Why are there escape velocities for gravity/

[Ferraridude]

I know that for a rocket ship to escape Earth's gravity, it must be going at least 17,500 mph, or else it won't go into orbit.
For a black hole, light can't escape because a black hole's escape velocity is faster than c.

But why is there an escape velocity for objects with mass?

Why could I not theoretically get in a space suit and climb an infinite ladder to mars?

 

[Jimmy]

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero. It is the speed needed to "break free" from a gravitational field without further propulsion.

Why could I not theoretically get in a space suit and climb an infinite ladder to mars?

A rocket moving out of a gravity well does not actually need to attain escape velocity to do so, but could achieve the same result at walking speed with a suitable mode of propulsion and sufficient fuel. Escape velocity only applies to ballistic trajectories.

http://en.wikipedia.org/wiki/Escape_velocity

ETA:

I know that for a rocket ship to escape Earth's gravity, it must be going at least 17,500 mph, or else it won't go into orbit.

Firstly, something which is in orbit hasn't escaped Earth's gravity. Secondly, to maintain an orbit, an object needs to be moving at a certain velocity which is a bit less than the escape velocity for a given orbital radius. See:
http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html

Slowly climbing out of the gravity well to a low Earth orbit won't work. You would immediately start falling back to the planet when you stop accelerating. I'm talking about a ship/rocket. If you were climbing a ladder, then the ladder would be supporting you and you wouldn't fall of course.

 

[cepheid]

I know that for a rocket ship to escape Earth's gravity, it must be going at least 17,500 mph, or else it won't go into orbit.
For a black hole, light can't escape because a black hole's escape velocity is faster than c.

But why is there an escape velocity for objects with mass?

Why could I not theoretically get in a space suit and climb an infinite ladder to mars?

Yeah, what Jimmy said. If you just throw a ball up, it has only gravity acting on it, and it comes back down due to this gravity, unless if you throw it fast enough that it can get far enough away that the gravitational pull of Earth is negligible. This has to occur *before* the ball slows to a stop, because if it slows to a stop too close to Earth (theoretically at any finite distance) then gravity will start to pull it back down. So the ball has to have an initial amount of kinetic equal to the amount of potential energy it gains in moving away from the Earth to a large distance, so that it is able to reach this large distance before it comes to a stop (i.e. before all its kinetic energy gets converted to potential).

In contrast, if you climb a ladder into space, then obviously gravity is not the only force acting on you! Every step of the way, a rung of the ladder is supporting your weight. So, you don't get pulled back down to Earth, because this downward pull is always canceled by an equal upward push throughout the journey.

 

[rbj]

because the gravitational pull down decreases in strength as you get farther away from the planet or star or black hole or whatever body (the inverse-square relationship), then it turns out that the energy required to move an object from the surface of the planet (or anywhere outside of the event horizon of the black hole) out to a point at infinity, that energy is finite. also, that finite energy required to move the object from the surface to infinity is proportional to the mass of the object (it takes twice as much energy to move two identical objects than one). from a Newtonian POV (so this doesn't apply to black holes) this energy is

[tex] E = G \frac{M m}{r} [/tex]

where M is the mass of the planet and r is the radius of the planet sphere and m is the mass of the object.
so then, if you launch a projectile (and it doesn't have to be straight up) at a speed where the kinetic energy (which is proportional to mass) equals or exceeds the required energy, then you put enough energy into it to lift the object off the planet and take it all the way to a point at infinity (it never stops and falls back). we know what kinetic energy looks like:

[tex] E = \frac{1}{2} m v^2 [/tex]

if you equate the two (the m cancels out) and solve for v, you get

[tex] \frac{1}{2} m v^2 = G \frac{M m}{r} [/tex]

[tex] v = \sqrt{2 G \frac{M}{r}} [/tex]

that's the escape velocity.

 

[Good4you]

I had a similar question on here; which was answered well.
https://www.physicsforums.com/showthread.php?t=320913

As i understand it:
The short answer is that you can escape the gravity of an object by "climbing" at less than the escape velocity. (Though i suppose without some velocity, at any distance the gravity would have an effect pulling you back until you get closer to another source of gravity.)
-Black holes are different, and an escape velocity of greater than "C" is not the best way to describe why you cannot "climb" your way out. The reasoning can be explained by light cones, and is fairly well explained in the thread at the link above.