What is escape velocity?
Answers
hi
In physics, the rate of recovery is the rate at which the kinetic energy of any object in the gravitational field is equal to its potential energy relative to the area in question. It refers to the speed that the object, which is usually in a three-dimensional space, must reach in order to get rid of the gravitational field that affects it.
Answer:
In physics (specifically, celestial mechanics), escape velocity is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a massive body, that is, to achieve an infinite distance from it. Escape velocity is a function of the mass of the body and distance to the center of mass of the body.
A rocket, continuously accelerated by its exhaust, need not reach ballistic escape velocity at any distance since it is supplied with additional kinetic energy by the expulsion of its reaction mass. It can achieve escape at any speed, given a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape.
The escape velocity from Earth's surface is about 11,186 m/s (6.951 mi/s; 40,270 km/h; 36,700 ft/s; 25,020 mph; 21,744 kn).[1] More generally, 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] an object which has achieved escape velocity is neither on the surface, nor in a closed orbit (of any radius). With escape velocity in a direction pointing away from the ground of a massive body, the object will move away from the body, slowing forever and approaching, but never reaching, zero speed. Once escape velocity is achieved, no further impulse need be applied for it to continue in its escape. In other words, if given escape velocity, the object will move away from the other body, continually slowing, and will asymptotically approach zero speed as the object's distance approaches infinity, never to come back.[2] Speeds higher than escape velocity have a positive speed at infinity. Note that the minimum escape velocity assumes that there is no friction (e.g., atmospheric drag), which would increase the required instantaneous velocity to escape the gravitational influence, and that there will be no future acceleration or deceleration (for example from thrust or gravity from other objects), which would change the required instantaneous velocity.
For a spherically symmetric, massive body such as a star, or planet, the escape velocity for that body, at a given distance, is calculated by the formula[3]
{\displaystyle v_{e}={\sqrt {\frac {2GM}{r}}},}v_e = \sqrt{\frac{2GM}{r}},
where G is the universal gravitational constant (G ≈ 6.67×10−11 m3·kg−1·s−2), M the mass of the body to be escaped from, and r the distance from the center of mass of the body to the object.[nb 2] The relationship is independent of the mass of the object escaping the massive body. Conversely, a body that falls under the force of gravitational attraction of mass M, from infinity, starting with zero velocity, will strike the massive object with a velocity equal to its escape velocity given by the same formula.
When given an initial speed {\displaystyle V}V greater than the escape speed {\displaystyle v_{e},}{\displaystyle v_{e},} the object will asymptotically approach the hyperbolic excess speed {\displaystyle v_{\infty },}{\displaystyle v_{\infty },} satisfying the equation:[4]
{\displaystyle {v_{\infty }}^{2}=V^{2}-{v_{e}}^{2}.}{\displaystyle {v_{\infty }}^{2}=V^{2}-{v_{e}}^{2}.}
In these equations atmospheric friction (air drag) is not taken into account.