# Gravitational potential energy and center of mass

I've been reading about the Nordtvedt effect, and how Gravitational Binding Energy (GBE) affects total mass. According to the WP article, experimental evidence rules out the existence of this effect.

I read this result as meaning that the gravitational potential energy is accounted for, both as gravitational mass and inertial mass.

The problem I am attempting to understand is the direction of the force created by GBE.

Gravity is a force, which has a location and a direction in space oriented toward where the gravity field is created. If GBE creates a gravitational force, what is its direction?

Consider an example of two gravitationally bound masses: the Earth-Moon system. I am on the surface of the Earth, so I feel mostly the gravity of the Earth, but I also feel the gravity of the Moon. The GBE of the Earth itself will be accounted for as part of the Earth's mass, so it is nothing special. A similar situation accounts for the mass and GBE of the Moon. But what about the gravitational potential energy of the Earth-Moon system? According to my calculations, that energy is on the order of $10^{28} J$, which is equivalent to a mass of about $10^{11} kg$, which is quite significant.

Where will this extra gravitational force pull me? I can think of two reasonable possibilities:

1. This extra gravity will exert a force of attraction proportional to the masses that participate in the system. The Earth would feel slightly more massive than it would be without the Moon. The same for the Moon, which will feel proportionally more massive because it is gravitationally bound to the Earth.

2. This extra energy will attract to the center of mass of the system. That extra force could be felt, but it is not available for experimental validation because the center of mass of the Earth-Moon system is well under the Earth's crust.

About possibility 2, we could check another bigger gravitationally bound system: the Galaxy!

The mass of the galaxy is about: $6\times 10^{42} kg$, and its radius is about $4.8\times 10^{20} m$. So its GBE, assuming a spherically uniform distribution (quite a gross estimation), would be:

$$U = \frac {3GM^2} {5r}$$ $$U \approx 3 \times 10^{54} J$$

Converting that $U$ to the equivalent mass, and measuring it in solar masses (1 solar mass $M_\odot = 1.98\times 10^{30} kg$)

$$U \approx 1.7 \times 10^7 M_\odot$$

So, if we look at the center of mass of the Galaxy, we should find a gravitational force equivalent to that of an object of several million solar masses.

Now the question: What does the mainstream physics community think of this problem? What is the direction of the gravitational force produced by gravitational binding energy?

• The formula for GBE you are using is valid for an object of uniform density, which is obviously very wrong for a galaxy. A few other remarks: (1) there is no unit in your values. (2) It would be much more readable (and correct) if you wrote $U\simeq 1,68\times 10^8\,M_\odot$. – Tom-Tom Sep 17 '14 at 12:12
• @V.Rossetto: Obviously my GBE calculation is a gross oversimplification. A big error in the final value is expected, but here I'm interested in the order of magnitude of the result: is the result comparable to a moon or to a million suns?). About the others: (1) do you mean the missing $J$ (added)? (2) What do you mean exactly? the $\simeq$ vs $\approx$ or the scientific notation? – rodrigo Sep 17 '14 at 12:22
• You use 8 significant digits for a result that is based on values with one or two significant digits. It is wrong to do it because in physics, the number of significant digits provides an information concerning the accuracy of the value. – Tom-Tom Sep 17 '14 at 12:27
• @Tom-Tom Ah, that. Very true. Corrected – rodrigo Sep 17 '14 at 12:29
• @Floris: Ops! Corrected, thanks! Actually $10^{17}$ came simply from $c^2$. I forgot to do the division. – rodrigo Sep 17 '14 at 13:14