Talk:Escape velocity

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Multiple sources[edit]

I find this section confusing. Let me give an example of why I don't understand this. Let's pretend that the earth is farther away from the sun than it really is, so that the escape velocity from the sun at this distance is 11.2 km/s, just as for the earth. So then we get that the escape velocity from sun/earth, with earth orbiting at this distance, would be sqrt(11.2^2 + 11.2^2) = 15.8 km/s. Now earth's (circular) orbital speed at this distance will be .7071*11.2 = 7.9 km/s, so we predict that we need to go at a speed of 15.8 - 7.9 = 7.9 km/s (relative to earth) in the direction of earth's orbit. But THIS IS LESS THAN EARTH'S ESCAPE VELOCITY! So, relative to the earth, the speed we need to start out with to go out to infinity is now LESS than it would be if the sun were not there?? Kier07 07:48, 4 September 2006 (UTC)

I didn't verify your number but I'll assuming they are correct. Then yes the velocity needed to escape Earth and Sun in your example is going to be less then the escape velocity of Earth alone. What the section doesn't say is, the object is allowed to "fall through" Earth. So your object starts up moving away from Earth at 7.9 km/s, eventually due to Earth's gravity, it will turn around and starting moving toward Earth. If the path doesn't collide with Earth, it will escape. If it's path collides with Earth, it's assumed to be able to fall throught it. The actualy path the object takes to escape is a swirl. It may swirl around Earth, or Sun depending on the initial conditions. Please also remember Earth is also moving so the object would not simply oscillate back and forth through Earth's center. --NYC 21:14, 14 March 2007 (UTC)
That is a good point, Kier07. I added some explanation why the original value is not enough. The newer version [1] is more plausible, but a source and/or an explanation is still needed.--Patrick (talk) 21:25, 31 January 2010 (UTC)
The current formula is also on, but the author hellfire says he/she is not sure.--Patrick (talk) 08:02, 1 February 2010 (UTC)

Firstly, what is this rubbish about the article needing 'multiple sources'? Maths doesn't need ANY sources, only logical correctness. Any Wikipedia article about a maths concept merely requires enough maths logic for the claims to be tested. Secondly, the article needs to begin with the CLEAR statement that 'escape velocity' is a mass-independent statement of the energy required for an object to escape a gravitational well. Escape velocity is a very clever statement of ENERGY requirements, not speed. In fact, an object cannot be given an instantaneous speed, so the speed referred to in an 'escape velocity' isn't even a special case, and the term confuses most people into thinking a rocket must reach at last this speed to leave the Earth. Of course, real rockets are based on the principle of diminished mass as the fuel providing thrust burns up. Since the term 'escape velocity' exists for practical reasons, shouldn't the article address the 'applied' aspects of the term as well? (talk) 16:31, 17 April 2013 (UTC)

Re "Maths doesn't need ANY sources, only logical correctness", have a look at our rubbish policy in wp:NOR. - DVdm (talk) 17:35, 17 April 2013 (UTC)
The term Multiple Sources means Multiple Gravitational Sources. The question is: What radial velocity would a body launched from Earth need to escape both the Sun's and the Earth's gravitational fields. I haven't checked the numbers but I assume that the escape velocity for the Sun is that computed at the Earth's orbit. Virgil H. Soule (talk) 07:28, 25 January 2014 (UTC)

A ladder[edit]

An astronaut using a ladder can escape at any speed. No matter how slowly. But no matter how slowly an astronaut climbs a ladder "out" of a black hole, all paths will lead inward. Even if that astronaut is very strong, there is no escape out of a black hole, because the gravity of the black hole makes all paths curve inwards. This has nothing to do with the escape velocity being larger than the speed of light. A black hole curves ALL paths inwards. Can this be made clearer in this article? (talk) 02:42, 27 January 2010 (UTC)

Escape velocities with respect to the sun[edit]

According to my calculation, with the below formula

v_e = \sqrt{\frac{2GM}{r}}

I got these values:

at Mercury 47.9 km/s
at Venus 35.0 km/s
at Earth 29.8 km/s
at Mars 24.1 km/s
at Jupiter 13.1 km/s
at Saturn 9.6 km/s
at Uranus 6.8 km/s
at Neptune 5.4 km/s

It is not the same as in the table in the article. What is correct? --BIL (talk) 14:11, 18 February 2010 (UTC)

BIL, all your values are out by a factor of \sqrt{2}. You just missed the 2 in your calculations. Cosmogoblin (talk) 21:14, 1 November 2010 (UTC)
Point of clarification: The escape velocities given in the article are Surface Escape Velocities for each of the planets in the list. The velocities listed above presumably were computed at the mean orbital distance of each planet in the list. The two are completely different. Virgil H. Soule (talk) 07:28, 25 January 2014 (UTC)

Removal of Misconception Section[edit]

This section has multiple issues. First, there's no source verifying this misconception even exists (I've never heard of it). Second and more seriously, the section is wrong. An object in orbit does need to reach escape velocity to "escape" orbit. While any additional increment in speed will cause it to leave its current orbit, if the velocity isn't great enough, it will simply assume a different orbit. Fell Gleaming(talk) 01:27, 10 May 2010 (UTC)

