<p>About 70,000 km. If the Earth orbited Jupiter (or flew by) at a closer distance, not just we would leave the surface, but the whole Earth would disintegrate since all its mass will be leaving too.</p>

<p>70,000 km is Jupiter's <a href="https://en.wikipedia.org/wiki/Roche_limit" rel="noreferrer">Roche limit</a> (although its actual value varies a lot depending on the other involved body), the radius where tidal forces (already explained in uhoh's answer) overwhelm gravitational forces and any orbiting body can't hold itself by its own weight. In that context, people on the surface don't behave differently than rocks.</p>

<p>Btw., this scenario is also explored <a href="https://www.youtube.com/watch?v=ym7bDBg9jQ4" rel="noreferrer">in a youtube video</a>. I wouldn't say it's very good but it may be helpful to explain the Roche limit to a 7 year old.</p>

CC BY-SA 4.0
2019-05-27T09:52:53.923Z
I didn't even think to mention the Roche limit; very good point! `+1`
uhoh
CC BY-SA 4.0
2019-05-27T11:03:07.040Z
Note that 70,000 km also happens to be Jupiter's radius, so the planets would need to be touching. (And there's not a "the" Roche limit; it depends on the density of the secondary object, that is, Earth in this example).
hmakholm left over Monica
CC BY-SA 4.0
2019-05-27T11:14:12.443Z
@HenningMakholm - Starting the answer with "about" was intended to deal with variation of Roche limit due to a lot of factors. However I edited the answer to make it more clear.
Pere
CC BY-SA 4.0
2019-05-27T13:44:33.870Z
People aren't held together by gravitational forces, so being inside the Roche limit won't pull them apart like it would for a planet. In that regard, people on the surface *do* behave differently than rocks - a pile of gravel will separate into individual pebbles as it gets sucked into Jupiter's gravitational well, but a person will remain intact.
Nuclear Hoagie
CC BY-SA 4.0
2019-05-27T13:59:25.940Z
@NuclearWang But... people *are* held to the earth purely by gravitational forces.  Being within the roche limit would not pull humans apart but we would still act like those pebbles in the sense that individual humans would not be stuck to the earth by gravity any longer.
J...
CC BY-SA 4.0
2019-05-27T21:13:54.583Z
To this layman it seems obvious that Jupiter's gravity would make life unpleasant (if not impossible) on Earth at a much greater radius than 70Mm.
EvilSnack
CC BY-SA 4.0
2019-05-28T04:10:19.713Z
At *just* inside the Roche limit, the math in @uhoh's answer shows that loose objects aren't literally ripped off the surface.  I think the mechanism is more gradual even for a totally non-rigid aggregate of gravel: with its own gravity being unopposed in the other directions, it would elongate in the tidal-force direction.  This puts the ends farther and farther from the centre of mass, and increases the distance for the gravity gradient. This *eventually* leads to it being torn apart, but one quick pass wouldn't rip loose objects off the surface (esp for a stiff / viscous object like Earth)
Peter Cordes
CC BY-SA 4.0
2019-05-28T04:18:26.067Z
The wikipedia article shows the elongation as part of the torn-apart effect right at the Roche limit.  See also video animations like https://www.youtube.com/watch?v=tGS_gZm6HxI  and https://www.youtube.com/watch?v=6e5yyIcq40o.  And remember, it's the radius where a rigid or fluid body can't exist *long term* in a stable circular orbit.  An elliptical orbit that just dips in at perigee could conceivably let an aggregate body re-form itself when farther out.  **See [How fast does Roche limit disintegration proceed?](//space.stackexchange.com/q/10113) on space.SE**
Peter Cordes
CC BY-SA 4.0
2019-05-28T04:35:59.000Z
@HenningMakholm: I think this answer accidentally took Jupiter's actual radius as its Roche limit!, because the [Wikipedia page has a table of density and radius](https://en.wikipedia.org/wiki/Roche_limit#Selected_examples) for objects in our solar system at the top of the section for Roche limits for pairs of bodies.  The actual rigid-body Roche limit (where objects are pulled off the surface by tidal forces) is $R_m * (2  \rho_M / \rho_m) ^ {1/3}$ = `71493000 * (2 * 1326/5513)^(1/3)` = **56 018 km  using the "fully rigid-satellite" formula from Wikipedia.**
Peter Cordes
CC BY-SA 4.0
2019-05-28T04:49:47.513Z
Or the "more accurate" formula for a tidally locked body (including the body's rotation) gives us 62 709 km.  But Earth isn't rotating that fast.  (The orbital period for [a circular orbit just skimming the surface](https://en.wikipedia.org/wiki/Orbital_period#Orbital_period_as_a_function_of_central_body's_density) of Jupiter is ~2.86 hours, which is a lot faster than 24 hours, and why this Roche limit formula predicts a body can be torn apart so much farther away than if it's not rotating.
Peter Cordes
CC BY-SA 4.0
2019-05-28T04:54:50.537Z
BTW, in my first comment about being deformed before coming apart, I was referring to the "fluid" Roche limit.  Once I looked up the wiki article, I realized that the "rigid" Roche limit is exactly what we want here.  That's the assumption of rigid but no tensile strength, which works perfectly for loose objects pulled off the surface.  For the integrity of Earth itself: it's squishy long-term (molten rock), and in a close orbit would deform like for the "fluid" Roche limit (which has a much larger radius).  Tidal drag speedup up our rotation = catastrophic tsunamis!
Peter Cordes
CC BY-SA 4.0
2019-05-28T08:54:35.870Z
The actual Roche limit for Earth over Jupiter would be way larger, because such a big body like Earth wouldn't behave like a rigid body.
Pere
CC BY-SA 4.0
2019-05-28T21:48:04.380Z
@mistertribs - Interestingly, that article states that "If the planet passes within the Roche limit at periastron, however, mass can be stripped from it". Likely, that mass would include some people standing on the planet.
Pere
CC BY-SA 4.0
2019-05-29T14:42:11.567Z
"The Roche Limit" does not exist for a single body. One could define "A Roche Limit" for a Jupiter-Earth system, but not for a Jupiter. And it would be different for a Jupiter-Mercury system etc.
David Tonhofer
CC BY-SA 4.0
2020-04-20T08:21:09.523Z
That's the reason the answer starts with "about".
Pere
CC BY-SA 4.0
2019-05-27T09:16:05.037Z
2019-05-27T12:53:06.653Z
2019-05-27T12:53:06.653Z
Pere
<p>As other answers point out, Jupiter is not quite massive enough to pull a planet of Earth's density apart. But we can use a slightly heavier object instead -- say, a small cold brown dwarf massing 13 Jupiters, or about 4000 Earths. According to the rigid-body Roche formula, its Roche limit is then <span class="math-container">$\sqrt[3]{2\cdot 4000}=20$</span> earth radii, or 130,000 km. The radius of the brown dwarf is not much larger than Jupiter's (since they're both made of compressible gas), so smaller than 100,000 km, and there's some room for Earth to be destroyed without actually colliding with it.</p>

