# Explain Escape Velocity **URL:** \ **Category:** Factual Questions\ **Created:** [December 12, 2006, 6:38pm UTC](https://boards.straightdope.com/t/explain-escape-velocity/383847 "2006-12-12T18:38:03Z")\ **Posts on this page:** 6\ **Page:** 2
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [December 12, 2006, 11:37pm UTC](https://boards.straightdope.com/t/explain-escape-velocity/383847/21 "2006-12-12T23:37:39Z")
[QUOTE=Chronos] This is actually correct, if by “extremely eccentric” you mean “eccentricity equal to 1”. Of course, a conic section with eccentricity greater than 1 isn’t an ellipse any more, it’s a parabola, which, in fact, never returns to Earth. If your speed is greater than escape speed, meanwhile, your eccentricity will be greater than 1, and your orbit will be a hyperbola. [/QUOTE] Another way to look at this (which is somewhat more abstract but doesn’t involve messing about with any conic sections) is that in order to escape the gravity field of a planet, you need a certain level of kinetic energy relative to the planet. Once you achieve that level, you will be in an escape trajectory, regardless of what direction you are pointing (assuming, of course, that your trajectory doesn’t take you back through the planet). Such energy methods are quite useful for determining required velocities and momentum to achieve a particular orbit or transfer from one to another, but if you want specifics on where and what direction, you have to consider actual geometry. Stranger ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [December 12, 2006, 11:41pm UTC](https://boards.straightdope.com/t/explain-escape-velocity/383847/22 "2006-12-12T23:41:27Z")
[QUOTE=msmith537] If I fire that projectile at 12 km/s, it won’t decelerate to zero velocity until it has passed a point where gravity no longer has an effect [/QUOTE] Properly speaking, there is no such point - the force of gravity is greater than zero at any finite distance. But it’s also true that a projectile that starts at or above the appropriate escape speed never slows to zero. So it never falls back to earth. Note that escape speed is basically equal to the speed that an object would have at the earth’s surface if it started falling from zero velocity and an enormous distance. ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [December 12, 2006, 11:47pm UTC](https://boards.straightdope.com/t/explain-escape-velocity/383847/23 "2006-12-12T23:47:56Z")
[QUOTE=scr4] The Shuttle _does_ need to reach a speed of about 8km/s. It doesn’t matter how quickly it gets up to that speed, but it does need to get there. Otherwise, once the engines are shut off, the Shuttle in an elliptical orbit that intersects the earth’s atmosphere. If it achieves 8km/s, it’s in an almost circular orbit, and the entire orbit is above the atmosphere. [/QUOTE] Well, it matters to the occupants how fast it goes; 20Gs of acceleration (particularly in the inverted position in which the Shuttle ascends to orbit) would definitely be detrimental to the health and well-being of the crew. 😉 Pedantic addition, but after MECO (Main Engine Cut-Off), the [OMS](http://en.wikipedia.org/wiki/Orbital_maneuvering_system) (Orbital Maneuvering System, the small pods on the aft of the Shuttle straddling the SME housing) perform an orbital insertion burn to circularize the orbit, otherwise the Shuttle reenters the atmosphere somewhere around the Indian Ocean or Pacific Ocean, depending on trajectory plan. Stranger ---
**Author:** ![Will\_Repair](https://avatars.discourse-cdn.com/v4/letter/w/a698b9/32.png) [@Will\_Repair](https://boards.straightdope.com/u/Will_Repair)\ **Post date:** [December 13, 2006, 3:21pm UTC](https://boards.straightdope.com/t/explain-escape-velocity/383847/24 "2006-12-13T15:21:27Z")
Thanks all. This is the obit of my ignorance. ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [December 13, 2006, 3:58pm UTC](https://boards.straightdope.com/t/explain-escape-velocity/383847/25 "2006-12-13T15:58:05Z")
[QUOTE=Xema] Properly speaking, there is no such point - the force of gravity is greater than zero at any finite distance. [/QUOTE] I missed this one yesterday. Yes, technically speaking, the gravitational attraction for a specific body approaches zero asymptotically (but never quite reaching it); in practice, as least insofar as plotting orbital or celestrial trajectories, once an object is past the [sphere of influence](http://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29) (SOI), a small body (one for which it’s gravitational field is essentially negligible) it outside the effect of the planet and will never return or remain in orbit in absence of any other influences. The SOI for the Earth is ~145 radii or 925,000km. Anything that reaches that distance (regardless of what instantaneous speed it attains in the interim) has effectively escaped the Earth’s gravitational influence. Of course, at that point, escape speed is essentially zero, so as long as you keep going, at some point you’ll attain escape speed regardless of how slow you are moving. Stranger ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [December 13, 2006, 7:06pm UTC](https://boards.straightdope.com/t/explain-escape-velocity/383847/26 "2006-12-13T19:06:32Z")
[QUOTE=Stranger On A Train] The SOI for the Earth is ~145 radii or 925,000km. Anything that reaches that distance (regardless of what instantaneous speed it attains in the interim) has effectively escaped the Earth’s gravitational influence. [/QUOTE] It would be more accurate to say that it has reached a place where other gravitational influences (most notably, the Sun’s) are stronger. [Previous page](https://boards.straightdope.com/t/explain-escape-velocity/383847.md?page=1)