LOL. So you're saying that if you have a vehicle doing 1m/s in orbit around the sun really high up in the Kuiper belt, that it needs solar surface escape velocity (618km/s) in order to escape from the Sun's gravity???- Wolfkeeper 01:44, 10 May 2010 (UTC)
It needs the full amount of energy as defined by (escape velocity KE) - (current KE), yes. I realize the original text is trying to state that, but it does so in an obfuscatory, confusing manner that is essentially incorrect as written. Also, in practical terms, one cannot escape earth's gravity at "any speed" ... the bare minimum of achieving NEO is required. (without invoking beanstalk-like methods). Furthermore, your reinsertion hasn't addressed the second point, which is that there is no source citation that such a misconception even exists.
Sorry, but you keep proving that the section is right about it being a misconception, by insisting that it's wrong!- Wolfkeeper 02:34, 10 May 2010 (UTC)
No, you've simply misread what I wrote. Escape KE - current KE = additional KE. Or, if you prefer, Current + Additional KE = Escape KE. Just as I said. Do you have a source that verifies this 'misconception' exists? And FYI, NEO and LEO are interchangeable terms...or were, at least, until asteroid mapping projects started preempting the NEO acronym. Fell Gleaming(talk) 02:58, 10 May 2010 (UTC)
Fell Gleaming is incorrect about the energies. An object in a high orbit does not need the same kinetic energy to escape as an object in a low orbit. It needs the same total energy. So for example, let's call GM/R the SEKE ("surface escape kinetic energy"). An object in a circular orbit at a distance of n times the planet's radius has a potential energy of −SEKE/n and a kinetic energy of SEKE/(2n). It needs to add SEKE/(2n) of energy in order to escape. This is not equal to SEKE minus current kinetic energy (unless n=1).
But I agree with him that the section was not clear and I have rewritten it a bit.Eric Kvaalen (talk) 09:56, 12 February 2013 (UTC)
Good job. It has become much better now, but let's find some good source(s) for this. Otherwise this section will remain unstable. I have tagged the section. - DVdm (talk) 15:17, 12 February 2013 (UTC)
First, I think you mean LEO, not NEO, but you don't need to achieve LEO velocity to escape either; a rocket with enough stages can escape arbitrarily slowly; it's just horribly inefficient, but you can do it. There's absolutely no minimum speed to escape, you can leave the planet at 1 mile per hour, eventually the escape velocity comes down to 1 mph, and then you can turn the engine off. There's a minimum speed you need so you can escape with the engines *off* though, which depends on position in space (*local* escape velocity).- Wolfkeeper 02:34, 10 May 2010 (UTC)
Basically, an object needs sufficient total energy. The point is that an object high up in the Kuiper belt would have nearly enough (potential) energy already to escape, and it then needs very little kinetic energy to do so.- Wolfkeeper 02:34, 10 May 2010 (UTC)
If you want the section left in, and can find a source verifying the misconception exists, I'll be happy to rewrite it so its clear. Fell Gleaming(talk) 02:16, 10 May 2010 (UTC)
FWIW, maybe you can do a very little something with this source (under the header The Great Escape). It is of course merely a letter by a certain Paul A. Delaney, Md (!) to some popular magazine, and a reply by Stuart F. Brown. DVdm (talk) 08:43, 10 May 2010 (UTC)

1st and 2nd cosmic velocity[edit]

Although both these two entries redirect to this page, this page does not explain what it means 1st and 2nd cosmic velocity. I think it should. --Pavel Jelinek (talk) 15:15, 9 July 2010 (UTC)

Yes, good catch. Useful information can be found here, here and here. I have added a little sourced remark. Please do feel free to reword or elaborate, and move it to a more appropriate place, because I think it belongs a bit higher up in the article, perhaps somewhere in the overview section? DVdm (talk) 16:10, 9 July 2010 (UTC)

Misconceptions section: "leave the Earth's gravity"[edit]

This is addressed to Gr8xoz regarding my recent edit. I removed the paragraph because the statement

In fact a vehicle can leave the Earth's gravity at any speed less than the speed of light.

is ill-defined. What does it mean to "leave" gravity? This brings up the discussion of gravitational spheres of influence, which are unrelated to the escape velocity. The next sentences contradict the first one:

At higher altitudes, the local escape velocity is lower. But at the instant the propulsion stops, the vehicle can only escape if its speed is greater than or equal to the local escape velocity at that position.

So you can "leave" Earth's gravity at any speed you want, but you can only "escape" Earth's gravity at escape velocity or higher. This doesn't clear up misconceptions, it only introduces more confusion. Finally, the fact that the last sentence,

As is obvious from the equation, this speed approaches 0 as r becomes large.

declares itself to be "obvious" means it is superfluous and can be deleted. It's not a misconception either. Thoughts? MarcusMaximus (talk) 21:01, 14 December 2010 (UTC)

I see your point, an object can never leave the Earth's gravity since it never becomes zero. I do not see how the discussion of gravitational spheres of influence is relevant, it is just the volume where the gravity from a body dominates over the gravity of other bodies. I would suggest that

In fact a vehicle can leave the Earth's gravity at any speed less than the speed of light.

is changed to

In fact a vehicle can leave the Earth at any speed less than the speed of light.

This also solves the contradiction. I have made this change, is this solving the problem?
--Gr8xoz (talk) 21:47, 18 December 2010 (UTC)

What misconception is this paragraph trying to address? Maybe it would help to state it first. MarcusMaximus (talk) 06:09, 20 December 2010 (UTC)

Less than the speed of light?[edit]

It says in misconceptions "In fact a vehicle can leave the Earth at any speed less than the speed of light" As nothing can travel faster than the speed of light why is it necessary to qualify the sentence in this way? Also, as it stands, the sentence suggests that if a faster than light propulsion was possible you couldn't use it to leave the Earth.

