# Loss of gravitational potential energy in an infinitely far object

Now consider I went into a rocket which goes an infinite distance far from earth. By infinite I mean very far. The gravitational attraction between me and the earth will significantly decrease and after a certain distance it will cease to act as the inertia of my rocket would be bigger than the gravitational force between my rocket and the earth. So where will the gravitational potential energy be lost? It is no more stored in the rocket since it does not get affected by that force. I know to recover it I can go back closer to the earth and my rocket will automatically start getting affected by the gravitational force again. So my question is, where is the gravitational potential energy stored for an object or a massive body that goes very very far from earth or any massive object?

When the rocket is at earth the total energy is

$E = E_m = mc^2$

When it takes off, the mass of the rocket decreases due to the fuel consumption but this is compensated by the increase in kinetic energy and potential energy so the total energy becomes

$E = E_m + E_k + E_g$

But when it reaches a point where gravity no longer affects it then $E_g$ = 0 because there is no more potential energy.

So the energy of the system becomes

$E = E_m + E_k$

So where did $E_g$ vanish?

There are many fundamental concepts that both you and Trimok have misunderstood.

First of all, you can completely ignore the mass energy. The kinetic energy in no way "compensates" for the loss in mass energy because no mass energy is lost in the example that you gave. Sure, the mass of the rocket decreased as it ejected propellant to move forward, but that mass hasn't disappeared. The propellant is still out there, floating around in space. You can't just ignore its mass energy because it's not in the rocket anymore, it still exists. In other words your total $E_m$ term is always the same throughout this situation, at no point is matter converted into another form of energy, like kinetic energy as you seem to suggest.

Second, this is conceptually wrong :

The gravitational attraction between me and the earth will significantly decrease and after a certain distance it will cease to act as the inertia of my rocket would be bigger than the gravitational force between my rocket and the earth.

You can't compare the inertia of a system with a force (they're not even measured in the same unit). One can't be "greater" than another. Regardless of the mass of an object, however small a force you apply on it it will still have an impact. There's no cutoff. This doesn't matter much because indeed, you can get to a point where this is small enough that you consider yourself to be in a situation where the potentially energy is now maximal (you seem to think that because this is 0 the potential energy has now disappeared, but this is not the case, it is sometimes chosen to be 0 at infinity but keep in mind that this is still an increase because in such conventions this potential energy is negative at lift off). Potential energy increases as you get further from the earth. This is because, intuitively enough, now that you are further from the earth you now have the potential to gain more kinetic energy by falling back towards the earth through a longer distance, gaining more speed in the process.

Finally, let's work out the actual energy distribution throughout all this. Like I said we can ignore mass energy because the total mass is conserved. Instead what we actually need to consider is :

$E_k$ : the kinetic energy of the rocket.

$E_c$ : the chemical potential energy stored in the propellant. It is this energy that will propel the rocket, not mass energy as you seem to think.

$E_g$ : the gravitational potential energy.

In the beginning, the rocket is sitting on the surface of the earth, and $E_c$ is maximal, $E_g$ is minimal, and $E_k$ is just $0$. As you start burning propellant, you release the energy stored in your propellant and $E_c$ begins to decrease as $E_k$ increases. To completely compute $E_k$ you actually have to take into account both the kinetic energy of the rocket and the kinetic energy of the ejected propellant. If you do, you will notice that this is actually not enough to make up for the loss in $E_c$. Indeed, the difference between the two corresponds to the gain in $E_g$ as the rocket goes further and further away from the earth, and indeed we will always have conservation of $E_k + E_c + E_g$.