# What would happen if an object on Earth has an equal or greater gravitational force than the Earth?

So say that there is this ridiculously dense object that fits on Earth. It has a bigger gravitational force than the Earth, and for some reason, doesn't break it. It also has a smaller force than the Sun.

What would happen? Would the Earth suddenly start spinning differently? Would it start orbiting away from the Sun? What about the moon?

(I don't even know if this is a possible idea, so that's why I also put a reality-check tag.)

• Denser than the earth but small enough to sit on its surface would probably result in a singularity. – Cereza Jun 5 '17 at 21:45
• You would have seen gravitational effects from the object long before it came anywhere near Earth. – HDE 226868 Jun 5 '17 at 21:45
• @HDE226868 That too. – Cereza Jun 5 '17 at 21:45
• The additional mass changes the orbit of the earth, possibly drawing the moon into collision and eventually careening into the sun, since our orbit is currently (relatively) stabilized for our mass. – Isaac Kotlicky Jun 5 '17 at 21:46
• You didn’t explain the situation precisely and each answer made up a different interpretation! – JDługosz Jun 6 '17 at 11:58

Update: object is magic in nature :-)

there is this ridiculously dense object that fits on Earth. It has a bigger gravitational force than the Earth, and for some reason, doesn't break it.

Okay. The scenarios below do not qualify. Now the important thing is how does this object come to be. That is, does it appear out of nowhere, or does it come from enough afar, and starts "hovering" near the Earth? Does it start orbiting the Sun together with Earth?

Moreover we are handwaving things about Earth too - its crust should shatter due to the object's gravitational field. Somehow the Earth's crust behaves as if it was made up of scrith. Then, what else could behave in an unforeseen way?

In the simplest (!) case, the object appears near Earth and is initially at rest relative to the latter. Even so, most of the Earth's surface, its water, its atmosphere etc, flows towards this new "bottom", and abandons Earth. The gravity on the opposite side of the Earth gets increased by around 1 g or more.

But the fact that the Earth rotates makes it so its whole surface is exposed to the object inside of 24 hours, scouring the whole surface clean, while all this material - bodies, cars, trains, lakes, small mountains, cities, and so on - "falls" towards the object from a height varying from several kilometers to around eight thousand kilometers. Most of these burn on reentry (the object is now surrounded by an atmosphere denser than Earth's).

This new "worldlet" is uninhabitable by humans because its surface gravity is high enough to be lethal (a larger mass than Earth, in a far denser package).

Most Earth satellites obviously either fall on the object or, but it's quite unlikely, get grav-assisted and shot towards outer space (they haven't energy enough, and probably turn into short-period comets) or towards the Sun.

The Moon itself is probably slingshotted away, either entering solar orbit halfway between the Earth and Mars, or falling inward towards the Sun.

The center of mass of the Earth-Moon system shifts towards the object and actually probably enters the object itself, it being more massive than Earth.

The two bodies - a scoured, cooling ball of rock once called Earth and a superdense ball covered with several kilometers of mud and scrag with occasional traces of organic compounds, a saltwater ocean also kilometers deep, and a dense nitrogen atmosphere - go on rotating around the Sun. Depending on the initial orbital parameters of the object, the new orbit might be the same as Earth, or more oblate, either farther or nearer to the Sun.

It is unclear whether we're talking mass or gravitational pull.

# Same pull, but low mass

Gravitational pull (acceleration) is proportional to the mass of the object divided by the square of the distance from its center.

Imagine a sphere with a radius of one meter. Its surface gravity would (numerically) be 6.67 × 10−11 times its mass, so to have a surface gravity of 1 g (9.81 ms−2) it would need to weigh 1.4 × 1011 kg, or 140 million tons; about the weight of 1400 Nimitz-class aircraft carriers in a sphere two meters in diameter.

While the mass would never be enough to modify the Earth's orbit, its density would be more than enough to make it sink towards the center of the Earth, and actually probably overshoot it — it would receive, in proportion, the same buoyancy of a leaden ball in a bubble of air. The ball would bounce to and fro several times before beginning to drift slowly around the center of the Earth (where it would receive almost no gravitational pull).

If we could, in some way, suspend it above the Earth, it would generate a small area of strange gravity; on the surface (d=1 m) the pull would be 1 g, neutralizing Earth's own pull, and an object would briefly float. At one meter from the sphere (d = 2 m from center), double the distance, one quarter the pull; so you would get .75 g downwards.