I also have a problem with the sentence in the lead that says "The term escape velocity is actually a misnomer, as the concept refers to a scalar speed which is independent of direction." The sentence isn't complete as it doesn't say why the use of the term velocity is a misnomer. How about adding something like "whereas velocity is the measurement of the rate and direction of change in position of an object"? Richerman (talk) 22:07, 16 February 2011 (UTC)

As there has been no reply I've made the changes myself
I think this is a rather unimportant detail but "a vehicle can leave the Earth at any speed" could be interpreted as a statement saying that faster than light travel is possible at least if you are escaping from the earth. Gr8xoz (talk) 17:44, 20 February 2011 (UTC)
It is a small detail but adding unnecessary wording confuses the reader and makes them think they are missing something. I suppose you could interpret it that way but to follow that logic means that every time you talk about "any speed" you would have to qualify it "less than the speed of light". So, for instance, "there is no speed limit on this road so vehicles may travel at any speed less than the speed of light." Richerman (talk) 09:32, 21 February 2011 (UTC)
I agree with Richerman. The statement could also imply that if you travel at light speed you can't leave. The whole paragraph needs to be overhauled. MarcusMaximus (talk) 03:29, 23 February 2011 (UTC)

Vector or scalar[edit]

Velocity, by definition, is a vector. I've gone to some length to point out why escape velocity is also a vector. Please cite a reference when changing this. Thx Androstachys (talk) 13:54, 21 March 2011 (UTC)

You're the one claiming that the existing derivation (which is independent of direction) is incomplete and you're not citing any references.Rememberway (talk) 14:39, 21 March 2011 (UTC)
It's an energy conservation thing- once the vehicle has enough specific energy then it's leaving the gravity well- period. Energy doesn't care about direction.Rememberway (talk) 14:39, 21 March 2011 (UTC)

Androstachys, if you look at the "derivation" section of the article, you will see a mathematical proof that it's actually a scalar. Specifically, it's just a matter of whether the kinetic energy is at least as big as the potential energy at infinite distance. The whole thing follows from the fact that the sum of kinetic energy and potential energy remains constant. Kinetic energy is a scalar and depends in speed but not on direction. That proves that what matters here is only a scalar, not a vector. Michael Hardy (talk) 17:58, 21 March 2011 (UTC)

Unless your trajectory intersects the body from which you are escaping, or one of its moons. :) --agr (talk) 20:35, 21 March 2011 (UTC)


There seems to be an error in the condition of the formula

v_e = \sqrt{\frac{2GM}{r}}.

Consider an example where the object to escape is a very light torus. By letting the point from which to escape approach the centre of gravity, the escape velocity can be made arbitrarily large. This is not correct.

One extra assertion making the formula correct, is to require that the gravitational force F imposed by the the body on the escaping object satisfies F = mMG/r^2 directed opposite of the escape path, for all points along its escape route. Here r is the distance from the point to the mass centre. Note that this is note satisfied close to the mass centre of he torus in the Example above. This is for example satisfied if the body to escape is a homogenous Ball and the escape starts outside the ball.

Jacco77 (talk) 15:40, 17 May 2011 (UTC)

In the overview section of the article, it states:

For the sake of simplicity, unless stated otherwise, we will assume that the scenario we are dealing with is that an object is attempting to escape from a uniform spherical planet by moving straight away (along a radial line away from the center of the planet), and that the only significant force acting on the moving object is the planet's gravity.

Obviously the case of a very light torus (or any torus) violates this assumption, and is not relevant to the development presented in the article. Similarly, this equation fails to account for various other perturbations (e.g. light pressure), and should be treated as a simplified theoretical model. You could add a short section discussing the shortcomings of this definition and the ways of dealing with situations that otherwise are not accurately treated by it if you like, as well. siafu (talk) 18:52, 17 May 2011 (UTC)
Thanks, missed that formulation. Perhaps I reacted to the formulation:

where G is the universal gravitational constant, M the mass of the planet, star or other body, and r the distance from the centre of gravity.

The "other body" part here made me think of more general bodies, such as a torus. Regarding the possible shortcomings section, I am not a physicist and the only shortcoming I could formulate is of the "pure old school mechanics" type formulated above, perhaps I could initiate such a section when I find time for it.Jacco77 (talk) 10:25, 18 May 2011 (UTC)
I added "for a spherically-symmetric body".--Patrick (talk) 22:41, 18 May 2011 (UTC)
I would like to point out this equation is valid only for points outside the said body. (talk) 22:39, 8 October 2012 (UTC)

Escape Velocity of the Milky Way[edit]

I fear this has to be in error, as I would assume the gravitational pull of the entire galaxy would be more than the gravitational pull of the Sun (even from the surface of the Sol). Would this have to do with the Inverse Square Law perhaps? Due to you moving further and further away from the galactic core, the amount of "tug" exerted on you would decrease exponentially. Now that I think about it, it might be right, depending on where at you are in the Milky Way. Perhaps it needs to be worked out what the galactic escape velocity would be when near the center as opposed to our local region. I think that would help clarify the issue more than at current. Sardonicone (talk) 02:35, 21 May 2011 (UTC)

Escape velocity increases as you increase the mass of the object that you are "escaping" from (the "primary"), but decreases as you move further away from the centre of mass of that object. For a location outside of a spherically symmetric primary (on the surface of a planet, for example) we can calculate escape velocity exactly, and it turns our to be proportional to the square root of the primary's mass, and inversely proportional to the square root of distance from the centre of the primary. Calculating escape velocity relative to the Miky Way is much more complicated and uncertain because (a) we are inside the Milky Way; (b) it is not spherically symmetric; and (c) we don't even have a very accurate measure of the mass of the Milky Way. But according to astrophysicist and Nobel prize winner John C. Mather, the escape velocty from the centre of the Milky Way could be around 1,000 km/s [2]. Gandalf61 (talk) 11:12, 21 May 2011 (UTC)

Deriving escape velocity using calculus should be dropped.[edit]

This section seems superfluous and overly complicated. I suggest that it is cut from the article.

