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?
Asked
Active
Viewed 97 times
1
-
1Yeah 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 https://space.stackexchange.com/q/4688/4660 answer what you want to know? – kim holder Apr 24 '21 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 '21 at 17:45
-
1I 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. – Apr 25 '21 at 09:55