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According to Wikipedia:

In physics, escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body.

But, say, I have a spaceship in low Earth orbit, from which hang a long rope. If its is strong enough, using it, one can climb from the ground up to the spaceship at any speed, 1 m/s for example, which is much slower than the first cosmic velocity.

Thus, an object can "escape from the gravitational influence of a massive body" at a speed lower than its escape velocity, as long as there is a rope or something similar, so technically speaking, escape velocity is the not minimum speed required to escape. Am I right?

If I'm right, this also means we can shoot a sensor into the event horizon of a blackhole, and then pull it back after data are collected, if we can find a really strong rope. Sounds cool.

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marked as duplicate by sammy gerbil, JMac, StephenG, stafusa, Kyle Kanos Jan 7 '18 at 0:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The definition of escape velocity relies on there being no other forces on the object. The standard derivation comes from comparing the kinetic energy of an object at escape velocity to its gravitational potential energy. $\endgroup$ – ZachMcDargh Jan 5 '18 at 23:46
  • $\begingroup$ Nope. You're wrong. LEO is under the gravitational influence of the object. Also, a really long rope would work - if you could make it infinitely long. Sadly, infinitely long is much further than you think. $\endgroup$ – user121330 Jan 5 '18 at 23:51
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    $\begingroup$ The next paragraph in the wiki article explains : Note that the minimum escape velocity assumes that ... there will be no future sources of additional velocity (e.g., thrust), which would reduce the required instantaneous velocity. $\endgroup$ – sammy gerbil Jan 5 '18 at 23:51
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    $\begingroup$ Event horizons don't play nice like Earth gravity. Saying "this also means we can shoot a sensor into the event horizon of a blackhole, and then pull it back after data are collected, if we can find a really strong rope. Sounds cool." implies that we could ever possibly design such a rope, and that we could somehow pull the object back up. Unless we discover something dramatically different about the laws of physics or behaviour of black holes; neither of those will ever be possible. $\endgroup$ – JMac Jan 5 '18 at 23:58
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    $\begingroup$ Possible duplicate of Another layman blackhole question, pulling one end of a string out from behind the event horizon $\endgroup$ – sammy gerbil Jan 6 '18 at 0:08
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The definition you quote is a short introduction to the concept. A better definition comes later in the article:

For a given gravitational potential energy at a given position, the escape velocity is the minimum speed an object without propulsion needs to be able to "escape" from the gravity (i.e. so that gravity will never manage to pull it back).

That is, if an object is traveling at or above escape velocity for a certain planet or star, then it will never return to that planet or star with some other force intervening. The important detail is "without propulsion." The rope in your example would count as propulsion.

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IN measuring the escape velocity no further impulse should be applied on the craft so you will have to throw the craft with a force only once which helps it to overcome the gravitational potential energy .If I am climbing a rope with particular force then I am applying a force on the rope as well as on our body. So that disobey the rules of escape velocity .

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The correct definition is:

Escape velocity is the initial velocity associated to the energy to reach infinity (where the potential is zero) with null velocity.

Of course, in this definition, you are in a system where the energy of the particle is conserved.

For example: you are on a surface of a planet with radius $r_o$ and mass $m$. From this initial setup we want to reach infinity. To find the escape velocity we use the definition, so: $$E_i=\frac{1}{2}mv^2-\frac{Gm_om}{r_o} \qquad E_f=0$$ where $E_f=0$ since at infinity the velocity and the potential are zero. Using conservation of energy: $$v=\sqrt{\frac{2Gm_o}{r_o}}$$ which is independent on the particle but it depends only on the planet.

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