# Thought experiment regarding gravity

Let there be a mass $$M$$ which is a spherical shell of radius $$R$$.

Now the Gaussian flux about a enclosing sphere of radius $$r$$ where $$r>R$$ will be proportional to $$M$$ or $$GM$$ (where $$G$$ is the gravitational constant).

Now if this mass were distributed to R' where R'>R some work will have to be done so we can use some part of this mass M itself,convert to energy and be left with M' where M' < M.

However a person outside this whole arragement , say at a distance D where D> R and D > R' should not be able to feel any difference by this change.(This argument is similar to the hypotheses of there being no experiment to distinguish between free fall and uniform motion without any influence of gravity.) So the mass that this person sees will be different from both M and M'.

Does it imply that the mass one sees is dependent on the distance from that mass object? and also does it imply that gravitation field falls faster than 1/r^2?

• If an observer of mass m is at distance d from a mass M, and then he moves to a distance d'(d' > d), there has to be work done to raise his altitude and lets say this work or energy is sourced from M, leading to a drop in M to M'. However no local gravitational measurement would change for the observer is my hypotheses. So he still sees the original mass M but when getting closer to the other object we see that the mass is slightly lesser than M. Whats wrong with this argument?
– Ajay
Commented yesterday

Now if this mass were distributed to R' where R'>R some work will have to be done so we can use some part of this mass M itself,convert to energy and be left with M' where M' < M.

The key error in the reasoning lies here. In this statement you are trying to misuse the famous equation $$E=mc^2$$. It does not describe converting mass to energy; conversion would be $$\Delta E=-\Delta m c^2$$. The famous equation describes the equivalence of mass and energy. Mass does not need to be converted to energy, it already has energy. Similarly energy has mass, no conversion needed.

Note that the famous equation only applies for a system with no overall momentum. This system meets that criterion, so it applies. Therefore, no amount of internal rearrangement can change the mass or energy of the system. So $$M’=M$$.

The only way for $$M$$ to change would be for mass or energy to cross the boundary of the system. If it accretes some dust then $$M$$ will increase as will the energy. If it radiates some EM waves then the energy will decrease as will $$M$$.

Does it imply that the mass one sees is dependent on the distance from that mass object? and also does it imply that gravitation field falls faster than 1/r^2?

No. None of the subsequent reasoning holds once the error in the use of the famous $$E=mc^2$$ equation is corrected.

• I see that I downvoted this, but it must have been accidental. I tried to correct that, but apparently I can't change my vote unless/until this is edited. My apologies! Commented 8 hours ago
• @WillO no worries. Mis-clicks happen
– Dale
Commented 2 hours ago
• Corrected now!.. Commented 2 hours ago
• @WillO Thanks 😊
– Dale
Commented 2 hours ago

First of all, your gedankenexperiment is actually fairly realistic (if we consider spheres instead of shells). Imagine for example a star in the center of which occurs nuclear fusion. This is literally the conversion of part of the (rest) mass of the "fuel", e.g. hydrogen, into energy in the form of radiation. As this happens at the center of the star, the radiation can't escape immediately and exerts pressure. Usually, a star is in equilibrium, meaning that the radiation pressure together with the gas pressure stabilizes the star against gravitational collapse. However, in some situations (through very complicated processes), the internal pressure gets higher than the gravitational attraction such that the star expands, for example at the end of its life when it turns into a red giant. Some stars even change their radius periodically.

The solution to your problem lies in the fact that we have to be very careful which mass we are actually talking about. For one thing, there is the so called baryonic mass, which is just the sum of all the masses of the constituents of the star. Imagine taking the star apart into its atoms, measuring their individual masses and adding them. But this is not equal to the gravitational mass of the star, which is used in Newton's gravitational law to calculate the gravitational field that an outside observer measures. The discrepancy between those two values stems in fact again from mass-energy equivalence. The gravitational mass of the star corresponds to its total energy, which includes the rest mass of its constituents, but also its binding energy (and other forms of energy like the heat trapped inside and all other kinds of interaction energy between the constituents, but I will neglect this here). To put this into an equation: $$$$M = M_B + E_B$$$$ where $$M$$ is the gravitational mass, $$M_B$$ is the baryonic mass and $$E_B$$ is the binding energy (note that I choose the signs such that the binding energy is negative). In our example, the star expanded, which just means that it raised its binding energy. In other words, it converted part of its baryonic mass into binding energy. However, as we neglected any incoming or outgoing energy flux, the gravitational mass of the star (which, again, corresponds to its total energy content) stays constant! So an outside observer won't feel a difference, he will still measure the same gravitational mass.

Finally, let me note that we're talking about small effects here, at least considering "normal" stars. They stem from general relativity, which makes sense, as we used mass-energy equivalence, which is a relativistic effect. The situation we discussed is described by the Tolman-Oppenheimer-Volkoff equation. You can find a more mathematical discussion of this topic, including the different "kinds" of masses in the referenced Wikipedia article.

• I chose a shell so that there is symmetry across each point of the mass..in a sphere, the varying factor would be distance from the centre
– Ajay
Commented 15 hours ago

Let's look at this mathematically. Let $$\vec{g} \equiv \vec{g}(r)$$ be the gravitational field. Then Gauss' law in its differential form states

$$\operatorname{div} \vec{g} \enspace = \enspace - 4\pi G \rho$$

where $$G$$ is the gravitational constant and $$\rho$$ is the mass density. Integration of this equation over a ball of radius $$r$$ gives for the left-hand side

$$\int_{B(r)} \! dV \,\operatorname{div} \mathbf{g} \enspace = \enspace \oint_{S(r)} \! d\vec{A} \cdot \vec{g} \enspace = \enspace 4\pi r^2 \, g(r)$$

and for the right-hand side

$$-4\pi G \int_{B(r)} \! dV \, \rho \enspace = \enspace -4\pi G M$$

with $$M$$ being the total mass. Together one finds

$$g(r) \enspace = \enspace - \frac{GM}{r^2}$$

This means, assuming the observer is located at distance $$D > R' > R$$, that the radius $$R$$ of the shell has no influence on the field experienced by the observer. Therefore it does not matter whether the shell has radius $$R$$ or radius $$R'$$, as long as both are smaller than $$D$$. The only thing the observer would experience, is a change in mass. Assuming the system somehow reduces in mass from $$M$$ to $$M'$$, the observer will experience a field

$$g(r) \enspace = \enspace - \frac{GM'}{r^2}$$

so that indeed he sees the "new" mass $$M'$$ as it is.

I still think the mass of the sun would be different based on gravitational field of the Sun detected at say Mercury vs Pluto. This because the gravitational field has negative energy. So as one travels from Mercury to Pluto, the total Gaussian flux across an enclosing sphere would keep falling as the distance from the Sun increases. This is also expected if we consider conservation of energy. Maybe my original question was not rightly phrased. Apologies.