<p>Our brown dwarf, wandering sedentarily around in the galaxy, spots our sun, and decides to take a closer look. It comes screaming through the inner solar system on a hyperbolic orbit, which means that it will be moving at somewhat more than solar escape velocity when it narrowly misses Earth -- call it 100 km/s. Switching perspective, we can say that <em>Earth</em> comes at the brown dwarf at 100 km/s and just about misses. This speed allows us to spend a almost half an hour inside the Roche limit, if we almost touch the brown dwarf at the time of closest approach.</p>

<p>But that sounds like it will be unnecessarily dramatic, so it's instead have the brown dwarf give is a slightly wider berth such that at the time of closest approach, the acceleration of gravity at ground zero will be a more pedestrian <span class="math-container">$-0.1\;\rm m/s^2$</span>. That will be the case when our distance to the brown dwarf is <span class="math-container">$\sqrt[3]{\frac{9.82}{0.1+9.82}} = 0.997$</span> of the Roche limit, or 128,000 km. The length of our path through the Roche zone is then about 20,000 km, which means the encounter takes 200 seconds. Call it three minutes.</p>

<p>(The sharp-eyed reader will notice that these numbers mean that the far side of the earth is actually never inside the 130,000-km limit, but what really counts is the first derivative of the brown dwarf's gravitational field, so if you're standing on the antipodal point you'll still have the Earth's center-of-mass pulled away under you even if you yourself is outside the limit. The numbers are all approximate anyway).</p>

<p>(On the other hand, a few minutes clearly not enough time for the molten inside of the Earth to flow into a hydrostatic equilibrium in the new situation, so using the rigid-body formula is appropriate).</p>

<p><strong>What happens then?</strong></p>

<p>First, of course, it is an <strong>awesome sight</strong>. The brown dwarf dominates the sky with an angular diameter of somewhere between 60° and 100°.</p>

<p>Then, it may be getting <strong>uncomfortably hot</strong>. Not necessarily "the mountains are melting" hot or even "the seas boil away" hot. But <a href="https://www.space.com/22659-brown-dwarf-stars-hot-as-oven.html" rel="noreferrer">this</a> says that the coldest brown dwarves have about the temperature of a baking oven, and having a significant part of the sky at 150 °C can make anyone sweat. No worries, though -- it will all be over in a few hours, so just go inside and crank up the AC; that'll deal with it fine.</p>

<p>Right at ground zero <strong>gravity decreases smoothly</strong> while we approach Roche. When it passes zero G you're in free fall and start floating gently upwards. Except that everything around you -- cars, houses, trees, the soil itself -- is also in free fall since the only thing that kept them down was gravity. So to a first approximation your <em>local</em> experience is not about bring ripped off Earth, but just of weightlessness. (<em>Or is it?</em> See below.)</p>

<p>Ditto at the antipodal point.</p>

<p>One problem that shows up here is that <strong>the atmosphere is escaping</strong> into space. Since there is no gravity to keep it down, it escapes rather faster than the gentle floating of cars, trees, and people, propelled by its own pressure. Even before we reach Roche, the air may have become too thin to breathe. On the other hand fresh air will rush in from the surrounding areas to fill the void, creating the great-great-grandmother of all hurricanes. (And a great-great-grandfather around the antipode, of course).</p>

<p>On a great circle 90° from ground zero, gravity increases to about 1.7 G. You feel heavy. Ho hum.</p>

<p><em>Between</em> these areas dramatic things happen. At about 45° from ground zero (or the antipode) the tidal force is at right angles to vertical, so the <em>strength</em> of gravity is about what we're used to -- but its direction is different. It's as if <strong>the world is tilted by tens of degrees</strong>, rather like how bad sci-fi movies pretend "entering a gravitational field" works. Tall buildings tip over; many not-so-tall ones just collapse. Lakes and seas do things that make the word "tsunami" pack up and go home, hopelessly outclassed. What the water doesn't get, unstoppable rockslides will. And don't forget the hypercane-force gales as the <em>atmosphere</em> slides "downwards" almost unimpeded.</p>

