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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.

nalzok
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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 .

Time rub
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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 without some other force intervening. The important detail is "without propulsion." The rope in your example would count as propulsion.

Mark H
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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.

Bellem
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