Attaining equilibrium between two forces going like r−2 is impossible unless one employs some technological tricks; it is a consequence of Earnshaw's Theorem, the same reason why you cannot gently float an object using a magnet or a charged plastic stick (active control is a tech trick and using gyrostabilization introduces an additional force).

So, no "gravity free" areas beneath the sphere.

# What if the object has the same mass of the Earth?

Then it either has a comparable density, or we're again in the "compressed matter" scenario.

In the first, more natural scenario, the two planets crunch together. Moreover, they have a gravitational potential energy in respect to their rest position (a sphere about 25% larger than the Earth) that's simply monstruous, and that energy would be converted into heat while the two planets grind together. Unless the second planet has a very cold inside, the Earth would be converted into a boiling ball of lava in a matter of hours.

The second scenario is, if possible, even worse. The dense ball of matter has a mass equal to the Earth, but a much smaller radius. Let's say 500 km. That's 13 times less than the radius of the Earth, and the gravitational pull would therefore be 132 = 169 times greater. At a distance of 500 km, the acceleration would still be around 40 g, which more than a human being can tolerate. Things would fall laterally — the sphere would be "down" for everything in a radius of thousands of kilometers, and a crushing death for anyone nearer than a couple thousand of kilometers.

But the same attraction would act on the Earth's mass — its crust, and the lava beneath. The Earth and this Death Star would rush towards one another, the tidal forces literally tearing the Earth apart. You can see something similar, albeit with a liquid way less viscous than lava, here.

# There is, however, one catch...

How it is that the sphere has such a density? The densest packing of protons in ordinary matter is osmium. Even the pressures at the center of the Earth cannot change the density of iron of more than a factor of two (less, actually: from around 8 to around 13 g/cm3).

It stands to reason that the dense sphere could not be kept at such a fantastically higher density by its own gravity.

Or in other words, our Death Star would not be stable. The minimum mass required to achieve some sort of stability is estimated around 10% of a solar mass. Beneath that level, there is no known process that could allow compressing matter inside its Schwarzschild radius, achieving black hole stability (it is theorized that such "micro black holes" could have formed during the Big Bang).

Therefore, the Death Star would simply inflate explosively, freeing its pent-up compression energy and smashing the Earth in the process. For the same reason, the famous "tea spoon of neutron star matter" poured on Earth would never sink to its center — it would cause a massive explosion. Followed by a considerable neutron activation, possibly followed by an appreciable nuclear "fizzle" as most materials near ground zero get transmuted into unstable and unlikely isotopes.

• A little bit of vodka, a bucket of water, and some salt and you could even touch it. – Draco18s no longer trusts SE Jun 6 '17 at 13:47

1. Object has equal to grater gravitational force than Earth.
2. Object is ridiculously dense object.
3. Object fits on Earth.
4. Earth does not break.
5. Object has a smaller gravitational force than the Sun.

Given your description doesn't specify the object being acted upon, I'm assuming by Force you meant Mass.

# Would the Earth suddenly start spinning differently?

Yes, the barycenter (center of mass) of the earth-object system would move to a point between them. They would then spin around this point. The greater the altitude of the object with respect to the surface of the Earth, the greater the effect. The barycenter of a 2 object system is defined as:

$$r_1 = \cfrac{a}{1 + \cfrac{M_1}{M_2}}$$

where $$r_1$$ is the distance from $$M_1$$ to the center of the system, $$a$$ is the distance between the two objects, and $$M$$ is the respective mass of each object. If in our case we assume their masses are identical and that $$M_1$$ is the earth, then:

$$r_1 = \cfrac{r_e + h}{1 + \cfrac{M_e}{M_e}} = \cfrac{1}{2}\left(r_e + h\right)$$.

Therefore, if the object is sitting on the surface of the Earth ($$h = 0$$), then the Earth will begin orbiting around $$r_1 = 3185.5 \space\text{km}$$ which is just below the edge of the outer core. If you allow for magnetohydrodynamics, then the change in angular momentum would alter the flow of iron in the outer core and create turbulence. This would in turn destabilize the magnetosphere of the Earth.

If the height is higher, than the barycenter would move into the mantle.

Furthermore, given Kepler's Laws of Motion, the period ('day' length) of this orbit would be: $$T = \sqrt{\cfrac{4 \pi^2 a^3}{G\left(M_1 + M_2\right)}} = \sqrt{\cfrac{4 \pi^2 \left(r_1 + h\right)^3}{2GM_e}}$$

## If the object were at an altitude of 0 meters, the orbital period of The Object and Earth would be 21 minutes. At an altitude of 35,768 km (Geosynchronous Orbit), the period would be 15 hours. How's that for a shortened day?

Note: I ignored the gravitational effects of such an object this close to The Earth since OP specified:

and for some reason, doesn't break it.

# Would it start orbiting away from the Sun?

## No, it'll drop MUCH closer to the Sun.

This answer is not quite as simple due to orbital mechanics. The mass of the 'Earth' system will has doubled, therefore, the force of gravity upon the system from the sun would also double:

$$F_{G_2} = G\cfrac{2M_1 M_2}{r^2} = 2F_{G_1}$$ where $$F_{G_1} = G\cfrac{M_1 M_2}{r^2}$$;

but, since newton's 2nd law states $$F = ma$$ then the combination yields:

$$F_G = 2ma = G\cfrac{2m M}{r^2} \Rightarrow a = G\cfrac{M}{r^2}$$

which is the normal acceleration of the Earth.

Instead, the sudden increase in mass on earth ($$m \rightarrow 2m$$) would create a change in momentum (impulse). Now, several assumptions must be made, all of whom are heavily dependent on the object's origin. I assume the following:

1. The Object 'appeared' out of nowhere.
2. The Object and The Earth may be considered a single system with respect to The Sun.

Given conservation of momentum:

$$P_i = P_f$$

$$m_1i v_1i = m_1f v_1f + m_2f v_2f; m_1 = m_2 = m_e; v_1f = v_2f ; v_1i = v_e$$

$$m_e v_e = 2m_e v_f \Rightarrow v_e = 2v_f \Rightarrow v_f = \frac{1}{2} v_e$$

### Therefore, after the object appears, the combined Earth-Object System would lose half its velocity under conservation of momentum.

Using the Vis-Viva equation, we may then calculate what the semimajor axis of the elliptical orbit would become (it won't but bear with me here):

$$v^2 = GM \left(\cfrac{2}{r} - \cfrac{1}{a}\right) = \cfrac{2GM}{r} - \cfrac{GM}{a}$$

$$\cfrac{GM}{a} = \cfrac{2GM}{r} - v^2 \Rightarrow \cfrac{1}{a} = \cfrac{2}{r} - \cfrac{v^2}{GM}$$

$$a = \cfrac{1}{\cfrac{2}{r} - \cfrac{v^2}{GM} }$$

Great, now we can figure out the semimajor axis of the new orbit. Since Earth has an elliptical orbit, our orbital velocity, $$v_e$$ varies over the year. At aphelion (when we're farthest from The Sun in July), $$v_e = 29.29 \space\text{km/s}$$. At perihelion (when we're closest to The Sun in January) $$v_e = 30.29 \space\text{km/s}$$ (source: NASA). Since they have a 3.3% difference, we'll simplify our calculations and just go with $$v = 30 \space\text{km/s}$$. We'll also take $$r = 150 \times 10^6 \space\text{km}$$, $$M = 2 \times 10^{30} \space\text{kg}$$).

This yields $$75 \times 10^6 \space\text{km}$$ or 0.5 AU. For comparison, Venus orbits around $$108 \times 10^6 \space\text{km}$$ from the sun, and Mercury orbits between $$46\times 10^6 \space\text{km}$$ and $$70 \times 10^6 \space\text{km}$$ from the Sun. Since this would be the new stable orbit, the Earth-Object system would naturally tend towards this orbit; however, it's still at 1 AU.

## Therefore Earth would swing into orbit 5 million kilometers above Mercury. Needless to say, this would be BAD. World on fire BAD.

This is BAD for a multitude of reasons:

1. At this distance, all water on Earth would boil. The oceans, the seas, etc.
3. While swinging into this orbit, the orbits of Mecury and Venus would also be altered. If The Earth and Mercury were to get close enough, a collision would be a significant possibility.
4. Since Earth is swinging into this orbit (0.5 AU) from its original orbit (1AU), it'll swing back out of it. Then it'll swing back in, and repeat until the orbit has stabilized.

Note: I used Kepler's laws and convservation of momentum to make simplifying assumptions. The reality is that I'd need to simulate this to come up with an accurate representation.

