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Re: How Far Can A Locomotive Jump? And, with enough speed
Author: BOB2
Date: 08-06-2017 - 17:04
In physics, escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body.
The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 25,020 mph)[1] at the surface. 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 return.[2] 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 gravititational influence, and that there will be no future sources of additional velocity (e.g., thrust), which would reduce 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]
v e = 2 G M r , {\displaystyle v_{e}={\sqrt {\frac {2GM}{r}}},} v_e = \sqrt{\frac{2GM}{r}},
where G is the universal gravitational constant (G ¡Ö 6.67¡Á10−11 m3¡¤kg−1¡¤s−2), M the mass of the body to be escaped, 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 mass body M. Conversely, a body that falls under the force of gravitational attraction of mass M, from infinity, starting with zero velocity, will strike the mass with a velocity equal to its escape velocity.
When given a speed V {\displaystyle V} V greater than the escape speed v e , {\displaystyle v_{e},} {\displaystyle v_{e},} the object will asymptotically approach the hyperbolic excess speed v ¡Þ , {\displaystyle v_{\infty },} {\displaystyle v_{\infty },} satisfying the equation:[4]
v ¡Þ 2 = V 2 − v e 2 . {\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. A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape. Escape velocity is only required to send a ballistic object on a trajectory that will allow the object to escape the gravity well of the mass M.
Known to Newton, NASA, and Kin Jun Un.... and my fellow rocket scientist's on Altamont, now let's see if that mathematical notation comes through on David's pages......
The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 25,020 mph)[1] at the surface. 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 return.[2] 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 gravititational influence, and that there will be no future sources of additional velocity (e.g., thrust), which would reduce 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]
v e = 2 G M r , {\displaystyle v_{e}={\sqrt {\frac {2GM}{r}}},} v_e = \sqrt{\frac{2GM}{r}},
where G is the universal gravitational constant (G ¡Ö 6.67¡Á10−11 m3¡¤kg−1¡¤s−2), M the mass of the body to be escaped, 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 mass body M. Conversely, a body that falls under the force of gravitational attraction of mass M, from infinity, starting with zero velocity, will strike the mass with a velocity equal to its escape velocity.
When given a speed V {\displaystyle V} V greater than the escape speed v e , {\displaystyle v_{e},} {\displaystyle v_{e},} the object will asymptotically approach the hyperbolic excess speed v ¡Þ , {\displaystyle v_{\infty },} {\displaystyle v_{\infty },} satisfying the equation:[4]
v ¡Þ 2 = V 2 − v e 2 . {\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. A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape. Escape velocity is only required to send a ballistic object on a trajectory that will allow the object to escape the gravity well of the mass M.
Known to Newton, NASA, and Kin Jun Un.... and my fellow rocket scientist's on Altamont, now let's see if that mathematical notation comes through on David's pages......
