# The impossible possibility - weightless mass

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.

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Re your edit: I would then pose the question this way: 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. – Alfred Centauri Aug 8 '12 at 12:32

## 3 Answers

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:

1. My answer above only works initially. Once the mass is moving, you have to account for the Coriolis force.
2. By "immune to gravity," I'm taking you to mean that it no longer interacts with the earth's gravitational field. So I'm simply dropping the real force from the equation for the effective force in a noninertial reference frame. If you meant "has no gravity," the effective force is zero, because all the pseudo forces are proportional to the mass. But the acceleration is still what I quoted above, because that just comes from the noninertal nature of the reference frame.
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So...What exactly does this mean in Layman's terms? – Arlen Beiler Aug 8 '12 at 12:55
@ArlenBeiler The rock flies out into space unless you tied it down. – Alan Rominger Aug 8 '12 at 12:59
@ArlenBeiler in layman's terms it means it would appear to accelerate off into space. This is because it keeps on moving in a straight line, whereas objects on the Earth's surface move in circles due to the Earth's rotation. I would add that it would also experience a buoyancy force due to being effectively lighter than air, so it would rise into the air for that reason as well. – Nathaniel Aug 8 '12 at 13:01
@ArlenBeiler The false upward acceleration is equal to 0.3 % of normal gravity. The buoyant force will be a little bit less than that, since if it's the density of water, the buoyancy will be about 1/1300 of its normal weight. So if its mass is $1000 kg$, it would have a "negative weight" of $3.37 kg$. The force you need to tie it down would be $33 N$. – Alan Rominger Aug 8 '12 at 13:32
@ArlenBeiler the buoyancy force would depend on its volume, just as for a helium balloon, rather than its original weight. If it's very big in size then you'll be able to use it to lift yourself. On the other hand, the "false acceleration" force is kind of interesting. The rate of acceleration won't change depending on its original mass, but its intertial mass will determine how easy it is to change its motion. If it's light then you can hold it down (forcing it to move in a circle along with everything else), but it's very heavy then yes, you'll be able to ride it into space. – Nathaniel Aug 8 '12 at 13:39

I like Alfred's answer because it gets to the point that any answer is equally as valid. If we are ignoring what is theoretically possible, the answer "it turns into an unsuspecting bowl of petunias" is perfectly valid and every bit as meaningful as any other answer.

What I expect you would like to hear, however, is the answer to the question "If the gravitational mass of the iron piece suddenly became effectively zero, what would happen?"

In that case, the answer is, the inertial mass of the block would cause it to appear to fly at a path tangential to it's path of rotation about the axis of the Earth. Depending on where you were, that would look very different; on the poles, for example, the block would appear to start spinning. Additionally, there would be an force tangential to the path of the Earth's orbit about the Sun. Due to the relatively low angular velocity (one rotation per year), this force would be significant smaller than the force causing spinning (one rotation per day). A similar force would apply for the tidal force caused by the moon and indeed for every other object in the universe. The big hitters, however, are the rotation of Earth and the orbit about the Sun.

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Interesting.

The obvious answer is that it will float as if it was in vacuum. It will hit other objects and they will either crash to small bits or will normally defer in different directions.

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Will it only float, and continue circling the earth like it was? If so, why on earth? (pardon the pun) – Arlen Beiler Aug 8 '12 at 1:25