• The new object will need to be orbiting the sun already, to match speeds with Earth and gently come into contact. So the orbit will not change. If the mass appeared without orbital velocity it would splat which is not described. – JDługosz Jun 6 '17 at 11:54
• For divisions involving values that themselves involve divisions, \cfrac{}{} typesets much nicer than \frac{}{}. Also, I recommend \left( and \right) instead of plain parenthesis. I've edited your answer for this. – a CVn Jun 6 '17 at 14:55
• And of course, you do realize that the atmosphere is already boiling, right? Oxygen boils at around -200°C (I'm too lazy to look up the exact value) and nitrogen at a slightly lower temperature IIRC. There are very few solids suspended in the atmosphere, and certainly very few that are suspended in the atmosphere more than very briefly. So "atmosphere boiling BAD" is the exact situation we're already in. – a CVn Jun 6 '17 at 14:56
• Besides, I'm not sure you can assume Earth's (relatively circular) orbit's average orbital speed of 30 km/s when applying to a perihelion of 75 million km, as you seem to. In such an elliptical orbit, Earth would be moving substantially faster near perihelion than it would on average, let alone near aphelion, as per Kepler's second law of planetary motion. – a CVn Jun 6 '17 at 15:06
• – KareemElashmawy Jun 6 '17 at 18:25

Neutron star material, if it could be contained by means other than its own gravity; would require a cube about 390 meters on a side to equal the mass of the Earth, if it could somehow be 'placed' on the Earth.

As LSemi says, it would simply sink to the core, accelerating, and bounce back and forth for some time. (Before that, it would suck all the atmosphere from the far side of the planet.)

The bouncing would likely cause violent bending and shaking of the Earth, unpredictable randomized havoc and extreme sloshing of the crust and mantle (like a liquid), of all tectonic plates and all bodies of water. Expect massive volcanic eruptions. The disruptions will cause friction heating of massive proportions that will probably liquefy the crust, boil the oceans, and sterilize the planet. I believe all this gravitational jerking about would also throw the Earth and Moon out of their orbits.

There would be no time to admire or wonder about this; the havoc and death start pretty much immediately at the site of deposit.

Remember, it is 4000 miles to the Earth center, nothing is going to stop this thing from punching a hole through the Earth, coming out the other side, and diving back in to repeat that many times before settling down. Do not think the Earth is stiff enough to maintain its shape; it is going to be more like an anvil falling through air; the Earth material will be moving like the chaotic air turbulence you see in a wind tunnel. Nobody that saw this thing appear would live more than a few seconds to tell about it.

Well, for starters, in the least-apocalyptic scenario where the object's mass is comparable to Earth's rather than the Sun's, everything closer to the object than to the Earth's center of mass is going to start falling towards the object (this is a massive oversimplification, but it should give you an idea of what sort of magnitude of bad news this is). "Down" for these people, houses, oceans, magma pipes, and so on is suddenly more like "sideways" shading to "up".

Everything else is also going to fall towards the object, just slower. So yes, per LSerni's answer, it will end up close to the center of the Earth - one way or another.

Now, that aside, I'm guessing you'd rather handwave in some infinite structural integrity for the Earth for the duration of this, in which case I'm inclined to say yes, eventually the Earth+object duo would spiral into the sun, probably some time after the moon spiraled into the Earth, the "eventually" depending on how close the object's force is to the "sun" rather than "Earth" end of the spectrum: you've effectively doubled the Earth's mass at the low end, meaning that what was once an equilibrium of 30 km/s (Earth's orbital velocity) "forward" for every 30 km/s sunward turns into, roughly, 30 km/s^2 sunward acceleration.

(Or I could be totally botching my orbital mechanics, in which case someone please correct me)

• «everything closer to the object than to the Earth's center of mass is going to start falling towards the object » you are assuming some specific mass and size that the OP did not explain. Can you elaborate on how you interpreted it? – JDługosz Jun 6 '17 at 11:56
• The OP specified "on Earth", which I took to mean "on the Earth's surface"; "bigger gravitational force than the Earth" and "smaller force than the Sun" I took to mean "somewhere in the Earth/Sun range", and figured I'd give the least-apocalyptic end of things (the Earth-level force) a brief description before getting into the rest of it. – Stephen Voris Jun 6 '17 at 17:52
• So edit that into the Answer. (then we clean up the comments) – JDługosz Jun 7 '17 at 3:18