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The escape velocity on Earth is around 11km/s, and this is the speed required to escape Earth's gravitational field. However nothing would ever get far enough away to be completely free of it.

At which distance from the Earth would a rocket have to be before the effects are negligible?

Do space agencies ever stop taking the gravitational field strength of the Earth into account when launching probes?

This leads onto my main question which is, if we were to build a launch platform on the Moon, would we have to take Earth's gravitational field strength into account when taking off?

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  • $\begingroup$ Probably useful: Lagrange points Wikipedia entry; especially the "Earth-moon" sub-section. $\endgroup$ – Kyle Kanos Dec 4 '16 at 11:18
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    $\begingroup$ Any influence is neglegible if and only if it does not matter for the purposes you have in mind when doing a calculation. I'm not sure what more can be said about this conceptually without just saying what is neglegible in specific situations. $\endgroup$ – ACuriousMind Dec 4 '16 at 13:42
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Do space agencies ever stop taking the gravitational field strength of the Earth into account when launching probes?

The answer is "yes" during mission planning. During mission operations, the answer is "it depends".

Consider a vehicle to be sent to Mars. During the mission planning stage, the amount of fuel needed to get to Mars will be computed using a patched conic approximation. The trajectory is split into three parts: third stage firing to exit from the Earth's gravitational sphere of influence, where only the Earth's gravity is of interest, then to entry into Mar's gravitational sphere of influence, where only the Sun's gravity is of interest, and finally from there to Mars entry interface, where only Mars' gravity is of interested. The goal of this planning exercise is to determine the best times for performing the launch and how much fuel will be needed.

One way to solve this problem is to look at the leg of the trip between Earth and Mars. Since we're ignoring everything but the Sun, the trajectory is a conic section. Assume we know when the vehicle will exit Earth's sphere of influence and when it will enter Mar's sphere of influence. This gives two points in space and time that need to connected with a transfer orbit. Solving for this transfer orbit is Lambert's problem. (Note: The link is to the wikipedia article, which isn't perfect. It is however more or less permanent. You can find much better by using asking your favorite search engine about Lambert's problem.)

That gives us one data point. The goal is to search for the optimal dates. So we'll solve Lambert's problem over and over and over again, each time with different Earth exit and Mars entry dates. The result is a porkchop plot. See When sending a probe to Mars, how is the optimal travel path calculated and Why is there a gap in porkchop plots for details.

What if we didn't ignore the gravitation from the planets during that transfer orbit? Now you have a very hard to solve boundary value problem, with unsolvable numerical equations of motion. The only choice is to numerically integrate those equations of motion -- and we need to do this many, many times over to find that sweet spot. Solving Lambert's problem thousands of times over is hard but doable. Solving this shooting problem is orders of magnitude harder, and given that the mission planners are going to up the ante on the fuel estimate by at least 10%, adding the complication of gravitation from other bodies is completely unnecessary.

This leads onto my main question which is, if we were to build a launch platform on the Moon, would we have to take Earth's gravitational field strength into account when taking off?

That's a tough one. One doesn't need to model the Moon's gravitational influence on a vehicle launching from the Earth. Note well: The Earth's Moon is the most significant third body perturbation with regard to launching from the surface of the Earth when analyzed from a frame with its origin at the center of the Earth.

From the perspective of an Earth-centered frame, the Sun's influence is about half that of the Moon, and the contribution from every other planet is miniscule. Suppose one ignores the gravitational influence from both the the Moon and the Sun during the eight to twelve minutes that it takes to go from the surface of the Earth to above the Earth's atmosphere. The result is a sub-millimeter error in position. The problem is that even the very best navigation sensors will result in a meter-level error. There is no point in chasing after millimeter level errors when meter-level errors exist.

What makes this your question a tough question is that when launching from the Moon, the relative third body perturbations due to the Earth will be about 250 times greater than those due to the Moon when launching from the Earth. The perturbations alone will be about 81 times stronger, and the Moon's surface gravity is about 1/6 that of the Earth's. 6*81 is about 250. Quadrupling that for safety means we're at a factor of 1000, and now you can't ignore the gravitation of the Earth, even during launch.

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  • $\begingroup$ The nice thing about the earth's gravity on moon is that it is mostly coming from the same direction relative to your launch location. This might make calculating it easier. $\endgroup$ – Paŭlo Ebermann Dec 5 '16 at 18:48
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ACuriousMind is right, but the OP has asked about a specific situation. Yes you do have to take into account the gravitational field of other planets when launching rockets. Not just the Earth when launching from the Moon. But the Moon and Jupiter when launching from Earth.

Suppose you launch an airplane and fly it to another city. Then gravity of the moon deflects the airplane a small amount. And so does the wind and perhaps other things. The pilot is continually guiding the airplane. He adjusts for the wind and everything else. The effect of the Moon is much smaller than the wind, and is not noticeable as a separate contribution. It is negligible. On the other hand, the wind is strong enough that it is worth planning for.

A rocket is different. The engine burns at launch, and then it coasts. The engine is used for a few small course corrections during the flight. It may be used again at the end to slow down and enter orbit around another planet. It is important to calculate the trajectory as precisely as possible. This keeps course corrections small, which saves fuel, which saves weight. Weight is important.

Also, a rocket takes months or years to reach its destination. A small force acting continually has a large effect. If the effect of other planets was not taken into account, the rocket would completely miss its target.

For example, consider the orbit of the Earth. Without the Sun, the Earth would travel in a straight line. A year of the Sun's gravity makes it travel roughly in a circle, completely different from a straight line. This question shows that the Sun accelerates the Earth at 0.0058 $m/s^2$. This is tiny, about 0.0006 G's.

Forces from other planets is even smaller than that for most of a trajectory. But they cannot be neglected.

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