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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".)

[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?

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EDIT: "U" should be "U/c²"

**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?

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EDIT: "U" should be "U/c²"

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