# Gravitational potential [closed]

I understand how we calculate gravitational potential and how we set it to be zero at infinity, and I understand its value at he Earth's surface is $$-64\,\mathrm{MJ/kg}$$ by using $$r$$ as Earth's radius. However, how is it we can escape the earth's gravitational field if we leave the surface with that equivalent energy per unit mass even though the field goes out to infinity?

• Yeah I don't know what you mean by 'Cale at he Earth's surface'. I think I understand what you're asking, and it's a good question. Does space.stackexchange.com/q/4688/4660 answer what you want to know? – kim holder Apr 24 at 17:42
• But this might be a separate question that deserves answering, I couldn't find a near duplicate. Please edit it though James. Brief help for you - beyond a certain distance the force of gravity acting on an object is so small that even a tiny amount of velocity overcomes it. And it's the velocity that's the key, not an amount of energy. – kim holder Apr 24 at 17:45
• I have edited this to try to improve the wording, and I'm about to cast a vote to reopen. I'm not sure it is on-topic here (may be it should be on Physics SE where it is probably a duplicate), but I would answer it if it was reopened, so I'm doing this out of self-interest as I like thinking about this stuff! I won't be upset if it remains closed of course. – user21103 Apr 25 at 9:55