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Starting with $E=mc^{2}$, I solved for $m$ and plugged the result into $W=mg$.

Which was very confusing, as objects at rest had no weight:

$$W = (e/c²)\cdot g \implies W = (0/c²)\cdot g \implies W = 0$$

I would hypothesize that I am either applying a specific-use formula to applications for which it was not intended; or making other false assumption(s) based on ignorance. Hopefully gravity is not a Matrixesque illusion, I only have 2 months of physics under my belt. Anyway, there's a question in there somewhere.

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    $\begingroup$ Watch out with the interchanging between $E$ to $e$ as the same thing. Confusing matters. $\endgroup$ – Steeven Jan 28 '16 at 10:29
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The mistake is that bodies at rest don't have zero energy. The relativistic energy is given by

$$ E^2=p^2c^2+m^2c^4 $$

where $p$ is the body's momentum and $m$ its rest mass. Hence, if the object is at rest, $p=0$, so

$$ E=mc^2 \rightarrow m=\frac{E}{c^2}\neq 0 $$

and

$$ W=mg=\frac{E}{c^2}g\neq 0 $$

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In physics, you always have to know the meaning behind the formula. The formula itself is meaningless, it's just a way of writing the meaning in a short and universal form. Even if you have two variables that describe the same quantity, but in a different context, equating them will probably lead to nonsense.

In this case, E is precisely the REST energy of an object: the energy contained in its mass. So this is the meaning here. Objects not at rest aren't even described by your formula - you have to add a factor $\gamma=(1-(v/c)^2)^{-1}$, and in the first approximation for small velocities, the difference between the rest energy and the full energy gets you to the expression for kinetic energy that you probably know.

Your first formula therefore just converts the rest energy into mass (basically, just different units for the same thing - in particle physics, masses are even specified in units of energy). It's a conversion of units, nothing more.

The second formula just talks about weight-mass conversion for a given gravitational acceleration. While on earth, both tell approximately the same information and people even use them interchangeably, even though mass is a property of the object (how much matter is there), independent on its location, while the force of gravity is a vector that wants to set the mass into motion.

Plugging mass from first equation to the other means just: how much a stationary object weighs on earth, if you know its rest energy. Nothing but conversion from Joules (or electron volts) to newtons. If you do the reverse, you again just weigh the object on a standard scale and compute how much rest energy it contains.

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There's a new equation which serves as a correction to the previous one,

${ E }^{ 2 }={ { m }_{ 0 }^{ 2 } }{ c }^{ 4 }+{ p }^{ 2 }{ c }^{ 2 }$

Or as previously stated in a comment, objects at rest do not have $E=0$ or $m=0$. Misunderstandings including this have made people reluctant to talk about relativistic mass.

If the given system is at rest, $p=0$, $E=mc^{2}$ and $m \neq 0$ and $mg \neq 0$.

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I think it would help for you to first understand what does this E in E=mc^2 means. Yes it means energy. But more specifically, E in this equation is the energy you can generate by converting m amount of rest mass into energy. So the idea this equation describe is that you can convert rest mass into energy, or you can convert energy into rest mass.

It is common for someone who has just begun learning physics to get confused of the meaning of all these letters. Letters in physics equations are quite often reused, but they could mean complete different things.

For example W=mg and W=Fs. Both equations contain the letter W, but one refers to weight while the other refers to work done. So during your study of physics, make sure you understand the meaning of the physical quantities in a physics equation to avoid confusion!

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  • $\begingroup$ Let me know if you still need further clarification if you can't think it through. Happy to help! $\endgroup$ – Eric Jan 28 '16 at 12:02
  • $\begingroup$ also when you say a resting object has 0 energy, that is true if you mean the kinetic energy of the object, it is indeed 0. well depends on the frame. but the object may contain other kinds of energy as well, for example, internal energy, gravitational potential energy if it is on earth. $\endgroup$ – Eric Jan 28 '16 at 12:05
  • $\begingroup$ No, that was great. All the answers were both enlightening and appreciated. $\endgroup$ – Paul Tatman Jan 28 '16 at 12:28
  • $\begingroup$ Glad to hear! have fun learning physics! $\endgroup$ – Eric Jan 28 '16 at 12:29
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Without introducing any extra details, I would try to resolve your confusion in the simplest manner possible.

The key is that $E=mc^2$ means that the energy content of an object at rest that exhibits an inertia of $m$ is equal to $mc^2$. In a more explicit notation, one would write $E_0=mc^2$ instead of $E=mc^2$. $E_0$ means the energy content of an object at rest. This means that even an object at rest has some non-zero energy content - precisely equal to $mc^2$. Now, for an object at rest, $W = mg = \dfrac{E_0g}{c^2}$ $\neq 0$.

But one should keep in mind that Special Relativity (the theory that gives us $E_0=mc^2$) is not a theory compatible with Newtonian gravity. We modify Newtonian gravity and formulate a theory of gravity consistent with Special Relativity - we call it General Relativity. But, it turns out that for objects at rest, the formula $W=mg$ is still valid in General Relativity and thus, you can apply it for objects moving with sufficiently slow velocities with good enough accuracy.

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