The method for deriving the escape velocity is discussed sufficently in earlier sections where it states the escape velocity is derived from the kinetic energy at the surface being equal to the potential energy difference from the surface to infinity. This gives the formula

v_e = \sqrt{\frac{2GM}{r}}

From the definition of g

g = \frac{GM}{r^2}

you get the formula in terms of g and r only.

v_e = \sqrt{2gr}

This is basic maths and doesn't require it's own section of derivation with a complicated argument. Similary for the rest of this section.

Neilljones (talk) 02:00, 13 March 2012 (UTC)

I agree. The energy deriv .ation is simpler and better because it shows that escape velocity is independent of initial direction or trajectory. The "Deriving escape velocity using calculus" section adds nothing to the article and should be removed. Gandalf61 (talk) 09:25, 13 March 2012 (UTC)
Agreed. Worse than not being useful, the section is also too far removed from the overarching "assume object moving away radially unless stated otherwise" statement at the beginning of the article. The statement being necessary at least in the second part of the article. This produces a section of derivations where important assumptions are not made clear. The section may be confusing, or worse wrong for someone preparing say a presentation. Zogwarg (talk) 05:53, 16 October 2013 (UTC)

Reversion of rewrite[edit]

I have reverted a rewrite of the lead and Overview sections by Zedshort (talk · contribs) because it contained these sentences

"Escape velocity is a velocity in that it is the radial speed an object must have as it is directed radially away if it is to escape another mass's) gravitational pull. The escaping object may have an additional component of velocity normal to its radial velocity but in order to escape the planet's gravitational pull, it must have a speed in the radial direction at least as great in magnitude as the mass's "escape velocity"

which are simply wrong. The original Overview is correct when it says

"Escape velocity is actually a speed (not a velocity) because it does not specify a direction"

Escape velocity is independent of direction (ignoring effects of atmosphere, mountains etc.) - it is the same for a horizontally launched projectile (such as Newton's cannonball as for a vertically launched projectile. Gandalf61 (talk) 08:35, 2 July 2012 (UTC)

You are incorrect. Escape velocity is not just a speed but has a direction making it a vector quantity. The component of the speed in the radial direction must be equal to the escape velocity in magnitude. The direction of flight may be in any radial direction. I am speaking in terms of polar coordinates in which all radial lines away from a mass's center are in the radial direction. The escaping body may also have components of velocity in the phi and theta directions but those are at right angles to the radial direction and are of no consequence. It is the radial velocity magnitude that is of consequence. If a cannon ball is launched with a speed equal in magnitude to escape velocity and in a tangential direction it will lack the speed to "escape" the pull of the planet by a very small quantity. It will come close but be off by a small amount and will not escape the planets pull. If you don't understand polar coordinates you need to learn to use them. In addition your wholesale reverting of the other edits were inappropriate as they unrelated to the particular complaint about the use of the term "radial direction" and so were inappropriate.
The first sentence of a introduction should be easily comprehended but as it stood it was too advanced for 99% of the readers.
Be aware none of us own this article and if you are squatting on this and claim ownership you need to make room for others. I thoroughly understand the subject and don't make edits without giving it serious thought. The article is poorly written, full of redundancies and needs to be cleaned up and written in a clearer fashion. I am tempted to revert your reversion as you have changed my edits in a wholesale and inappropriate fashion. — Preceding unsigned comment added by Zedshort (talkcontribs) 14:12, July 2, 2012
The mistaken idea that escape velocity is a vector and depends on direction has been discussed before - see "Vector or scalar" above. If escape velocity were a vector then it couldn't be derived from a conservation of energy argument, since energy is a scalar.
Anyway, that was such a fundamental error on your part that it made me doubt the quality of the rest of your rewrite, and I didn't want to spend time disentangling the good from the bad. However, since you are now accusing me of "claiming ownerhsip" here, I will raise this issue at Wikipedia talk:WikiProject Physics, and see if other editors will step in and deal with it. Gandalf61 (talk) 14:53, 2 July 2012 (UTC)
No, Gandalf61 is correct, escape velocity is independent of direction, you simply have to reach that speed to be able to escape. Well caught.Teapeat (talk) 15:49, 2 July 2012 (UTC)
I agree with both the previous editors. Consider the pathalogical case - a satellite, height h above a planet has an escape velocity v 'towards the planet, not away from it. Assume that the planet has a hole through it and the satellite successfully passs through that hole and out the other side of the planet. When it gets to a height h on the other side, it will be travelling with a velocity v but away from the planet - it will escape in the wrong [?] direction. Martinvl (talk) 16:18, 2 July 2012 (UTC)

────────────────────────────────────────────────────────────────────────────────────────────────────I assume that those who believe that the escape speed depends on direction do so on the ground that the object's angular momentum requires it to have some amount of tangential motion which uses up some of the energy. However, they fail to realize that as the object moves further and further from the gravitating mass, the amount of tangential motion and energy needed for that purpose (conserving angular momentum) declines towards zero. JRSpriggs (talk) 17:26, 2 July 2012 (UTC)

