What is the third cosmic velocity? I have been studying Gravitation chapter and there I found one term:
Third cosmic velocity which is also known as interstellar speed. So what is it? What it really tells about? I tried to gather some information and I guess its something about velocity at infinity but still I am not sure what is it.
 A: Cosmic Velocity has nothing to do with infinity.
A cosmic velocity is the minimum speed directed in the necessary direction to escape the gravitational attraction of a cosmic body such as a planet, a star, or a galaxy.
Here is a paper which a student wrote about the four cosmic velocities.  I don't know if his exact classifications are in common usage, but here it is for what it's worth:
The cosmic velocities (Krzysztof Mastyna) (NB: PDF)
Here is a much more reliable analysis of the Third Cosmic Velocity:
Controversies about the value of the third cosmic velocity (NB: PDF)
You might also google Escape Velocity.
A: *

*It's probably too late to change custom & convention now, but technically it is a speed rather than a velocity, since it refers to the magnitude of a velocity vector. The third cosmic speed $v_3\approx 16.7 km/s$ is the initial escape speed needed for an un-powered projectile launched from a (non-rotating but orbiting) Earth to leave the gravity well of the solar system. 

*The simplest derivation is probably to consider the specific kinetic energy of the projectile relative to an instantaneous inertial frame for Earth. If Earth have been massless, then the specific kinetic escape energy would have been
$$ \frac{1}{2}(v_{2,S} -v_{1,S})^2 ,\tag{1}$$
where 
$$  v_{1,S} ~:=~\sqrt{\frac{GM_S}{R_{SE}}}\tag{2}$$ 
is the orbital speed of Earth, and where
$$ 
 v_{2,S}~:=~\sqrt{2} v_{1,S} \tag{3}$$
is the escape speed from the Sun's gravity, starting from Earth's position. (Both speeds $v_{1,S}$ and $v_{2,S}$ are measured relative to the Heliocentric reference frame.)

*Taking the Earth's gravitation into account, with escape speed (=the second cosmic speed)
$$ v_2~:=~\sqrt{\frac{2GM_E}{R_{E}}}, \tag{4}$$
the specific kinetic escape energy becomes in total
$$ \frac{1}{2}v_3^2~=~\frac{1}{2}v_2^2+\frac{1}{2}(v_{2,S} -v_{1,S})^2,\tag{5} $$
from which we can deduce the third cosmic speed $v_3$. Above we have used various simplifying assumptions. See Refs. 1 & 2 for more details. 
References:


*

*S. Halas & A. Pacek, Controversies about the value of the third cosmic velocity, Ann. Phys. Sec. AAA 68 (2013) 63. (Hat tip: Ernie.)

*http://hirophysics.com/Study/first-second-third-cosmic-velocities-blackholes.pdf
A: In physics, escape velocity is the speed at which the sum of an object's kinetic energy and its gravitational potential energy is equal to zero.[nb 1] It is the speed needed to "break free" from the gravitational attraction of a massive body, without further propulsion, i.e., without spending more fuel.
For a spherically symmetric massive body such as a (non-rotating) star or planet, the escape velocity at a given distance is calculated by the formula[1]
$$v_e = \sqrt{\frac{2GM}{r}}$$
A: Third cosmic velocity is the velocity required to throw an object out of the solar system from the surface of earth to infinity with considering the gravitational effects of only sun and earth and ignoring the gravitational effects of other planets.
