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<sup>−11</sup> times its mass, so to have a surface gravity of 1 g (9.81 ms<sup>−2</sup>) it would need to weigh 1.4 × 10<sup>11</sup> kg, or 140 million tons; about the weight of 1400 [Nimitz-class][1] 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<sup>−2</sup> is impossible unless one employs some technological tricks; it is a consequence of [Earnshaw's Theorem][2], the same reason why you *cannot* gently float an object using a magnet or a charged plastic stick *([active control][3] is a tech trick and using [gyrostabilization][4] 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 13<sup>2</sup> = 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][5]. 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/cm<sup>3</sup>). 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. [1]: https://en.wikipedia.org/wiki/Nimitz-class_aircraft_carrier [2]: https://en.wikipedia.org/wiki/Earnshaw%27s_theorem [3]: https://www.youtube.com/watch?v=Km6N8PXpj2c [4]: https://en.wikipedia.org/wiki/Levitron [5]: https://www.youtube.com/watch?v=ntQ7qGilqZE