If it were in a retrograde orbit, it would at least be capable of decreasing its distance from the Earth over time, since then the frame-dragging effect would be decelerating it, instead of accelerating it.

My actual answer is more down-to-earth -- :)

Since the Earth has a large mass, and since one fairly well established property of mass is inertia, I'm willing to go half-way, and say that the Earth doesn't accelerate toward us: because it doesn't move at all. If I perform the jumping experiment (with a mass of 180 lbs x 1), or even if I get all the men in China to (180 lbs x 1 billion), the Earth is held in place in spacetime by the inertia associated with its mass.

It's approximately equivalent to throwing a tennis ball at an approaching freight train and expecting to derail it, or to halt or delay the train, even temporarily: mathematically, a calculation might be done that demonstrates there is a calculable effect (as some on here have done); but such calculations tend to ignore the inertial component (quite large, for a body of planetary mass).

If I had some significant fraction of the mass of the Moon, and then jumped, I might reasonably expect that a measurable effect would result. But the Earth is massive enough for its position to remain unaffected below a limit determined by Kepler's laws of planetary motion.

Again, I have a memory about Newtonian conservation of momentum, but which I suspect won't apply within a closed system: me plus the Earth.

But in relation to does the Earth accelerate towards the Moon, well the answer is it does! Well, it does in part, at least. It's called the tide, and at any point on the Earth's equator that has open sea, you'll experience high tide once a day when the Moon is more or less overhead.

This is due to planetary inertia! If the Earth had no inertial component to its mass, when the tidewater moved 4 ft closer to the Moon at local noon, so would the Earth: in that case you would not notice a change in the tidal level, because both ocean and seashore would have moved by an equivalent amount.

The fact that you do notice the tide rising and falling each day is a proof of the existence of inertia (the Earth has it, so the seashore has it), but the sea has very much less of it.


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