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There is a finite amount of energy needed to be spent on a rocket or a rock that is taken out of the Earth's gravitational field. Lets suppose the rock is taken far enough to have no effect of Earth's gravity. What happens to this energy? Who gets this?

Where does this energy go?

Assume that the rock went just outside the gravitational field with zero velocity.

Now.

Lets say at the gravitational boundary ( place were the gravitational effect is very small) you decide to convert the mass of the rock to pure energy. What will be the energy released ? $$ E = m c ^ 2 $$ or $$ E = m c ^ 2 + mgd $$ where d is the distance of the mass from Earth.

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  • $\begingroup$ There is no such thing as "just outside the gravitational field". The potential energy is always there were the object to fall back to Earth. Your question is somewhat related to en.wikipedia.org/wiki/Gravitational_binding_energy $\endgroup$ – Brandon Enright Feb 18 '14 at 17:15
  • $\begingroup$ Lets say at the gravitational boundary ( place were the gravitational effect is very small) you decide to convert the mass of the rock to pure energy. What will be the energy released ? $ E = m c ^ 2 $ or $ E = m c ^ 2 + mgd $ where d is the distance of the mass from Earth. $\endgroup$ – Deepak Nath Feb 18 '14 at 19:07
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The energy is still there in the form of gravitational potential. Think of leaving the earth as a process similar to riding a bicycle up a hill. (When you ride a bike up a hill, you're moving against Earth's gravitational field thereby gaining potential energy, just like what happens to the rock when it moves up away from the surface of the earth.) If the rock is eventually so far away that the effect of Earth's gravity is negligible, this is a similar situation to you reaching a flat spot on the top of the hill that you just rode your bicycle up. Now you can just sit there on your bike for as long as you like, but if you turn around and approach the crest of the hill, you will start rolling back down again, with your potential energy that you gained while climbing the hill now being converted to kinetic energy as you coast back down. This is the same as what would happen to the rock if someone nudged it ever so slightly back toward the earth (it would fall back down).

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  • $\begingroup$ What happens if you convert the entire bicycle to energy at the top of the mountain? Will it be $E = mc^2 or E = mc^2 + mgh $ $\endgroup$ – Deepak Nath Feb 18 '14 at 19:04

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