Talk:Escape velocity

Could someone write the escape velocities into units of mach as well as km/s? Thanks. -- Astudent 04:52 Apr 30, 2003 (UTC)

Umm, you do realize that mach pretty explicitly refers to a speed in an atmosphere, and when you're in an atmosphere, atmospheric drag is going to completely skew the escape velocities anyway, right? -- John Owens 06:13 Apr 30, 2003 (UTC)

---

This makes no sense....

(One complication is that virtually all astronomical objects rotate. The frame of reference must not rotate for that statement to be correct. Moreover, the gravitational slingshot effect sometimes involves transfer of energy to the projectile from the slingshot body that depends on the spatial relationship between projectile and body. The body loses some angular momentum -- possibly rotational as well as orbital -- adding kinetic energy to the projectile. Therefore, the complete gravitational field of the slingshot body must be included in the overall field, which then can no longer be approximatively treated as symmetric. Moreover, not only do escape velocities vary from place to place, they vary with time in such cases. Even moreso, they may sometimes depend on direction.)

Again, this makes no sense.....

(For reasons given above, computers may often have to be used to compute solar-system escape velocities to some desired precision.)

(One complication is that virtually all astronomical objects rotate.

Planets and stuff spin.

The frame of reference must not rotate for that statement to be correct.

Refers to "so that "velocity" is a misnomer; it is a scalar quantity and would more accurately be called "escape speed"". If you have a rotating frame of reference, then it will be a vector quantity rather than scalar, and it would be appropriate to call it a "velocity" after all. E.g., in computing the escape velocity for the solar system from a point on the Earth, or in its near vicinity, you would get different escape speeds depending on where in the Earth's gravitational field the object was, even if you calculate the velocity relative to the Sun's position rather than the Earth's.

Moreover, the gravitational slingshot effect sometimes involves transfer of energy to the projectile from the slingshot body that depends on the spatial relationship between projectile and body.

The Voyager spacecraft didn't have enough fuel to get out of the solar system under their own thrust; because they were given trajectories that gave them a significant boost from the slingshot effect at Jupiter & Saturn, that speed became greater than the escape velocity in that direction. If they'd just been given that much speed in a random direction, they would almost certainly still be orbiting the Sun.

The body loses some angular momentum -- possibly rotational as well as orbital -- adding kinetic energy to the projectile.

In the slingshot effect, some of the larger body's (call it a planet) momentum is transferred to the spacecraft. Because the planet has such greater mass, it's not measurable how much the planet is slowed down (or sped up, if you approach from the other direction). Mechanically, a slingshot can be considered much the same as a totally elastic collision between the planet and spacecraft; it's just that long-range gravity is the medium of the interaction, instead of close-up electromagnetic forces (Van der Waals, I think?). There can also be a minor component of acceleration from the tidal effects, which will affect the planet's rotation.

Therefore, the complete gravitational field of the slingshot body must be included in the overall field, which then can no longer be approximatively treated as symmetric. Moreover, not only do escape velocities vary from place to place, they vary with time in such cases. Even moreso, they may sometimes depend on direction.)

So for calculating an escape velocity from the Sun, you need to consider "lumps" in the gravitational field, such as moving planets, which will alter the speed needed to escape the system in different directions.

(For reasons given above, computers may often have to be used to compute solar-system escape velocities to some desired precision.)

I hope that from the above, it should be obvious (and not even in the way mathematicians say "it's obvious") why a computer would be helpful.

--John Owens 22:32, 2004 Feb 1 (UTC)

---

This thing has way, way too many parenthetical remarks. If the information is important, just work it into the body of the article. --24.156.119.50 16:14, 28 Mar 2004 (UTC)

I'm inclined to agree. I just added a section; now maybe I'll go back and do a bit of rearranging of the earlier sections. Michael Hardy 22:40, 6 Sep 2004 (UTC)

---

I figure this article would really benefit from a brief table showing the escape velocities of major astronomical bodies (earth, luna, sun, mars, saturn, maybe our galaxy). -- Finlay McWalter | Talk 22:47, 25 Jul 2004 (UTC)

These two sentences need work

Someone added these two sentences in the first paragraph:

It is a theoretical quantity, because it assumes that an object is fired into space like a bullet. Instead propulsion is almost always used during the first part of the flight, and to "escape", at no time the escape velocity need to be attained.

