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LSerni
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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.

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

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.

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JDługosz
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Imagine a sphere with a radius of one meter. Its surface gravity would (numerically) be 6.67 *× 10-11−11 times its mass, so to have a surface gravity of 1 g (9.81 ms-2−2) 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-2−2 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).

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.

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.

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).

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.

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.

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).

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.

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.

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LSerni
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Same pull, but low mass

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.

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).

Same pull, but low mass

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.

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LSerni
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