<p>This assumes that the ground below is rigid, of course. It isn't quite, though it probably has enough structural integrity that the preceding paragraph is still true. In any case, the entire crust of Earth starts sliding "down" towards ground zero (or, as always, the antipode). Different parts of the crust slide at different velocities, though. Near the "ho hum" zone the crust is stretched; at ground zero or antipode, crust piles up. Nothing actually has time to move more than (very roughly) some tens of kilometers from its starting position at best, but that is quite sufficient to get <strong>cataclysmic hyper-earthquakes</strong> at every tectonically active zone on earth. Where there is no active zone to take up the stress, new ones open up.</p>

<p>I'm not entirely sure what the mantle is doing, but it probably isn't something nice.</p>

<p>One thing the mantle <em>is</em> doing happens around ground zero. Without any net gravity to keep the crust down, hydrostatic pressure in the lower lithosphere drops towards zero. Dissolved <strong>volatiles in magmas everywhere</strong> attempt to outgas, forming bubbles and expanding the magma until the sheer inertia of the overlying rocks resists it. The effect is to push the crust upwards <em>faster</em> than it is being pulled by the tides. So standing at ground zero you may not get to experience weightlessness after all. Instead you get to stand right on top of the greatest volcano eruption in the history of the planet. Truly the experience of a lifetime.</p>

<p>Then the three minutes are up and the brown dwarf recedes again.</p>

<p>At ground zero you're now at least a kilometer higher than you started out, together with everything around you, and still moving upwards at tens of kilometers an hour. That's far less than escape velocity, so what goes up must come down again. Except "down" is now most likely a boiling volcanic inferno. Aren't you glad you didn't get roasted by the brown dwarf to start with?</p>

<p>There's still time for the sliding tectonic plates to slide to a halt, and for the new rifts in the "ho hum" zone to start rivaling the ground zero volcano. Unless the interior of the earth deformed <em>elastically</em> so everything now tries to slide <em>back</em>.</p>

<p>The <em>planet</em> still exists, though. No mass was actually lost. On the other hand, the encounter did change our <em>collective</em> velocity by several tens of kilometers per second, which is broadly comparable to our usual orbital motion. That's going to wreak total havoc on the seasons.</p>

<p>Oh well. It's not as if any of us would be around to complain about that.</p>

<hr>

<p>(The sharp-eyed reader from before will note that most of these calamities would happen even without getting all the way to the Roche limit. So if the world has to end, Jupiter's gravity field <em>might</em> be capable enough, after all.)</p>

CC BY-SA 4.0
2019-05-28T23:07:46.570Z
2019-05-28T23:07:46.570Z
hmakholm left over Monica
<blockquote>
  <p>TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface.</p>
</blockquote>

<p><a href="https://astronomy.stackexchange.com/questions/32059/if-a-massive-object-like-jupiter-flew-past-the-earth-how-close-would-it-need-to/32061?noredirect=1#comment58605_32061">thanks to PeterCordes</a></p>

<hr>

<p>Jupiter's gravity will pull on the Earth itself, as well as everything on it. </p>

<p>It's not like a vacuum cleaner that selectively lifts small and light objects, the gravitational force will scale with the mass of each object; if the Earth is a zillion times more massive than we are, then Jupiter's gravitational force will also be about a zillion times larger.</p>

<p>What that means is that Earth will accelerate towards Jupiter, and we will accelerate along with it, and so we won't "feel the tug" anywhere near as strongly as one might suspect.</p>

<p>Instead, let's think about the size of the Earth, and the fact that people on the near side will be closer to Jupiter than the center of mass of the Earth, and people on the far side will be farther away.</p>

<p>Since people nearer to Jupiter will feel a <em>slightly stronger</em> acceleration than the center of mass of the Earth, they will feel a quite gentle tug. We'll calculate that in a minute. </p>

<p>But believe it or not, people on the far side of the Earth, feeling less of a tug than the Earth's center of mass, will believe they are being pulled in the opposite direction! They won't really be pulled away from Jupiter, but they will not accelerate towards Jupiter as fast as the Earth, and so it will feel like they are being repelled.</p>

<p>This kind of force is called a <a href="https://en.wikipedia.org/wiki/Tidal_force" rel="noreferrer">tidal force</a> and this is the picture that's often used with the concept:</p>

<p><img src="https://i.sstatic.net/Hr0OE.png" alt="enter image description here"> <a href="https://en.wikipedia.org/wiki/File:Field_tidal.svg" rel="noreferrer">Source</a> Replace "Satellite" with "Jupiter"</p>

<p>The acceleration we feel due to gravity is expressed as</p>

<p><span class="math-container">$$a_G = \frac{GM}{r^2}$$</span></p>

<p>where <span class="math-container">$G$</span> is the gravitational constant and equal to about <span class="math-container">$6.674 \times 10^{-11}$</span> m^3/kg s^2 and M is each mass that's pulling on you.</p>

<p>If you put in 6378137 meters and the mass of the Earth (<span class="math-container">$5.972 \times 10^{+24}$</span> kg) you get the familiar 9.8 m/s^2.</p>

