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The mass of a helium nucleus is less than the mass of two isolated protons and two isolated neutrons. When the component hadrons are assembled, this mass is lost as energy ($E=mc^2$). This makes it (at least sort of) seem like mass is a kind of potential energy. Matter and antimatter can annihilate each other, releasing energy ($E=mc^2$). This also makes it (at least sort of) seem like mass is a kind of potential energy.

If I have a mass at approximately sea-level on Earth, will flying it away from Earth increase its mass? (If so, is this related to relativistic mass?)


The original question doesn't really ask what I mean to ask. Suppose that, after the mass has been flown into space, it is stopped relative to Earth. (I think that would require a geosynchronous orbit.) Would its mass be different than it was before it left Earth?

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It is all about nuclear forces and those act only on very short (nuclear size) distances. So, no, you don't see those effects when moving macroscopic objects. – Pygmalion Jun 5 '12 at 18:06
Mass can be an elusive quantity in GR, but I think there is a sense in which the effective mass of an object increases as it falls into a Schwarzschild metric. But as with special relativity, it's not a real change. It's because you and the mass are using different co-ordinates: Schwarzschild co-ordinates for you and shell co-ordinates for the mass. I haven't put this in as an answer because it's unhelpful at best and actively misleading at worst :-) – John Rennie Jun 5 '12 at 20:00
Mass has the same mass at high altitude, whether it was lifted there, or thrown there. Mass of a mass that has just left the hand of the thrower has been increased by the throwing. Mass of a mass that a hand has just started lifting, has not changed. Either thrown mass must lose mass, or lifted mass must gain mass, during the travel. – kartsa Jun 6 '12 at 19:02
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Yes. Suppose you have a plate at ground level. If you lift that plate up to a certain height, then the mass of the system Earth-plate is increased.

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Do you have any sources or justification for why this is? – yakiv Jun 6 '12 at 19:16
It's easy to understand why. Imagine that lifted plate is falling to earth. Just before it hits the ground the plate is moving at a some velocity. So at ground level the total energy of the system Earth-moving_plate more than total energy of the system Earth-motionless_plate. – voix Jun 6 '12 at 20:04
If you were to measure the mass of the Earth-plate system for cases in which the plate was in the air versus on the ground, where would you find the extra mass in the first case compared to the second? That is, would you find it in the Earth, the plate, the spacetime in which they live, or none of the above? Also, if the answer is that the mass is somewhere else other than the plate, then strictly speaking wouldn't the answer to the OP's question be 'no'? – kleingordon Jun 6 '12 at 21:59
I think the confusing part of this question is that it was posed in terms of the mass of the plate alone, not the mass + plate system. In general relativity, the mass-energy contributions of components of the system depend on the reference frame. I've posted an answer to elaborate on this, with a link to my source. – kleingordon Jun 6 '12 at 22:24
@voix: If the energy used to lift the plate above the Earth came from the Earth-plate system as it was before the lifting occurred (say, from fossil fuel), would the mass of the Earth-plate system be the same after the lifting as it was before? – yakiv Jun 11 '12 at 5:29

It depends on your reference frame. In the frame of the plate, its mass (strictly speaking, its mass-energy in its local freely falling frame, when you think about the problem in terms of general relativity) will be the same no matter what height it is at. However, for an observer located tens of thousands of kilometers above the Earth's surface, the mass-energy contribution of the plate to the total mass-energy of the Earth + plate system will be greater when the plate is elevated than when it is on the surface of the earth.

To lend credence to this answer, I direct you to Lubos Motl's response to a question of mine that is similar to this one: What is the mass of individual components in a gravitationally bound system? Some of Lubos' most important explanations come in the comments to his answer. In particular: "the locally measured mass/energy in a gravitational field isn't the same thing as the contribution of this mass/energy to the total mass/energy as seen from infinity. Roughly speaking, the two quantities differ by the multiplicative constant $\sqrt{g_{00}}$, related to the gravitational potential..."

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No. Mass is a kind of energy, but it is a different kind of energy than gravitational potential energy. As you move an object away from the earth, the gravitational potential energy changes, but the mass does not.

One caveat though: in the preceding paragraph, as in general, it is implicitly assumed that "mass" means "rest mass," namely the mass as measured at rest with respect to the object and very close to the object. If you make the same measurement when you are moving with respect to the object, or when you are reasonably far away, you may get a different result.

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You can gain some insight for this problem by first considering the Einstein Equivalence principle. Imagine that the mass is at a height $r=R+h$ and move into a freely falling frame whose origin coincides with the mass. From your reference frame it simply looks like the mass is accelerating away from you at a rate of $g\,\text{m}/\text{s}^2$. Just before you hit the ground at $r=R$ and $t = \sqrt{\frac{2h}{g}}$ the mass is moving away from you with velocity $v= g\times\sqrt{\frac{2h}{g}}$ and thus according to special relativity the total mass-energy of the body is $$E = \frac{m_0c^2}{\sqrt{1 - \frac{2gh}{c^2}}}$$ Now let's expand this to first order in $\frac{gh}{c^2}$ to get $$E \approx m_0c^2\left(1 - \frac{gh}{c^2}\right) = m_0c^2 + m_0 g h$$ or that $$m_\text{rel} = m_0 + \frac{m_0 gh}{c^2}$$ We see that the mass-energy of the body is directly proportional to its height measured from the Earth. It looks like the Newtonian potential energy is contributing to the relativistic mass of the body, but this statement is incompatible with general relativity as the concept of gravitation potentials are non-existent. Rather the true reason for this difference is gravitational redshift. The intimate connection between energy and time evident in the definition of the four-momentum result in gravitational time dilation being responsible for the measured difference in energy.

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People are mistaking mass for force. An object moving does not have more mass, it has inertia, and moving it away from the earth weakens the gravitational attraction between the earth and the object, causing a weight change. However, weight is a measurement of gravitational force on between two bodies containing mass. Mass, however, cannot and does not change. It can be added to or taken from by adding or removing matter, but even then, it is only located elsewhere.

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But its mass changes as well, due to relativistic effects (moving away from earth means we give him potential energy, this energy increases its mass). – peterh Jul 10 '15 at 21:02

Yes. If you take a mass and accelerate it it's mass will be its relativistic mass. But remember with relativity everything is "relative". If you say give a velocity (relative to earth) to a rocket, but you are on the rocket measuring its weight you'll measure the rest mass since the rocket is at rest relative to you. Relative to an observer on the ground thought he mass of the rocket will have changed.

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I wasn't meaning that sort of increase in mass, though I guess my questions doesn't indicate that. I'll edit my question to specify what I mean. – yakiv Jun 5 '12 at 18:25

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