For what it's worth, the recently published physics textbook that I'm using (Sears and Zemansky's University Physics with Modern Physics, Thirteenth Edition, by Young and Freedman) doesn't even use the term escape velocity, instead saying "escape speed, the speed required for a body to escape completely from a planet." It goes on to talk about gravitational potential energy, satellites, and Newton's canon. Nowhere that I see does it say anything that would support Zedshort's claims. 786b6364 (talk) 17:54, 2 July 2012 (UTC)

The problem with this article is its lack of sources. For the escape velocity being independent of angle, there's for instance this one on pages 82 and 83:
  • The projectile will escape to infinity provided that E >= 0; or equivalently that its velocity exceeds the escape velocity ve = sqrt(2Rg). Note that this condition is independent of the angle of launch α (so long as we neglect atmospheric drag).
So this is just a matter of sourcing. - DVdm (talk) 18:38, 2 July 2012 (UTC)
I stand corrected. Thanks for the reference. I always though there would be a very small difference depending on the direction of launch. But I still believe the article contains redundancies and the first line should be one that draws the reader in with a simplified explanation rather than the pedantic one used at present. Zedshort (talk) 00:57, 3 July 2012 (UTC)
  • I checked a couple of references: the Penguin Dictionary of Mathematics has a definition for Escape speed (escape velocity) and then defines it in terms of speed. The Oxford Concise Science dictionary has "escape velocity The minimum speed needed ...". Generally it seems that "escape velocity" is a misnomer and it is actually a scalar quantity, the direction is almost never specified. I'd be tempted to rename the page to be "escape speed" and add a redirect from "escape velocity" to the newly moved page, although maybe it isn't worth the effort. Anyway I think that you were correct to undo the changes as far as the direction goes, although some of the other changes that you reverted, which simplified the language, could have been kept. CodeTheorist (talk) 20:46, 2 July 2012 (UTC)
WP:COMMONNAME. While some sources are clearly using escape speed, I think escape velocity is still the more common term. Unless you can provide evidence to the contrary, I wouldn't move the page. 786b6364 (talk) 02:28, 3 July 2012 (UTC)
Agreed, I think that "escape velocity" is more common than "escape speed", although I have no hard evidence either way. So unfortunately we'll have to leave the page name as it is. CodeTheorist (talk) 12:28, 3 July 2012 (UTC)
It is called escape velocity because it clearly is a velocity. The escape 'speed' might theoretically work in any set direction but any real escape path is clearly limited to a single vector at any one moment and is therefore a velocity. Lucien86 (talk) 06:15, 15 August 2012 (UTC)
In real life it is a velocity, if you were to move at escape speed straight towards the surface of a planet, you would not escape. However mathematically it is a speed and not a velocity because 'escape speed' is just talking about the force of gravity, while hitting into the ground involves the electromagnetic force. So if you were to ignore all forces other than gravity, no matter what direction you move at escape speed, you will still escape. (talk) 15:34, 7 January 2013 (UTC)
No, in real life it really is a speed, since it's derived from the energy equation, so travelling in any direction with the escape velocity will allow you to escape. If, hypothetically, there were a hole in the planet you're pointing at that you could pass through, you would still escape once you came out the other side because your total energy would be the same. Obviously this is ignoring all non-conservative effects (like atmospheric drag), but it really is a speed in all senses of the word, and the use of the word "velocity" is just by tradition. siafu (talk) 17:14, 7 January 2013 (UTC)
That's pretty much what I was saying, I was just explaining to all the people that say it's a velocity why it's not. — Preceding unsigned comment added by (talk) 20:34, 7 January 2013 (UTC)

Centres of Mass?[edit]

The equation Ve=root(2*G*M/r) doesn't work very well for trajectories that go or start near centres of mass. As it treats objects as point masses it frequently shows escape velocities crossing the speed of light- for instance at r= 9 cm for Earth, or r= 2.9 km for the sun. I know an equation for a distributed mass would solve this problem but is obviously far more complicated, I would love to see an equation that worked from mass centre - or even for non-spherical objects. :) Maybe there should be a section on centres of mass? Lucien86 (talk) 06:22, 15 August 2012 (UTC)

It's not the equation that breaks down, but the simplifying assumption of representing objects as point masses. A radius of 9 cm for an object the mass of the Earth, for example, is less than the Schwarzschild radius for that mass, so a value greater than the speed of light is entirely appropriate. In reality, this isn't much of an issue as all practical applications would already recognize that such a small radius would create a subterranean orbit. I don't think that there's much need for a new section; a sentence pointing out the shortcomings of the point mass assumption is more than sufficient. siafu (talk) 17:12, 15 August 2012 (UTC)

Luna I attained escape velocity?[edit]

I think the caption on the image of Luna I is at best misleading, if not flat-out wrong:

Luna 1, launched in 1959, was the first man-made object to achieve escape velocity from the Earth (see below table).[1]

And the table says for Earth V-sub-E is 11.2 km/s. So at what point did Luna I attain that speed? I would submit that it did not. As explained elsewhere in the article, it was not put into orbit in one, ballistic, push - it attained LEO (relatively) slowly, and from there attained the (local, lesser) V-sub-E.

robcranfill \a\t

-- (talk) 18:01, 2 October 2012 (UTC)

"As explained elsewhere in the article, it was not put into orbit in one, ballistic, push - it attained LEO (relatively) slowly, and from there attained the (local, lesser) V-sub-E."
Can you please post this section as I cannot find anything verifying this claim, especially not in relation to Luna 1. As the Luna 1 article states, it had no propulsion of its own. It was pushed to its final velocity by the R-7's third stage, almost certainly itself on a ballistic trajectory. There is no evidence that the third stage was in LEO prior to this or of a "slow" raising of Luna 1's orbit. What with, there's no EDS?!
Besides, the statement that Luna 1 achieved escape velocity is inline cited to a NASA archive, frankly I suspect they would know more than most the truth of the matter... ChiZeroOne (talk) 19:23, 2 October 2012 (UTC)
(edit conflict)Per the article on Luna 1,

Luna 1 became the first ever man-made object to reach the escape velocity of the Earth (what is also known as the second cosmic velocity), when it separated from its 1472 kg third stage.

