gravity - Why does my apple not weigh 500 tons?

Related to this question: What is potential energy truly?

$E=mc^2$ - energy equals mass. So, if an object has gravitational potential energy relative to another object, it should have additional mass to represent that extra energy.

So, if I have an apple in my hand, it should have quite a large amount of gravitational potential energy relative to a black hole at the edge of the observable universe. I did a quick calculation of how much this gravitational potential energy would be (using Newton's Law of Gravitation) and it came out to be the equivalent of around 500 tons (and that's just for one black hole!).

So, why does my apple not weigh a lot more than it appears to?

EDIT: Here's my calculation of the GPE:

Assumptions:

1. The apple has mass $m_a=0.3 kg$


2. The black hole has a mass equal to 1000 solar masses, $m_b=2\times10^{33} kg$


3. Distance between the apple and black hole is 60 billion light years

Obviously $GPE=mgh$ can't be used, because the gravitational acceleration is not constant and varies with distance, so it has to be integrated. I used Newton's Law of Gravitation to provide the acceleration:

$$ F=\frac{G{m_1}{m_2}}{h^2} $$

A small change in height produces a small change in energy, which can be used to integrate:

$$ \begin{eqnarray*} \delta E&=&F \delta h\\ \int_0^EdE&=&G{m_a}{m_b}\int_1^R\frac{1}{h^2}dh \end{eqnarray*} $$

I'm integrating the force starting at 1m from the center of the black hole, as the force goes to infinity at $h=0$. It's a little crude, but it should provide a conservative estimate.

$$ \begin{eqnarray*} E&=&G{m_a}{m_b}(1-\frac{1}{R})\\ E&\approx& G{m_a}{m_b}\;\;\;\;\;\;(R>>1)\\ G&=&6.673\times 10^{-11}\;\;N(\frac{m}{kg})^2\\ G{m_a}{m_b}&=&4\times 10^{22}\;\;Nm\\ E&=&mc^2\\ \Delta m&=&\frac{4\times 10^{22}}{(3\times 10^8)^2}\\ &\approx&444,000\;\;kg \end{eqnarray*} $$

So, a little short of 500 tons, but still pretty big!

Answer

You are reading the formula the other way round. With $$\tag{1}E = mc^2$$ we mean that the energy associated to the mass $m$ of a particle at rest is $E=mc^2$. In oher words, (1) strictly holds for a particle at rest, not interacting with anything else.

What about all the potential energy that a particle has? That must be included on the right hand side of (1). So the version of (1) including potential energy would be $$ \tag{2} E = mc^2 + E_g,$$ with $E_g$ denoting all the gravitational potential energy contributions that you want. So you can now replace $E_g$ with your calculated gravitational potential energy (and then of course all the gravitational energy contributions coming from the interaction of the apple with the rest of the universe... good luck with that).

Now it is crucial to understand that (as also pointed out by Phoenix in the comments) energy is always defined up to a constant, and what physically matters in the dynamic of a particle (or an apple) is the change in the energy between two states (say the apple at point A and the apple at point B). So $E_g$ can be as big as you want (it can be infinite for that matter), and that would not change anything in the dynamic of the apple. This is way it is not generally included, unless the gravitational energy plays a role in the calculations. You could think of a more general version of (1) as being something like $$ \tag{3} E = \sqrt{m^2c^4 + c^2p^2} + E_{gravitational} + E_{em} + E_{whatever} + C, $$ where $E_{em}$ keeps count of the electromagnetic interaction of the object with the rest of the universe (if there is one), $E_{whatever}$ keeps count of all the other possible interaction you can think of, and $C$ is a constant that you can always choose arbitrarily. Obviously, writing (1) like (3) would be in most cases useless, and hence only the relevant contributions are included.

So to conclude, the mass in $F=ma$ keeping count of relativistic corrections is just $$ \tag{4} m = \gamma m_0 \equiv \frac{m_0}{\sqrt{1-v^2/c^2}},$$ where $\gamma$ is the Lorentz factor, $m_0$ the rest mass used in (1), and $v$ the velocity of the particle. Einstein's formula is meant to tell you the amount of energy that you would obtain by completely disintegrating the mass of a particle, not as a way to calculate the inertial mass of an object.