<p>If Jupiter were 114,000,000 meters or 114,000 kilometers away, the Earth would accelerate at 1 g towards it, but people on the close and far side would accelerate very differently. On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 greater, so they would feel that they weighed 12% less. And people on the far side would also feel about the same amount lighter because they felt less acceleration than the Earth.</p>

<p>If Jupiter were so close that it were practically touching the Earth, it still wouldn't pull is off of Earth, assuming that Earth remained intact. <strong>But that wouldn't last very long!!!</strong> Earth would be accelerating towards Jupiter at about 20.9 m/s^2, and people on the near side would feel acceleration of 24.8 towards Jupiter, but relative to Earth that's only 3.9 m/s^2, so not enough to overcome Earth's gravity of -9.8 m/s^2.</p>

<p>On the far side of Earth it's similar; the acceleration towards Jupiter would be 17.8 m/s^2 but minus Earth's acceleration of - 20.9 it's -3.0 m/s^2 away, but that's also not enough to overcome the attraction to Earth of in this case +9.8 m/s^2.</p>

<p><strong>When Earth touches Jupiter, we will feel about 40% lighter on the near side and 31% lighter on the far side of Earth, but we would not leave the surface.</strong></p>

<p>However, in just minutes we'd be pulled so deep into Jupiter that we would be crushed by Jupiter's internal atmospheric pressure.</p>

<p>It would certainly be fun, but it wouldn't last long!</p>

CC BY-SA 4.0
2019-05-27T17:33:18.880Z
@ShakesBeerCH it looks like your edit was rejected, but there was indeed an error in the arithmetic. $GM_J/(R_E+R_J)^2=20.9$ m/s^2, etc. Can you check again, thanks!
uhoh
CC BY-SA 4.0
2019-05-27T18:35:05.543Z
Isn't the earth's momentum a zillion times larger, so its motion won't be perturbed as much as a person?
Barmar
CC BY-SA 4.0
2019-05-27T18:49:02.650Z
@Barmar double check the second paragraph that begins with: "It's not like a vacuum cleaner that selectively lifts small and light objects..." The Earth's momentum does not affect the Earth's *acceleration*, and the acceleration of objects of different sizes near Jupiter will be (almost) the same because the force depends on the mass; $F_E=M_EM_J/r^2$ so the acceleration $a_E=F/M_E=M_J/r^2$ is independent of the mass. I say "almost" because the Earth is so big that it's gravitational effect on Jupiter is very small but can't be completely neglected.
uhoh
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2019-05-27T22:56:22.030Z
TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface.
Peter Cordes
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2019-05-28T00:06:38.800Z
@PeterCordes that's so much better than I could have done I've just quoted you, thank you. Please feel free to edit the answer further!
uhoh
CC BY-SA 4.0
2019-05-28T04:07:10.817Z
Glad I could help, thanks for doing the math and writing it up, this is an interesting Q&A. :)  I thought about adding in the phrasing "having the Earth pulled out from under them (even faster than the extra pull of Earth + Jupiter)" for the people on the far side, but I don't see a place to put it without being redundant or rewriting a whole chunk.
Peter Cordes
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2019-05-28T05:04:47.917Z
Lighter on the far side?
mckenzm
CC BY-SA 4.0
2019-05-28T05:24:38.790Z
@mckenzm yes. Earth's center of mass is closer to Jupiter than the people on the far side, so *Earth experiences a greater acceleration* than the people do. That's why the diagram showing outward-pointing arrows on *both sides* of the Earth is always confusing at first.
uhoh
CC BY-SA 4.0
2019-05-30T06:03:40.897Z
"On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 less" <- 1.2 m/s^2 _more_?
Logan Pickup
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2019-05-30T06:10:57.400Z
@LoganPickup yep I think you're right, changing now. Thank you!
uhoh
CC BY-SA 4.0
2019-05-27T01:06:21.783Z
2019-05-30T06:11:28.157Z
2019-05-30T06:11:28.157Z
uhoh
<p>I understand this is a silly hypothetical but I'm asking for a 7 year old so please bear with me.</p>

<p>Imagine an interstellar stray gas giant comes flying through our solar system. </p>

<p>If we were not concerned that it would also steal our atmosphere and create tidal forces that destroyed everything... How close would it need to come to us to exert enough gravity to lift people off the ground and pull them into its own orbit?</p>