Another way to look at the escape velocity is the minimum relative velocity that will allow travel beyond the gravitational influence of the body in question; i.e., it is not possible to exit the gravitional sphere of influence (even given infinite time) of the object without at some point reaching escape velocity. You note that the V_{E} will be less at high altitudes, but this does not mean that it is not a proper escape velocity. Escape velocity is defined by the current radius, so a theoretical escape velocity from a radius near the core of the Earth will be higher than it will at sea level, and it will be even lower in near the equator or from points at the top of mountains. siafu (talk) 19:25, 2 October 2012 (UTC)


I removed the following sentence from the lead:

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.

This is not true. In order to escape the gravitational pull of a body, a particle (or point-sized elephant) will at some point need to exceed the escape velocity somewhere along the trajectory. As the distance increase (r goes to infinity) the velocity will decress (v goes to zero); that is, you don't need to achieve the escape velocity for the orbital radius at which you started, but you will eventually cross the escape velocity or else fall back to the gravitational body, in some sort of orbit, once the opposing force (thrust) is removed. This restriction is only for constant values of escape velocity, and therefore only useful for ballistic trajectories, but it's not true to say that it does not apply to continuous (or at least non-instantaneous) thrust trajectories as well. siafu (talk) 00:07, 9 October 2012 (UTC)

I have undone that edit. Escape velocity is an initial velocity, allowing escape without further propulsion. The statement was obviously correct. In order to avoid this happening again, it would be nice to have a reference for it, but, as it is so trivially obvious, a ref may be hard to find. - DVdm (talk) 08:39, 9 October 2012 (UTC)
The statement is still incorrect; in order to escape, this velocity will be attained, otherwise escape simply won't happen. Escape velocity is also the velocity at which a continuous-propulsion craft can stop thrusting and coast to infinity, and if it does not ever cross this velocity threshold, it will remain in an orbit around the central body. I think you are confusing the situations in which escape velocity is a meaningful and useful quantity, with what is strictly speaking true. You can call it an initial velocity now, but there's no precondition on the equation derivation that says that this is required as calculating an escape velocity can be done for any value of r, whether the particle is "at rest" (wrt the central body) or moving at high relative speed. You'll have a hard time finding a reference for something that is incorrect, I'm afraid. siafu (talk) 14:53, 9 October 2012 (UTC)
Re "in order to escape, this velocity will be attained": It doesn't, because "this velocity" changes with location. As soon as you go a bit further out, the escape velocity gets less. When you change your location, "this velocity" is irrelevant and no longer exists. When you have unlimited propulsion, the limiting escape velocity is zero. And of course what you call "a continuous-propulsion craft" is essentially a propulsionless craft when it does "stop thrusting".

Input from others? - DVdm (talk) 15:34, 9 October 2012 (UTC)

Exactly; this is what I explained in the first place. Escape velocity is dependent on location, and this is exactly why the statement as written is incorrect-- "walking speed" will eventually be the escape velocity as you get far enough away. If you stop applying thrust just short of the point where your velocity exceeds the local escape velocity, you will not escape. I'm not advocating making the article more confusing with a complicated explanation of this, but it should not include assertions that are incorrect. siafu (talk) 15:47, 9 October 2012 (UTC)
I still think the statement is correct. If "could achieve the same result" refers to "could escape to infinity", then for all practical purposes the concept "escape velocity" is useless and irrelevant. With propulsion one can escape at any speed, and "walking" stands for "any" here. I see no problem with the statement, but obviously I would support replacing "walking" with "any" in the statement. - DVdm (talk) 16:00, 9 October 2012 (UTC)
The problem is that this statement is based on the choice of starting point to be a priveleged one. Starting from a very high orbit, the escape velocity is very low, and could even be walking speed. Starting from the surface of the Earth, it's very high. It's not providing any useful information to point out that you could do it ballistically or not. The use of "could escape to infinity" comes from the mathematical derivation itself, as the article (and any source describing escape velocity) says:

At its final state, it will be an infinite distance away from the planet, and its speed will be negligibly small and assumed to be 0.

I do not disagree that escape velocity is only meaningful when considering ballistic trajectories, since in a non-ballistic trajectory one need be safe in the knowledge that at some point your velocity will happen to equal the escape velocity. This is different from saying that it doesn't apply, because nothing in the mathematical assumptions involved in the escape velocity equation is violated in a non-ballistic case-- at least, not anything that isn't already untrue in reality for the ballistic trajectory. siafu (talk) 16:09, 9 October 2012 (UTC)
Yes, you are correct where you say that starting from a very high orbit, the escape velocity is very low, and could even be walking speed. So the "walking" should indeed be replaced with "any". No problem. - DVdm (talk) 16:26, 9 October 2012 (UTC)

Strictly speaking, the escape velocity for any body is zero at infinity. Virgil H. Soule (talk) 07:28, 25 January 2014 (UTC)

Name change[edit]

I suggest that the name of this page be changed to 'Escape speed'. As is disscussed above, escape speed is a speed and not a velocity. KingSupernova (talk) 15:40, 9 January 2013 (UTC)

