Talk:Escape velocity

Multiple sources

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 http://www.physicsforums.com/showthread.php?t=16708, but the author hellfire says he/she is not sure.--Patrick (talk) 08:02, 1 February 2010 (UTC)

A ladder

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? 68.200.98.166 (talk) 02:42, 27 January 2010 (UTC)

Escape velocities with respect to the sun

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)

Removal of Misconception Section

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)

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

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"

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?

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

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)

Error

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)

Escape Velocity of the Milky Way

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)