Right at the beginning it says this is about objects that are not being propelled, so "the first part of the flight" during which propulsion is used is simply not relevant; this topic does not apply to it, but only to the later part where there is no propulsion. Moreover, it does not make sense to say "propulsion is almost always used", since a comet passing by the Earth that is not pulled into an orbit around the Earth escapes without propulsion, simply because it is always moving at a speed higher than the escape speed at whatever distance from Earth it's at. Moreover, there is no assumption about "being fired like a bullet", since the assumption, stated right in the first sentence, is that this whole thing applies only when there is no "propulsion". The last clause is ungrammatical; it says:

to "escape", at no time the escape velocity need to be attained.

What does that mean? Does it mean an object under propulsion can be lifted to an infinite height? If so, it's not relevant; this is only about objects not under propulsion. More precisely, it is about objects on which no forces act except the conservative gravitational field. How other objects behave, i.e., objects on which other, possibly non-conservative forces act, is simply not part of this topic. This topic treats only objects that are not propelled.

And what is meant by the words "It is a theoretical quantity"? I might have thought it means that in reality gravitational forces other than the Earth act on an object. But after the word "because" it then appears that something else was intended. Michael Hardy 14:43, 7 Sep 2004 (UTC)

Note that the escape velocity from the surface of Earth is not a speed that a spacecraft sent to outer space actually attains: while being propelled the height increases and the corresponding escape velocity decreases.

But if the spacecraft is not going to orbit Earth, but is being sent to photograph Uranus close-up, doesn't it actually go a lot faster than the escape speed at the surface? I seem to recall that the Apollo spacecraft going to the moon moved at something like 10 miles per second during the early part of the trip. Michael Hardy 22:45, 7 Sep 2004 (UTC)

I don't think so, [1] says 10.4 km/s (Earth Fixed Velocity).--Patrick 08:52, 8 Sep 2004 (UTC)

Urhixidur, please explain why this would not be correct:

(Considering the kinetic energies it may seem strange that a speed of 12 km/s (by itself corresponding to 72 MJ/kg) can increase the kinetic energy of the spacecraft from 450 to 900 MJ/kg, but the gain is at the expense of the Earth's kinetic energy, the launch will slightly slow down Earth.)

--Patrick 16:56, 8 Sep 2004 (UTC)

There is no mention of where the vehicle's energy is coming from, and it need not come from the Earth. Consider the Earth and the prepped vehicle as two distinct bodies, co-orbiting the Sun and in contact with each other. Assume the vehicle is at the Earth's far side from the Sun. The vehicle suddenly acquires the requisite extra 12 km/s of forward motion, possibly by ejecting part of its own mass at extremely high velocity in the opposite direction (I chose the vehicle's position so that the ejecta would not blast the Earth's surface and cause complications; here we assume the vehicle and its ejecta both "miss" the Earth itself). The vehicle will indeed eventually leave the solar system (its ejecta will do the same much faster, in all likelihood). In what way has this affected the Earth's momentum or energy? None whatsoever.

There is the matter of the gravitational influence of the vehicle on the Earth and vice-versa, but this can be neglected here. In fact, if you insist on not neglecting it, you'll notice that the vehicle, since it leaves *ahead* of Earth, will lose some of its energy to the Earth's benefit through gravitational interaction (it needs to escape the Earth as well as the Sun). Had the vehicle decided to leave in the opposite direction (which would require a 72 km/s delta-v!), its gravity would indeed slow the Earth very slightly.

The sentence as originally stated introduced a badly formed problem. Giving an object a velocity within a frame of reference gives it energy within that same frame (in this case 12 km/s translating to 72 MJ/kg). Switching to the Sun-centered frame, we see the object go from 30 km/s to 42 km/s and hence its energy go from 450 to 882 MJ/kg, a difference of 432 MJ/kg. Without specifying the rest of the system implicated (what imparted the extra velocity to our object?), we can't make any pronouncements.

Is this explanation clear enough?

Urhixidur 17:44, 2004 Sep 8 (UTC)

Interesting example. I have again considered the energies. Assume that the ejected mass is 1/n times the mass of the spacecraft, then it is ejected at a speed of 12n km/s relative to Earth. The apparent gain of 360 MJ per kg spacecraft can be explained by the fact that the kinetic energy, relative to Earth, of the ejected mass is 72n MJ per kg spacecraft, while relative to the solar system the energy only increases 72n - 360 MJ per kg spacecraft. With a large n as in your example it is clear that the large amount of energy to reach 42 km/s has to be produced anyway in the form of kinetic energy of ejected mass. I wonder whether there is a theoretical minimum amount of energy needed for the acceleration from 30 to 42 km/s; is it actually possible to use some kinetic energy of Earth, by "pushing it away"?.--Patrick 22:12, 8 Sep 2004 (UTC)