CC BY-SA 4.0
2019-05-27T09:37:08.540Z
The question would be more interesting with a rather small body (like a small, dense moon or even better, a small black hole) whose gravity field close by is stronger than the Earth's but farther away too weak to suck the Earth in.
Peter - Reinstate Monica
CC BY-SA 4.0
2019-05-27T12:27:53.917Z
@PeterA.Schneider I'm not quite sure what you mean. If in the answers below you replaced Jupiter with a black hole of the same mass, there'd be no significant difference (initially at least): the gravitational force is more or less the same.
Chappo Hasn't Forgotten
CC BY-SA 4.0
2019-05-27T14:16:09.717Z
@Chappo Not of the same mass but of a much smaller mass, and much closer, exploiting the inhomogeneity of its gravitational field. Imagine a black hole 10 km above us exerting 1g on us. (Its mass would be much smaller than Jupiter's.) The far side of the earth, being 12000 km away, would only experience (12000/10)^2 ~ 1.4E-6 g, i.e. almost no attraction. *That* black hole flying by at **9** km distance would suck us up, and some of the upper 1 km of earth's crust.
Peter - Reinstate Monica
CC BY-SA 4.0
2019-05-27T17:12:13.453Z
That would be a cute story if one day, we wake up and we find out that birds disappeared, "stolen" by a planet passing by
Taladris
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2019-05-27T18:32:47.880Z
I think you could get a more "fun" answer if you wrote to what-if.xkcd.com.
Barmar
CC BY-SA 4.0
2019-05-28T01:27:30.353Z
@Barmar: Assuming that's even still active - the last post there was months ago at least.
Vikki
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2019-05-28T02:02:23.773Z
This is the very definition of the Roche limit of the passing body.
Loren Pechtel
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2019-05-28T05:35:03.640Z
I've just asked [Can we  to have emoji in posts?](https://space.meta.stackexchange.com/q/1292/12102)
uhoh
CC BY-SA 4.0
2019-05-28T06:57:38.390Z
@LorenPechtel Which means that things like mountains and the seas get sucked up too.
Aron
CC BY-SA 4.0
2019-05-28T16:47:19.683Z
@Aron Correct.  The planet is destroyed.
Loren Pechtel
CC BY-SA 4.0
2019-05-28T20:22:41.960Z
@Barmar If Randall answered it, then, yes, it would. Sadly, Randall's last post on What If was a year ago last week.
reirab
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2019-05-28T21:21:29.540Z
@reirab Yes, someone else already pointed out that he seems to have abandoned What If.
Barmar
CC BY-SA 4.0
2019-05-29T20:40:33.643Z
@reirab No, it can't be! Surely he's just working on a new What-If book, keep the faith!
brichins
CC BY-SA 4.0
2019-05-30T05:21:16.630Z
@Aron If you were very, very careful, no. Mountains aren't held to the Earth's surface merely through gravity, while humans are. The same is true for oceans, though there the extra attractive force is much smaller. Of course, both planets would be rather disturbed anyway, but the OP explicitly said he doesn't want to consider those effects.
Luaan
<gravity>
If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?
Yevgeny Simkin
# If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?

(Asked by: username_0 on 2019-05-27)

I understand this is a silly hypothetical but I'm asking for a 7 year old so please bear with me.

Imagine an interstellar stray gas giant comes flying through our solar system.

If we were not concerned that it would also steal our atmosphere and create tidal forces that destroyed everything... How close would it need to come to us to exert enough gravity to lift people off the ground and pull them into its own orbit?

**Tags:** gravity

> **Comments:**
> **username_1**: The question would be more interesting with a rather small body (like a small, dense moon or even better, a small black hole) whose gravity field close by is stronger than the Earth's but farther away too weak to suck the Earth in.  
> **username_2**: @PeterA.Schneider I'm not quite sure what you mean. If in the answers below you replaced Jupiter with a black hole of the same mass, there'd be no significant difference (initially at least): the gravitational force is more or less the same.  
> **username_1**: @username_2 Not of the same mass but of a much smaller mass, and much closer, exploiting the inhomogeneity of its gravitational field. Imagine a black hole 10 km above us exerting 1g on us. (Its mass would be much smaller than Jupiter's.) The far side of the earth, being 12000 km away, would only experience (12000/10)^2 ~ 1.4E-6 g, i.e. almost no attraction. *That* black hole flying by at **9** km distance would suck us up, and some of the upper 1 km of earth's crust.  
> **username_3**: That would be a cute story if one day, we wake up and we find out that birds disappeared, "stolen" by a planet passing by  
> **username_4**: I think you could get a more "fun" answer if you wrote to what-if.xkcd.com.  
> **username_5**: @username_4: Assuming that's even still active - the last post there was months ago at least.  
> **username_6**: This is the very definition of the Roche limit of the passing body.  
> **username_7**: I've just asked [Can we  to have emoji in posts?](https://space.meta.stackexchange.com/q/1292/12102)  
> **username_8**: @username_6 Which means that things like mountains and the seas get sucked up too.  
> **username_6**: @username_8 Correct.  The planet is destroyed.  
> **username_9**: @username_4 If Randall answered it, then, yes, it would. Sadly, Randall's last post on What If was a year ago last week.  
> **username_4**: @username_9 Yes, someone else already pointed out that he seems to have abandoned What If.  
> **username_10**: @username_9 No, it can't be! Surely he's just working on a new What-If book, keep the faith!  
> **username_11**: @username_8 If you were very, very careful, no. Mountains aren't held to the Earth's surface merely through gravity, while humans are. The same is true for oceans, though there the extra attractive force is much smaller. Of course, both planets would be rather disturbed anyway, but the OP explicitly said he doesn't want to consider those effects.  

---

## 3 answers

### Answer by username_7 on 2019-05-27. Score: 101 | Accepted answer

> TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface.

[thanks to username_17](https://astronomy.stackexchange.com/questions/32059/if-a-massive-object-like-jupiter-flew-past-the-earth-how-close-would-it-need-to/32061?noredirect=1#comment58605_32061)

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Jupiter's gravity will pull on the Earth itself, as well as everything on it.

It's not like a vacuum cleaner that selectively lifts small and light objects, the gravitational force will scale with the mass of each object; if the Earth is a zillion times more massive than we are, then Jupiter's gravitational force will also be about a zillion times larger.

What that means is that Earth will accelerate towards Jupiter, and we will accelerate along with it, and so we won't "feel the tug" anywhere near as strongly as one might suspect.

