Does an object lose mass when raised against a gravitational field?

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When an object (say a ball) is pulled some distance away from another object (say the Earth) against the objects' mutual gravitational attraction, does the mass of each object decrease? (Assume that the energy expended in moving the Earth and ball away from each other comes from within the system, and not from some external "hand of God".)


My reasoning: Since no energy is added to or taken from the Earth-ball system, the total energy of the system is constant. The potential energy of the system increases. Therefore, some other type of energy within the system must decrease by an equal amount. I presume this to be its mass-energy. This would make "gravitational binding energy" analogous to nuclear binding energy.

Tentative conclusion: The total energy of the system remains constant. The potential energy of the system increases. The mass-energy of the system decreases by an equal amount. Is this correct?

Further question: Newton's law of universal gravitation states:

[tex]\vec F = - G\frac{{{m_1}{m_2}}}{{{r^2}}}\hat r[/tex]

In order to calculate the work done in moving the masses [itex]m_1[/itex] and [itex]m_2[/itex] from separation [itex]r_1[/itex] to [itex]r_2[/itex], I was taught to simply integrate Newton's force expression between the distance limits, like so:

[tex]W = - \int_{{r_1}}^{{r_2}} {\vec F \cdot d\vec r} = G{m_1}{m_2}\int_{{r_1}}^{{r_2}} {\frac{{dr}}{{{r^2}}}} = G{m_1}{m_2}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)[/tex]

This calculation treats the masses [itex]m_1[/itex] and [itex]m_2[/itex] as invariant. However, if I was correct in my earlier presumption, the masses [itex]m_1[/itex] and [itex]m_2[/itex] are not invariant, but vary inversely with [itex]r[/itex]:

[tex]{m_1} = {m_{1,0}} - U[/tex]
[tex]{m_1} = {m_{1,0}} - U[/tex]

where [itex]m_{1,0}[/itex] and [itex]m_{2,0}[/itex] are the masses of [itex]m_1[/itex] and [itex]m_2[/itex] at infinite separation, and [itex]U[/itex] is the gravitational potential energy between the two masses: [itex]U = - G\frac{{{m_1}{m_2}}}{r}[/itex]. This would seem to suggest that the equation [itex]W = G{m_1}{m_2}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)[/itex] is only an approximation.

Is any of this correct?

-------
EDIT: "U" should be "U/c²"
 
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Answers and Replies

  • #2
Drakkith
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The mass of something never increases or decreases unless you add or subtract part of the object. Merely moving it does nothing to its mass.

My reasoning: Since no energy is added to or taken from the Earth-ball system, the total energy of the system is constant. The potential energy of the system increases. Therefore, some other type of energy within the system must decrease by an equal amount. I presume this to be its mass-energy. This would make "gravitational binding energy" analogous to nuclear binding energy.
The decrease in energy you are talking about was using the energy required to put that object at whatever distance from the earth against gravity. You have to include whatever you used to move the object as part of the system.
 
  • #3
When an object (say a ball) is pulled some distance away from another object (say the Earth) against the objects' mutual gravitational attraction, does the mass of each object decrease? (Assume that the energy expended in moving the Earth and ball away from each other comes from within the system, and not from some external "hand of God".)
I think you mean the mass will increase, although I'm not sure if it is the objects or the mass of the gravitational field that will increase. This might be powered by the decrease in mass that happens in a nuclear reactor, for instance.
 
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Thank for you your reply, Drakkith. So I was wrong when I said:
The potential energy of the system increases. Therefore, some other type of energy within the system must decrease by an equal amount. I presume this to be its mass-energy.
The mass-energy is invariant, then?

Could you help me understand this in terms of the related phenomenon of nuclear binding energy? Suppose a system consists of one neutron and one proton, separated by some large initial distance, [itex]r[/itex], and approaching one another at a very slow speed. The initial mass of the system is [itex]m_p + m_n[/itex]. Kinetic energy is negligible. The initial mass-energy of the system is therefore [itex]({m_p} + {m_n}){c^2}[/itex]

The two particles continue to approach each other with approximately zero acceleration until, eventually, they come within range of the nuclear strong force. At this point I am not sure what happens. I would guess that they "snap" together, forming a deuterium nucleus.

Here is the part I don't understand. A deuterium nucleus has more mass-energy than the total mass-energy of a proton plus a neutron. But no energy was added to this system, and none was taken away. It would appear to me that the system's mass-energy has increased and its potential energy has decreased, but the sum of {mass-energy + potential energy} has remained constant. What am I missing?
 
  • #5
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T

Here is the part I don't understand. A deuterium nucleus has more mass-energy than the total mass-energy of a proton plus a neutron. But no energy was added to this system, and none was taken away. It would appear to me that the system's mass-energy has increased and its potential energy has decreased, but the sum of {mass-energy + potential energy} has remained constant. What am I missing?
Well, i'm not sure about deuterium*, but helium nucleus has less energy than rest mass of two proton and two (? i think there are two :p) neutrons. In process of "binding" I think (more-less kinetic) energy is taken away (probably in form of radiation). That is reason why we are trying to run a nuclear fusion to harvest energy, isn't it?

In your example with gravitational field the particle getting far from centre on its own (using it's kinetical energy) shouldn't change it's mass since it's energy is constatnt.

*Edit: forget it, I thought you wrote " total mass-energy of a proton plus a ELECTRON" :) i should get some sleep
 
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  • #6
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Suppose a system consists of one neutron and one proton, separated by some large initial distance, [itex]r[/itex], and approaching one another at a very slow speed. The initial mass of the system is [itex]m_p + m_n[/itex]. Kinetic energy is negligible.
I missed that (i should better quit anyway, it's quite late here).
Here I think your assumtion that speed of approach will be very small through entire process is wrong. Kinetic energy won't be negligible, at least surely not against binding energy of deuterium atom.
 
  • #7
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Moreover I think there is no mass-energy plus potential energy or anything. There is just energy, and whole energy of system equals it's mass (times c^2). I mean if you have closed system then sum of rest masses of all particles, their kinetic energy and energy of interactions should compose "mass of the system".
But this is one of things in which I feel I have huge holes in my knowledge. For example I think change of electrostatic potential energy is equal to change of energy of electric field, so you shouldn't count them both in sum of total energy. It seems confusing to me.
Takes my comments more like attempt for discussion between two people not knowing something, and not like definitive answers.
 
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  • #8
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Thanks, FroChro. Whoops, you are correct about the helium nucleus. And, after now looking up the mass of the deuterium nucleus, I see that it is indeed lighter than {proton + neutron}. A very bad oversight on my part to get that the wrong way round.

So you are most probably correct in supposing that, when the proton and neutron bind, they emit some energy as radiation.

I think I need to re-think my entire question. Something for the morning! Thanks again.
 
  • #9
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When an object (say a ball) is pulled some distance away from another object (say the Earth) against the objects' mutual gravitational attraction, does the mass of each object decrease?"
To my knowledge, mass "decrease" has never been shown to exist without the removal of actual matter.
 

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