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I think this one is pretty simple so excuse me for my ignorance. But since most planets in our solar system are very well tied to their orbit around the sun or orbit around their planet (for moons), I was wondering how can a really small spacecraft such as Voyager 1 avoid getting stuck into these orbits and avoid the gravitational force of these huge objects.

It's probably as simple as doing some math, but I imagine that it's because the spacecraft is so small compared to other objects that it makes it quite easy for it to escape gravity.

It's not easy for large objects (like moons) to escape their host planet gravity because they're much bigger, right?

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It is certainly possible to break the gravitational bind of a planets gravity. To do so you permanently you need to achieve escape velocity, which is an energetic statement that says a particle with enough kinetic energy will never 'come back down'. To calculate the associated velocity we equate the gravitational potential of a particle on the surface of a planet of mass $M$ and radius $r$ to the kinetic energy of the particle (mass $m$, velocity $v$): $$\frac{GMm}{r^2}=\frac{1}{2}mv^2$$ ($G$ is Newton's gravitational constant) and noting we cancel $m$ and rearrange for $v$ $$v=\sqrt{\frac{2GM}{r}}.$$

Wikipedia gives a list of these velocities for various solar system bodies: linky.

Note we don't need to instantly achieve this velocity unless unpowered. With constant acceleration we just need to reach $v$ for the radius $r$ we are away from the body we're trying to escape.

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You are absolutely right.

Considering $r$ constant, Newton's law of universal gravitation, $ F = \dfrac{Gm_1m_2}{r^2}$, says that the force of attraction between two objects with mass $m_1$ and $m_2$, depends on the product of their masses. Hence, though one object is having a larger mass (here planets), the force can be small as the other is having a small mass (here satellite).

Also, as the force depends on the square of the distance, even the satellite can feel a very strong force of planets, if it comes very near to the planet. But, before we set the whole course of journey of our satellite, we make sure that it does not give the planets the opportunity to swallow it.

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It is NOT possible to "break free of" gravitational attraction, as it is infinite in range. However, it is possible to get sufficiently far away from an object that its gravitational influence on other objects becomes negligible, but it certainly cannot be zero. Note that I am talking about only two objects here. In multi particle system, an object can experience a zero NET gravitational attraction.

Textbooks that use the language of "escaping a planet's gravity" do a huge disservice to students and beginners for two reasons. The first is that "gravity" does not exist. Over 19 years of teaching undergraduate physics, I have come to realize that the word "gravity" connotes, in the minds of students, a material substance. Objects that don't possess this substance don't exert gravitational attraction. The second is that, as I said above, gravitational forces have infinite range.

The term "escape velocity" is also a misnomer because "escape speed" doesn't depend on direction. The best way to think of this is in terms of energy. The correct framing of the question is this:

What kinetic energy must I give a spacecraft so that by the time the spacecraft is "very far away" from a planet, it (the craft) has no leftover kinetic energy?

Of course, a speed can be inferred from the initial kinetic energy. For objects launched from Earth's surface, the required "no kinetic energy when very far away speed" is about 11 km/s.

Note also that Nic's answer refers to "the gravitational potential energy of a particle" which constitutes another physics error. A particle cannot possess any form of potential energy because potential energy is a property of an interacting system. Instead, we should define a system to consist of particle plus Earth, and THEN talk about the interaction, or potential, energy of that system. Then everything makes sense and is internally consistent and correct.

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