Instead, let's think about the size of the Earth, and the fact that people on the near side will be closer to Jupiter than the center of mass of the Earth, and people on the far side will be farther away.

Since people nearer to Jupiter will feel a *slightly stronger* acceleration than the center of mass of the Earth, they will feel a quite gentle tug. We'll calculate that in a minute.

But believe it or not, people on the far side of the Earth, feeling less of a tug than the Earth's center of mass, will believe they are being pulled in the opposite direction! They won't really be pulled away from Jupiter, but they will not accelerate towards Jupiter as fast as the Earth, and so it will feel like they are being repelled.

This kind of force is called a [tidal force](https://en.wikipedia.org/wiki/Tidal_force) and this is the picture that's often used with the concept:

![enter image description here](https://i.sstatic.net/Hr0OE.png) [Source](https://en.wikipedia.org/wiki/File:Field_tidal.svg) Replace "Satellite" with "Jupiter"

The acceleration we feel due to gravity is expressed as

$$a\_G = \frac{GM}{r^2}$$

where $G$ is the gravitational constant and equal to about $6.674 \times 10^{-11}$ m^3/kg s^2 and M is each mass that's pulling on you.

If you put in 6378137 meters and the mass of the Earth ($5.972 \times 10^{+24}$ kg) you get the familiar 9.8 m/s^2.

If Jupiter were 114,000,000 meters or 114,000 kilometers away, the Earth would accelerate at 1 g towards it, but people on the close and far side would accelerate very differently. On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 greater, so they would feel that they weighed 12% less. And people on the far side would also feel about the same amount lighter because they felt less acceleration than the Earth.

If Jupiter were so close that it were practically touching the Earth, it still wouldn't pull is off of Earth, assuming that Earth remained intact. **But that wouldn't last very long!!!** Earth would be accelerating towards Jupiter at about 20.9 m/s^2, and people on the near side would feel acceleration of 24.8 towards Jupiter, but relative to Earth that's only 3.9 m/s^2, so not enough to overcome Earth's gravity of -9.8 m/s^2.

On the far side of Earth it's similar; the acceleration towards Jupiter would be 17.8 m/s^2 but minus Earth's acceleration of - 20.9 it's -3.0 m/s^2 away, but that's also not enough to overcome the attraction to Earth of in this case +9.8 m/s^2.

**When Earth touches Jupiter, we will feel about 40% lighter on the near side and 31% lighter on the far side of Earth, but we would not leave the surface.**

However, in just minutes we'd be pulled so deep into Jupiter that we would be crushed by Jupiter's internal atmospheric pressure.

It would certainly be fun, but it wouldn't last long!

> **Comments:**
> **username_7**: @ShakesBeerCH it looks like your edit was rejected, but there was indeed an error in the arithmetic. $GM_J/(R_E+R_J)^2=20.9$ m/s^2, etc. Can you check again, thanks!  
> **username_4**: Isn't the earth's momentum a zillion times larger, so its motion won't be perturbed as much as a person?  
> **username_7**: @username_4 double check the second paragraph that begins with: "It's not like a vacuum cleaner that selectively lifts small and light objects..." The Earth's momentum does not affect the Earth's *acceleration*, and the acceleration of objects of different sizes near Jupiter will be (almost) the same because the force depends on the mass; $F_E=M_EM_J/r^2$ so the acceleration $a_E=F/M_E=M_J/r^2$ is independent of the mass. I say "almost" because the Earth is so big that it's gravitational effect on Jupiter is very small but can't be completely neglected.  
> **username_17**: TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface.  
> **username_7**: @username_17 that's so much better than I could have done I've just quoted you, thank you. Please feel free to edit the answer further!  
> **username_17**: Glad I could help, thanks for doing the math and writing it up, this is an interesting Q&A. :)  I thought about adding in the phrasing "having the Earth pulled out from under them (even faster than the extra pull of Earth + Jupiter)" for the people on the far side, but I don't see a place to put it without being redundant or rewriting a whole chunk.  
> **username_19**: Lighter on the far side?  
> **username_7**: @username_19 yes. Earth's center of mass is closer to Jupiter than the people on the far side, so *Earth experiences a greater acceleration* than the people do. That's why the diagram showing outward-pointing arrows on *both sides* of the Earth is always confusing at first.  
> **username_20**: "On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 less" <- 1.2 m/s^2 _more_?  
> **username_7**: @username_20 yep I think you're right, changing now. Thank you!  

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### Answer by username_12 on 2019-05-27. Score: 20

About 70,000 km. If the Earth orbited Jupiter (or flew by) at a closer distance, not just we would leave the surface, but the whole Earth would disintegrate since all its mass will be leaving too.

70,000 km is Jupiter's [Roche limit](https://en.wikipedia.org/wiki/Roche_limit) (although its actual value varies a lot depending on the other involved body), the radius where tidal forces (already explained in username_7's answer) overwhelm gravitational forces and any orbiting body can't hold itself by its own weight. In that context, people on the surface don't behave differently than rocks.

Btw., this scenario is also explored [in a youtube video](https://www.youtube.com/watch?v=ym7bDBg9jQ4). I wouldn't say it's very good but it may be helpful to explain the Roche limit to a 7 year old.

