What is the energy required to create mass of m at a height of h above the Earth?

Is it $E= m c ^2$ or $E = mc ^ 2 + mgh$ ?

Let's reverse the process also.

If you convert mass $m$ at $h = 0$ to energy then

$$E=mc^2 \tag{1}$$

Now if you raise the mass to a height $h$ and convert it to energy which you are going to measure at the height $h$ then $$E=mc^2 + mgh \tag{2}$$

Is equation (2) correct?

If this is correct then

If you take a rock of mass $m$ on the Earth to very large distance or provide it with escape velocity so that it escapes the Earth's gravity (ignoring any other gravitational field), What is the energy contained in that rock?

Is it equation (1) or

$$ E = mc^2 + \dfrac{1}{2}mv^2, \tag{3}$$ where $v$ is the escape velocity?

If equation (3) is the accurate one according to the discussion above then once the mass has come out of the gravitational field the only way to store this extra energy will be by an increase in mass. So,

$$ dm = mv^2/(2c^2) $$


$$ dm = mgh/(c^2) $$

  • 1
    $\begingroup$ This is exactly why Einstein realized that gravity and mass had to be linked -- he imagined a perpetual motion machine where you created mass at an altitude, lowered it, extracted the kinetic energy, and then returned the initial energy to the original height. $\endgroup$ Commented Feb 18, 2014 at 18:18
  • $\begingroup$ Does that mean mass increases with height with $d m = mgh/(c^2)$ $\endgroup$ Commented Feb 18, 2014 at 18:22
  • $\begingroup$ There's a sense in which you coudl think that, but long before the amount of added mass would begin to matter, the $mgh$ formula would cease to be valid. $\endgroup$ Commented Feb 18, 2014 at 18:30
  • $\begingroup$ What about cases where m and g are sufficiently large? $dm = G * M * m/rc ^ 2$ $\endgroup$ Commented Feb 18, 2014 at 18:32
  • $\begingroup$ Please do not put block equation in comments. $\endgroup$ Commented Feb 18, 2014 at 19:07

2 Answers 2


Think logically. Assume that you want to create a mass on the earth, where $h=0$ (assumption). Therefore:


You as well need to consume some work to take the mass from $0 \to h$. So the energy needed is the energy you need to create it plus the one you need to "lift" it. So:

$$\sum E = E - W_{W(spent)} = E - (-mgh) = mc^2 + mgh = m(c^2 + gh)$$

Everyday example: Which state has more energy: a tidied or an untidies room? The answer is the tidied one because we've spent energy to tidy it

Since the gravitational field is a conservative one the work done to do this action is always $-\Delta U = -mg \Delta h$, so if you were already at $h$ then the change in height is 0. It may be a bit confusing but it has to do with your choice of zero potential energy level

  • $\begingroup$ Lets say the mass has to be created at a height h above the earth. This will use the same method or machine that would have created the mass on earth, But only do this at a height h above the Earth. $\endgroup$ Commented Feb 18, 2014 at 17:50
  • $\begingroup$ Suppose that you have two boxes of mass $m$, one standing on the earth and the other at height h above it. The total energy in two cases (suppose that the mass is already created) is in the first case is 0 and in the second case is $mgh$. So if we leave the system the first will remain the same while the second will fall down due to its initial energy. $\endgroup$
    – bolzano
    Commented Feb 18, 2014 at 17:58

It depends where your energy starts. Isolating two cases should give you the idea. The first is if the energy is already at height h, in the other we'll assume it starts ground level.

Case 1: $h_0 = h$ [already at height h]

$E = mc^2$. The gravitational potential energy was stored previously in energy. Energy is not immune to gravity. To create the mass at a height h, the energy has already fallen or risen to that point, and will contain the same gravitational potential energy in the form of energy or mass.

Case 2: $h_0 = 0$

$E = mc^2 + mgh$, because you need to lift the mass-energy to that height and then convert. Order of lifting and mass-energy conversion doesn't matter. You can lift the energy then convert or convert then lift the mass.

  • $\begingroup$ Lets reverse the process and try see. If you convert mass m at h = 0 to energy then $E=mc^2$. Now if you raise the mass to a height h and convert it to energy which you are going to measure at the height h then $E=mc^2 + mgh ?$ $\endgroup$ Commented Feb 18, 2014 at 18:14

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