Is it to think that the earth mass is somewhat less than sum of it's parts due to the increase in potential energy of the earth due to gravity?
If so, where does that potential energy "go" if it "goes" anywhere?
If so, where does that potential energy "go" if it "goes" anywhere?
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I think so, however you have to think about where that energy went. In atomic nuclei, the binding energy corresponds to a decrease in mass, but that's because nuclei won't stay in an excited state. When the nucleus forms, the binding energy is released to somewhere. Think about where the energy went when the Earth was formed. It went mostly to heat of formation. Much of it is still there, though some of it has been radiated.
Energy that's still there contributes a positive amount to the mass. So even though the negative potential energy detracts from the mass, some of it is added back by the remaining heat energy that resulted from that potential energy.
Let's get an idea of the magnitudes of energy involved.
I found this formula for the potential energy of a sphere of uniform density. The Earth does not really have uniform density, but I think the result of this formula will be off by less than a factor of 10, at most.
U = -(3/5)GM^2/R
G = 6.67428 * 10^-11 m^3/kg*s^2
Me = 5.9736 * 10^24 kg
Re = 6.371 * 10^6 m
U = -2.243 * 10^32 J
Quite a bit, but how about the rest energy of the Earth?
E = mc^2
Me = 5.9736 * 10^24 kg
c = 299,792,458 m/s
E = 5.3688 * 10^41 J
There are nine orders of magnitude between the two. The potential energy of the Earth is 0.0000001% of the rest energy.
When looking at this problem, I found it interesting that the formula for potential energy is proportional to the mass squared, whereas the formula for rest energy is proportional to mass, directly. That means for a fixed radius, if you keep increasing the mass, the potential energy will eventually reach the rest energy. Is that the mass for which that fixed radius is the Schwartzchild radius? Hmm.
Energy that's still there contributes a positive amount to the mass. So even though the negative potential energy detracts from the mass, some of it is added back by the remaining heat energy that resulted from that potential energy.
Let's get an idea of the magnitudes of energy involved.
I found this formula for the potential energy of a sphere of uniform density. The Earth does not really have uniform density, but I think the result of this formula will be off by less than a factor of 10, at most.
U = -(3/5)GM^2/R
G = 6.67428 * 10^-11 m^3/kg*s^2
Me = 5.9736 * 10^24 kg
Re = 6.371 * 10^6 m
U = -2.243 * 10^32 J
Quite a bit, but how about the rest energy of the Earth?
E = mc^2
Me = 5.9736 * 10^24 kg
c = 299,792,458 m/s
E = 5.3688 * 10^41 J
There are nine orders of magnitude between the two. The potential energy of the Earth is 0.0000001% of the rest energy.
When looking at this problem, I found it interesting that the formula for potential energy is proportional to the mass squared, whereas the formula for rest energy is proportional to mass, directly. That means for a fixed radius, if you keep increasing the mass, the potential energy will eventually reach the rest energy. Is that the mass for which that fixed radius is the Schwartzchild radius? Hmm.