> **Comments:**
> **username_7**: I didn't even think to mention the Roche limit; very good point! `+1`  
> **username_13**: Note that 70,000 km also happens to be Jupiter's radius, so the planets would need to be touching. (And there's not a "the" Roche limit; it depends on the density of the secondary object, that is, Earth in this example).  
> **username_12**: @HenningMakholm - Starting the answer with "about" was intended to deal with variation of Roche limit due to a lot of factors. However I edited the answer to make it more clear.  
> **username_14**: People aren't held together by gravitational forces, so being inside the Roche limit won't pull them apart like it would for a planet. In that regard, people on the surface *do* behave differently than rocks - a pile of gravel will separate into individual pebbles as it gets sucked into Jupiter's gravitational well, but a person will remain intact.  
> **username_15**: @NuclearWang But... people *are* held to the earth purely by gravitational forces.  Being within the roche limit would not pull humans apart but we would still act like those pebbles in the sense that individual humans would not be stuck to the earth by gravity any longer.  
> **username_16**: To this layman it seems obvious that Jupiter's gravity would make life unpleasant (if not impossible) on Earth at a much greater radius than 70Mm.  
> **username_17**: At *just* inside the Roche limit, the math in @username_7's answer shows that loose objects aren't literally ripped off the surface.  I think the mechanism is more gradual even for a totally non-rigid aggregate of gravel: with its own gravity being unopposed in the other directions, it would elongate in the tidal-force direction.  This puts the ends farther and farther from the centre of mass, and increases the distance for the gravity gradient. This *eventually* leads to it being torn apart, but one quick pass wouldn't rip loose objects off the surface (esp for a stiff / viscous object like Earth)  
> **username_17**: The wikipedia article shows the elongation as part of the torn-apart effect right at the Roche limit.  See also video animations like https://www.youtube.com/watch?v=tGS_gZm6HxI  and https://www.youtube.com/watch?v=6e5yyIcq40o.  And remember, it's the radius where a rigid or fluid body can't exist *long term* in a stable circular orbit.  An elliptical orbit that just dips in at perigee could conceivably let an aggregate body re-form itself when farther out.  **See [How fast does Roche limit disintegration proceed?](//space.stackexchange.com/q/10113) on space.SE**  
> **username_17**: @HenningMakholm: I think this answer accidentally took Jupiter's actual radius as its Roche limit!, because the [Wikipedia page has a table of density and radius](https://en.wikipedia.org/wiki/Roche_limit#Selected_examples) for objects in our solar system at the top of the section for Roche limits for pairs of bodies.  The actual rigid-body Roche limit (where objects are pulled off the surface by tidal forces) is $R_m * (2  \rho_M / \rho_m) ^ {1/3}$ = `71493000 * (2 * 1326/5513)^(1/3)` = **56 018 km  using the "fully rigid-satellite" formula from Wikipedia.**  
> **username_17**: Or the "more accurate" formula for a tidally locked body (including the body's rotation) gives us 62 709 km.  But Earth isn't rotating that fast.  (The orbital period for [a circular orbit just skimming the surface](https://en.wikipedia.org/wiki/Orbital_period#Orbital_period_as_a_function_of_central_body's_density) of Jupiter is ~2.86 hours, which is a lot faster than 24 hours, and why this Roche limit formula predicts a body can be torn apart so much farther away than if it's not rotating.  
> **username_17**: BTW, in my first comment about being deformed before coming apart, I was referring to the "fluid" Roche limit.  Once I looked up the wiki article, I realized that the "rigid" Roche limit is exactly what we want here.  That's the assumption of rigid but no tensile strength, which works perfectly for loose objects pulled off the surface.  For the integrity of Earth itself: it's squishy long-term (molten rock), and in a close orbit would deform like for the "fluid" Roche limit (which has a much larger radius).  Tidal drag speedup up our rotation = catastrophic tsunamis!  
> **username_12**: The actual Roche limit for Earth over Jupiter would be way larger, because such a big body like Earth wouldn't behave like a rigid body.  
> **username_12**: @mistertribs - Interestingly, that article states that "If the planet passes within the Roche limit at periastron, however, mass can be stripped from it". Likely, that mass would include some people standing on the planet.  
> **username_18**: "The Roche Limit" does not exist for a single body. One could define "A Roche Limit" for a Jupiter-Earth system, but not for a Jupiter. And it would be different for a Jupiter-Mercury system etc.  
> **username_12**: That's the reason the answer starts with "about".  

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### Answer by username_13 on 2019-05-28. Score: 18

As other answers point out, Jupiter is not quite massive enough to pull a planet of Earth's density apart. But we can use a slightly heavier object instead -- say, a small cold brown dwarf massing 13 Jupiters, or about 4000 Earths. According to the rigid-body Roche formula, its Roche limit is then $\sqrt[3]{2\cdot 4000}=20$ earth radii, or 130,000 km. The radius of the brown dwarf is not much larger than Jupiter's (since they're both made of compressible gas), so smaller than 100,000 km, and there's some room for Earth to be destroyed without actually colliding with it.

Our brown dwarf, wandering sedentarily around in the galaxy, spots our sun, and decides to take a closer look. It comes screaming through the inner solar system on a hyperbolic orbit, which means that it will be moving at somewhat more than solar escape velocity when it narrowly misses Earth -- call it 100 km/s. Switching perspective, we can say that *Earth* comes at the brown dwarf at 100 km/s and just about misses. This speed allows us to spend a almost half an hour inside the Roche limit, if we almost touch the brown dwarf at the time of closest approach.