While this is true, it's also true that the term is commonly referred to as "escape velocity" despite being a speed; for example, David Vallado's Fundamentals of Astrodynamics, probably the best-known and most popular textbook in the field, uses the term "escape velocity", as do the vast majority of sources. Per WP:COMMONNAME, policy is to use the common name for a thing, even if that name is less perfect than alternate, less common names. siafu (talk) 16:23, 9 January 2013 (UTC)


The Misconception section currently starts with the following statement: "Contrary to what is sometimes asserted, a powered vehicle, such as a rocket, does not have to attain the escape velocity in order to leave orbit and escape to outer space." Per discussion previously on this talk page, this statement is not true. Per the definition of escape velocity, the speed required for escape to infinity drops off with distance; if an object travels away from the central body at constant body, escape velocity will drop until it is equal to the object's velocity. If the object does not reach this point (i.e., if it stops adding energy), it will not escape. What might be a true statement is that it's not necessary to achieve any particular escape velocity (i.e., escape velocity for any particular r) to escape the gravity well, but I rather doubt that that's a common misconception. siafu (talk) 01:14, 14 March 2013 (UTC)

Also, this statement is a bit misleading: "The gain in energy is equal to the speed (in a non-rotating coordinate system centred on the planet) times the momentum gain, and since the momentum gain depends only on the mass of fuel ejected (times the exhaust speed relative to the rocket), it is best to fire the engine when the speed is high. (See Oberth effect.) So to escape the earth, for example, the fuel use is minimized if the rocket first attains low Earth orbit as quickly as possible and then (perhaps later) executes one or more burns, always at the perigee of whatever orbit it is in at that stage, until it acquires enough energy to escape." The mention of low Earth orbit is a bit weird-- obviously you want to reach LEO as quickly as possible executing further burns, but for a number of reasons that much more important than the Oberth effect (like getting above the atmosphere); the Oberth effect is only really invokable in the reasoning for executing an escape burn at perigee rather than elsewhere in the orbit. I suggest rewording this as: "So to escape the earth, for example, the fuel use is minimized if the rocket executes one or more burns, always at the perigee of whatever orbit it is in at that stage, until it acquires enough energy to escape." I'm not against removing the text altogether, though, as it's starting to bear less relevance to the topic of escape velocity when pared down. siafu (talk) 01:24, 14 March 2013 (UTC)

Sources, sources, sources... - DVdm (talk) 06:55, 14 March 2013 (UTC)
The section has none. Maybe it should just be removed. siafu (talk) 16:05, 14 March 2013 (UTC)
No problem for me. - DVdm (talk) 16:54, 14 March 2013 (UTC)
Done. Also, were you asking me to provide sources for challenging the veracity of unsourced statements before? That doesn't make sense... siafu (talk) 02:39, 15 March 2013 (UTC)
Nono... I was just asking to provide sources for your suggested rewording, of course. Otherwise this would never end. Cheers and thanks. - DVdm (talk) 12:03, 15 March 2013 (UTC)


Escape velocity is vector in the sense that it is always oriented away from the center of mass of a body in space. Gravitational potential energy is always oriented toward the center of mass and kinetic energy must be directed away to oppose it. Escape velocity is only of academic interest in defining the characteristics of a body in space. It has no real use in orbital mechanics. The escape velocities listed in the table are surface escape velocities for the bodies listed. In general, the Sun escape velocity is the only one that matters since vehicles launched on Earth-escape trajectories will typically enter an orbit around the Sun. A vehicle launched on a Mars tranfer orbit, for example, is in orbit around the sun. It should be noted that escape velocity can be defined at any radial distance from a body. The event horizon is effectively the surface of a black hole since anything interior to that is undefined.

Discussion of orbits or escape orbits should be separated from that of escape velocity. An escape orbit is a hyperbolic orbit with eccentricity greater than one. A parabolic orbit is unstable in that it may have eccentricity equal to one at some point but perturbations could cause it to become elliptical and closed at some other point. A body passing Jupiter, for example, can receive a gravitational assist that will insert it into a hyperbolic escape trajectory that would take it out of the solar system. A tangential launch or velocity-change maneuver could inject a vehicle into an escape orbit but usually is used to inject it into an elliptical orbit of some sort. The Apollo missions were launched on elliptical free-return trajectories. If a lunar injection maneuver wasn't performed on the Moon's backside, the vehicle would simply have looped around and returned to the Earth. To be captured by the Moon, the vehicle's velocity vector would have to be reduced in magnitude to the required Lunar orbital velocity.

The location of a launch site doesn't have much to do with escape velocity. An equatorial launch site is most desirable because it eliminates fuel that would be needed for plane-change maneuvers required to reach an equatorial orbit. Virgil H. Soule (talk) 16:40, 17 July 2013 (UTC)

