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If I make a rocket and want to reach a point in space above sea level where I don't need to burn fuel to rise, how far is that point? space or zero gravity area?

I am just trying to figure out distance where I can shut my rockets.

Put another way, how much fuel do I need, or how much distance do I need to cover so that I can orbit earth like ISS?

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closed as unclear what you're asking by JCRM, Rory Alsop, ReactingToAngularVues, James Jenkins, Manu H Dec 28 '17 at 22:19

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You should change your question to "Is there a point where gravity reaches zero?", or better yet; "At what point can I shut down my rockets?" Currently this question has no answer because gravity does not reach zero! $\endgroup$ – uhoh Dec 27 '17 at 21:29
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    $\begingroup$ Related, and can probably answer your question, but isn't exactly a duplicate: space.stackexchange.com/questions/7981/… $\endgroup$ – Nathan Tuggy Dec 27 '17 at 21:29
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    $\begingroup$ It's not a place, but a speed called escape velocity. Once you reach that speed, you won't fall back to Earth. Escape velocity is about 11km a second, so you gotta be going pretty fast. $\endgroup$ – zeta-band Dec 27 '17 at 21:39
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    $\begingroup$ In free fall, zero gravity may be just above the ground. $\endgroup$ – Uwe Dec 28 '17 at 8:18
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    $\begingroup$ Possible duplicate of Couldn't I escape Earth's gravity traveling only 1 mph (0.45 m/s)? $\endgroup$ – James Jenkins Dec 28 '17 at 17:30
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There is gravity everywhere. It doesn't work so that enough far away from the Earth, there won't be gravity any more.

In Low Earth Orbit (i.e. things orbiting Earth), they are in nearly the same gravitational field as we are. For example, the ISS orbits roughly 400 km above the Earth, which is only 400/6378 $\sim$ 6.3 % farther from the center of the Earth than the surface. Since gravity scales as $1/r^2$ that means at that altitude, gravity is still about 88% as strong as on the surface.

The people of the ISS are in weightlessness because they are also flying 7.8km/s speed (around 28000 km/h). This results in a centripetal force which compensates the gravity of the Earth. This is why the ISS doesn't fall down.

Farther away from the Earth, the gravity of the Moon, or other planets, or the Sun would affect more. These are typically far lesser effects as we have on the Earth, but they still exist. Planning the trajectories of space probes, all of them should be calculated.

Thus, if you simply fly upward and shut down the engines, you will fall down. If you fly faster than 11.2km/s, then your spaceship will leave Earth and never come back. If you want to orbit Earth, you have to leave the atmosphere (around 100km height) and accelerate to the speed of at least 7.8km/s sideways around the Earth (because there is a little air even there, 300km is more practical).

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    $\begingroup$ ISS altitude is pretty close to 400km, not 300km; the effect of gravity at ISS altitude is ~88.5% of that at sea level. $\endgroup$ – Russell Borogove Dec 27 '17 at 23:50
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    $\begingroup$ @RussellBorogove ISS fixed, 88.5% imho matches well "nearly the same". $\endgroup$ – peterh Dec 28 '17 at 1:07
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    $\begingroup$ I suggest adding the word 'sideways' to the last paragraph to emphasize that you need to accelerate orthogonal to "up". $\endgroup$ – Roman Reiner Dec 28 '17 at 12:20
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    $\begingroup$ To put it another way, in case they don't know what "centripetal force" means: The ISS is falling towards the Earth, but it is moving sideways so fast that it keeps missing. $\endgroup$ – MJ713 Dec 28 '17 at 19:54
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    $\begingroup$ The centripetal force is gravity It acts towards the centre. You mean "Centrifugal force", the term in a rotating frame of reference that balances the centripetal force of gravity $\endgroup$ – James K Jan 1 '18 at 19:07

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