But that sounds like it will be unnecessarily dramatic, so it's instead have the brown dwarf give is a slightly wider berth such that at the time of closest approach, the acceleration of gravity at ground zero will be a more pedestrian $-0.1\;\rm m/s^2$. That will be the case when our distance to the brown dwarf is $\sqrt[3]{\frac{9.82}{0.1+9.82}} = 0.997$ of the Roche limit, or 128,000 km. The length of our path through the Roche zone is then about 20,000 km, which means the encounter takes 200 seconds. Call it three minutes.

(The sharp-eyed reader will notice that these numbers mean that the far side of the earth is actually never inside the 130,000-km limit, but what really counts is the first derivative of the brown dwarf's gravitational field, so if you're standing on the antipodal point you'll still have the Earth's center-of-mass pulled away under you even if you yourself is outside the limit. The numbers are all approximate anyway).

(On the other hand, a few minutes clearly not enough time for the molten inside of the Earth to flow into a hydrostatic equilibrium in the new situation, so using the rigid-body formula is appropriate).

**What happens then?**

First, of course, it is an **awesome sight**. The brown dwarf dominates the sky with an angular diameter of somewhere between 60° and 100°.

Then, it may be getting **uncomfortably hot**. Not necessarily "the mountains are melting" hot or even "the seas boil away" hot. But [this](https://www.space.com/22659-brown-dwarf-stars-hot-as-oven.html) says that the coldest brown dwarves have about the temperature of a baking oven, and having a significant part of the sky at 150 °C can make anyone sweat. No worries, though -- it will all be over in a few hours, so just go inside and crank up the AC; that'll deal with it fine.

Right at ground zero **gravity decreases smoothly** while we approach Roche. When it passes zero G you're in free fall and start floating gently upwards. Except that everything around you -- cars, houses, trees, the soil itself -- is also in free fall since the only thing that kept them down was gravity. So to a first approximation your *local* experience is not about bring ripped off Earth, but just of weightlessness. (*Or is it?* See below.)

Ditto at the antipodal point.

One problem that shows up here is that **the atmosphere is escaping** into space. Since there is no gravity to keep it down, it escapes rather faster than the gentle floating of cars, trees, and people, propelled by its own pressure. Even before we reach Roche, the air may have become too thin to breathe. On the other hand fresh air will rush in from the surrounding areas to fill the void, creating the great-great-grandmother of all hurricanes. (And a great-great-grandfather around the antipode, of course).

On a great circle 90° from ground zero, gravity increases to about 1.7 G. You feel heavy. Ho hum.

*Between* these areas dramatic things happen. At about 45° from ground zero (or the antipode) the tidal force is at right angles to vertical, so the *strength* of gravity is about what we're used to -- but its direction is different. It's as if **the world is tilted by tens of degrees**, rather like how bad sci-fi movies pretend "entering a gravitational field" works. Tall buildings tip over; many not-so-tall ones just collapse. Lakes and seas do things that make the word "tsunami" pack up and go home, hopelessly outclassed. What the water doesn't get, unstoppable rockslides will. And don't forget the hypercane-force gales as the *atmosphere* slides "downwards" almost unimpeded.

This assumes that the ground below is rigid, of course. It isn't quite, though it probably has enough structural integrity that the preceding paragraph is still true. In any case, the entire crust of Earth starts sliding "down" towards ground zero (or, as always, the antipode). Different parts of the crust slide at different velocities, though. Near the "ho hum" zone the crust is stretched; at ground zero or antipode, crust piles up. Nothing actually has time to move more than (very roughly) some tens of kilometers from its starting position at best, but that is quite sufficient to get **cataclysmic hyper-earthquakes** at every tectonically active zone on earth. Where there is no active zone to take up the stress, new ones open up.

I'm not entirely sure what the mantle is doing, but it probably isn't something nice.

One thing the mantle *is* doing happens around ground zero. Without any net gravity to keep the crust down, hydrostatic pressure in the lower lithosphere drops towards zero. Dissolved **volatiles in magmas everywhere** attempt to outgas, forming bubbles and expanding the magma until the sheer inertia of the overlying rocks resists it. The effect is to push the crust upwards *faster* than it is being pulled by the tides. So standing at ground zero you may not get to experience weightlessness after all. Instead you get to stand right on top of the greatest volcano eruption in the history of the planet. Truly the experience of a lifetime.

Then the three minutes are up and the brown dwarf recedes again.

At ground zero you're now at least a kilometer higher than you started out, together with everything around you, and still moving upwards at tens of kilometers an hour. That's far less than escape velocity, so what goes up must come down again. Except "down" is now most likely a boiling volcanic inferno. Aren't you glad you didn't get roasted by the brown dwarf to start with?

There's still time for the sliding tectonic plates to slide to a halt, and for the new rifts in the "ho hum" zone to start rivaling the ground zero volcano. Unless the interior of the earth deformed *elastically* so everything now tries to slide *back*.

The *planet* still exists, though. No mass was actually lost. On the other hand, the encounter did change our *collective* velocity by several tens of kilometers per second, which is broadly comparable to our usual orbital motion. That's going to wreak total havoc on the seasons.

Oh well. It's not as if any of us would be around to complain about that.

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(The sharp-eyed reader from before will note that most of these calamities would happen even without getting all the way to the Roche limit. So if the world has to end, Jupiter's gravity field *might* be capable enough, after all.)


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