We shouldn't get too far off topic, but many of the statements above are incorrect. Escape velocity is derived from conservation of energy, and is thus just a speed, and not a vector in any sense of the word. It is not at all true that escape velocity need be directed in any particular direction-- the assumption behind its derivation is that all forces acting in the system are conservative, so as long as nothing is done to violate that assumption (e.g. adding air drag or physically contacting the central body), pointing the velocity in any direction will result in orbital escape. Moreover, energy is never "directed" as it is also just a scalar. Think on it this way: pointing the velocity vector in the direction of the gravity gradient (i.e. towards the central body) will cause the orbiter to convert potential energy into kinetic energy as it gets closer, and then the reverse, converting kinetic into potential as it moves away, but the total energy is constant.
The definition of a hyperbolic orbit is an orbit where e > 1, so "An escape orbit is a hyperbolic orbit with eccentricity greater than one" is completely redundant and irrelevant. Parabolic orbits are unstable, yes, but this is a more technical statement in the domain of stability theory because small pertubations in a parabolic orbit will tend to push the system away from e = 1 rather than revert towards it; again, completely irrelevant to this page. Finally, an equatorial launch site makes a big difference, and not just because of plane-changing maneuvers. The speed of an object on the surface of the Earth relative to the center of of mass of the Earth is much greater than the speed of an object sitting on the pole; this additional speed that is provided "for free" is reflected in the smaller amount of fuel required to achieve escape velocity from the equator than from the poles. siafu (talk) 15:58, 18 July 2013 (UTC)
You didn't read what I wrote! Gravitational acceleration is a vector directed toward the center of mass of a body, always. To escape a body's gravity, another body must be accelerated radially away from the parent body and attain a certain radial velocity. This velocity is called Escape Velocity. The direction requirement makes escape velocity a vector. Escape Velocity is a purely hypothetical concept. This whole discussion has gotten all muddled up with talk of orbits and space launches and all that, which have nothing to do with escape velocity. The article itself has suffered as a result of all these misconceptions. It's really a very straight-forward subject. Virgil H. Soule (talk) 07:28, 25 January 2014 (UTC)


Isn't this whole concept a bit...cocked up? The reach of gravity is infinite, isn't it? It just gets incredibly weak? If there were only two massive bodies in the universe (and no motive forces other than gravity), they would eventually collide, no matter how far apart they started. There is no "escape". There are simply greater forces and more prevalent gravity wells that counter gravity between distant masses. Applejuicefool (talk) 01:22, 2 September 2013 (UTC)

Perhaps a question for our wp:reference desk/science. This is where we discuss the article, not the subject — see wp:talk page guidelines. Good luck. - DVdm (talk) 07:24, 2 September 2013 (UTC)
If the kinetic energy of an object is greater than or equal to the gravitational potential energy, i.e. it has attained escape velocity, then no matter how long you wait the object has escaped and will keep going forever. In your specific (simplified) case, it is also possible that the two objects could never meet but orbit each other indefinitely. — Reatlas (talk) 06:30, 3 September 2013 (UTC)
It's difficult to understand why this proposition is wrong without a good understanding of calculus. For the same reason that Achilles will catch up with the turtle, there is indeed an actual limit in terms of speed away from each other that massive bodies will never be able to overcome. As they move away from each other, gravity will slow them down, but the force of gravity will also be getting weaker at the same time. siafu (talk) 14:51, 3 September 2013 (UTC)
I was taught in school that escape velocity is determined by the kinetic energy needed to lift an object from the surface of a body to "infinity". In a universe where gravity is an inverse-square force, this energy is finite. When gravity is an inverse-with-respect-to-distance force, as in the Planiverse, the energy is infinite. But that doesn't mean one can't travel from one planet to another.
This is a simple explanation (though, as pointed out, it helps to understand the integral calculus). Why the article doesn't start from such a simple explanation, I don't understand. This is rather a Rube Goldberg article -- explaining something simple in a complicated way. WilliamSommerwerck (talk) 18:08, 22 December 2013 (UTC)

Wrong hypothesis[edit]

There are a bunch of x-rays coming out of back holes. So the black holes escape gravity must be less than 299,792 km/s. Why don't you guys just learn the empirics first, and then the (often wrong) theories? -- (talk) 11:55, 11 September 2013 (UTC)

The X-rays don't come out of the black holes. They are generated outside the event horizon by infalling charged particles. For further questions you better go the wp:reference desk/science. Here we discuss the article, not the content — see wp:talk page guidelines. Good luck. - DVdm (talk) 12:01, 11 September 2013 (UTC)
Told ya to learn, not to critise: -- (talk) 12:03, 11 September 2013 (UTC)
DVdm is correct, and the article you have linked corroborates that. siafu (talk) 18:14, 11 September 2013 (UTC)
Oh yeah, let's quote it: "List of unsolved problems in physics - Why do the discs surrounding certain objects, such as the centersofactive galaxies, emit radiation jets along their polar axes?" But your answer was funny indeed, you just made my day. :D By the way DVdm just said another hypothesis, because we don't have any empirical data from where the jets are coming...they could also come directly out of the black holes core, so you may call it rather "drainage" than "hole". -- (talk) 13:26, 22 December 2013 (UTC)

launch pad and Earth[edit]

The article says "For a mass equal to a Saturn V rocket, the escape velocity relative to the launch pad is 253.5 am/s (8 nanometers per year) faster than the escape velocity relative to the mutual center of mass. When the mass reaches the Andromeda galaxy the earth will have recoiled 500 m." This is a very odd statement I think because it initially talks of the the rocket in relation to the launch pad and then at the end it brings in the earth. So is the statement concerned with the reaction of the earth or the launch pad to the rocket? (talk) 15:45, 10 October 2013 (UTC)

Since the launchpad is (presumably) attached to the Earth, the two are equivalent. siafu (talk) 20:16, 10 October 2013 (UTC)
The above may be correct but I think it needs rewriting so that it is clear. I think the reference to the launch pad is an indication that the launch is from the surface of the earth which has a lower escape velocity than the mutual center of orbit of the earth and the rocket ready for launch. I think the point is then confused by the recoil which is another issue. I am currently finding this difficult to describe and write because the original statement has many confusing references to different objects and so it should be simplified. (talk) 09:45, 11 October 2013 (UTC)