I have a hypothetical question about inertia.
Let's say I have an object with inertial mass a ton (2,000 lbs.), and it is sitting in my front yard, for instance. If it would suddenly become immune to gravity, i.e its gravitational mass becomes zero, what would happen to it? Now forget about whether this is actually possible, because science at its best tries to disprove itself. I just need to know what would happen to this block of iron sitting in my front yard if it would lose gravity.
Edit: By immune to gravity, I mean that it no longer is affected by gravity. It still has the same mass and everything else, it just isn't affected by gravity anymore.
I do have a reason for asking this. I said "ignore whether it's actually possible" because I know that everyone accepts it as fact that gravity can't be messed with (at least to my knowledge). However, I am the type that ask what if questions, so I asked "What if we could make an object not affected by gravity: What would happen to it?"
Translated into technical terms (from a comment): In the context of Newtonian mechanics, let there be a test particle with inertial mass $X$ but no gravitational mass, i.e., the test particle does not gravitate. The test particle is released from my front yard at t = 0. Describe the motion of the test particle.
Also, if you give technical answers (a good idea), please explain them in layman's terms, if that isn't asking too much. I more or less understand technical answer according to how they read in layman's terms.
Best Answer
Short answer: It would have an initial acceleration of $\omega^2R\cos\theta$, directed at an angle of $\theta$ down from the local azimuth. The direction will be towards the south if you're in the northern hemisphere, and towards the north if you are in the southern hemisphere. $\omega = 2\pi\mathrm{radian/day}$ is the angular rotation of the earth's rotation, $R = 6371 \mathrm{km}$ is the radius of the earth, and $\theta$ is your latitude.
Explanation: Because the earth rotates, your front yard is not an inertial reference frame. There are thus four pseudo forces that appear in that reference frame: the centrifugal force, the Coriolis force, a force due to any changes in the rotational speed of the earth, and a force due to the translational acceleration of the earth (along its orbit around the earth, for example). The last two terms are both small when applied to the earth, so we ignore them. The Coriolis force is proportional to the speed of the mass. The mass starts at rest, so that term is initially zero. When the mass starts moving, you will have to take it into account. That all leaves the centrifugal force, which is $-\vec\omega\times[\vec\omega\times(\vec r + \vec R)]$. Here, $\vec r$ is the displacement vector within your front yard,and $\vec R$ is the vector from the center of the earth to the origin you are using in your front yard. $r << R$ near the surface of the earth, so ignore $\vec r$. The result I quote above comes out of those assumptions. If you plug in the actual values, you get an effect "negative gravity" equal to about $0.3\%$ of what its weight is in the real world. That is the force you would need to tie the mass down.
Caveats: