# If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?

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?

• 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. May 27 '19 at 9:37
• @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. May 27 '19 at 14:16
• I think you could get a more "fun" answer if you wrote to what-if.xkcd.com. May 27 '19 at 18:32
• @Barmar: Assuming that's even still active - the last post there was months ago at least. May 28 '19 at 1:27
• This is the very definition of the Roche limit of the passing body. May 28 '19 at 2:02

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 PeterCordes

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 and this is the picture that's often used with the concept:

Source 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!

• @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
May 27 '19 at 17:33
• 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. May 27 '19 at 22:56
• @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
May 28 '19 at 0:06
• 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. May 28 '19 at 4:07
• "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? May 30 '19 at 6:03

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 (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.

Btw., this scenario is also explored in a youtube video. I wouldn't say it's very good but it may be helpful to explain the Roche limit to a 7 year old.

• 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). May 27 '19 at 11:03
• 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. May 27 '19 at 13:44
• @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...
May 27 '19 at 13:59
• 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) May 28 '19 at 4:10
• @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 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. May 28 '19 at 4:35

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 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.

(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.)