Assume, as you do, that the reaction mass (ejecta) is the fraction 1/n of the spacecraft's original mass. In the Earth frame, the initial kinetic energies are zero because nothing's moving. In the Sun frame, the kinetic energies are 450 MJ/kg for either parts (the reaction mass and payload parts). After the explosion, in the Earth frame we have the payload going one way with 72 MJ/kg (for the fraction (n-1)/n of the spacecraft's total mass) and the reaction mass going the other (at speed 12(n-1) km/s) with 72(n-1)² MJ/kg (note the square you missed). The overall (average) energy is 72(n-1) MJ/kg. This must have come from the explosives' potential chemical energy. In the Sun frame, we have the payload now going at 882 MJ/kg, and the reaction mass at 18(7-2n)² MJ/kg, average energy 18(21+4n) MJ/kg, delta 72(n-1) MJ/kg. Energy is conserved in translating between frames.

Urhixidur 22:53, 2004 Sep 8 (UTC)

OK, that seems the same; I used n for what you call n-1, and I expressed everything in kg payload (therefore I do not have the square).--Patrick 02:08, 9 Sep 2004 (UTC)

Say what?! Even if using a unit spacecraft mass, the square of the velocity is still in the picture. Assume n = 10; in that case the craft (9/10 of it) goes at 12 km/s whilst its reaction mass (1/10 of the original mass) goes at 12*9=108 km/s (conservation of momentum). The energy of the former is 72 MJ/kg whilst that of the latter is 5832 MJ/kg --its the square of the velocity, see...

Urhixidur 03:46, 2004 Sep 9 (UTC)

That is per kg reaction mass, for 1 kg payload we have 1/9 kg reaction mass.--Patrick 12:30, 9 Sep 2004 (UTC)

I'm not sure it's to the point, but exhaust speeds of 100 km/s are way, Way, WAY out there. See Spacecraft propulsion#Table of methods and their efficiencies. Normally it's the fuel that'd be 80–90% of the loaded ship.

Somewhere this needs the minimum escape speed from both the Earth & Sun = ${\sqrt {11.2^{2}+42.1^{2}}}-30=13.6\ \mathrm {km/s}$.

—wwoods 06:56, 9 Sep 2004 (UTC)

For rocket propulsion, yes, but this is a gedanken-experiment with "cannon propulsion". In combining Earth and Sun, you are quite right. Another approximation could be to work out the energy differential in moving to the edge of Earth's Hill sphere, and then let the Sun take over.

Urhixidur 11:39, 2004 Sep 9 (UTC)

speed vs velocity

what would be a better name for the phrase? escape velocity or escape speed? the article implies that speed is more accurate, as it is a scaler. i think this makes no sense, as escape velocity implies a speed to get off a planet, right? for the magnitute of the speed is to be at its minimum, doesn't the angle of launch need to be 90°? therefore it has a direction, and it is a vector quantity - so velocity is far more appropriate?

however i dont want to remove it from the article, as i am not sure i am right. comments? mastodon 22:20, 26 November 2005 (UTC)

The angle of lauch does not matter. That is why it is a scalar. (And scalar is the correct spelling. Michael Hardy 23:02, 26 November 2005 (UTC)

why not? if the angle was 45 degrees from the normal, a launched projectile would spend longer in the higher density part of the gravitational field, so would be more strongly attracted to the body, and would need more of a force, a greater magnitude of speed to escape. am i right? or am i not wrong?

Starting with the escape velocity, the velocity is a function of elevation (see parabolic trajectory). It is true that starting at an angle of 45 degrees the increase of altitude is slower, but also the decrease of speed, because in this case the gravity is not opposite to the velocity.--Patrick 01:07, 1 December 2005 (UTC)

also to be pedantic, as seems the trend (rolls eyes), if the angle was - say - 180° from the normal, the speed would need to be a little bit bigger to escape. comments? mastodon 23:12, 30 November 2005 (UTC)

It applies only when the parabolic orbit does not intersect the planet, hence, in the case of starting from the surface, if the direction is not downward (not lower than horizontal).--Patrick 00:29, 1 December 2005 (UTC)

I added some text about this in the orbit section.--Patrick 01:05, 1 December 2005 (UTC)

It's counterintuitive. The quickest way to understand this is to see that all you need is enough kinetic energy to lift you to infinite height, and that's finite because the field strength decreases fast enough as you go up. Michael Hardy 01:45, 1 December 2005 (UTC)

Question - shouldn't escape velocity = sqrt(2G/R) instead of sqrt(2GR)?

I'm still not convinced it can be called speed. It seems to me that since velocity is a vector, only the part of the vector that is directly opposing gravity would be helping the object escape. The other part of the vector would only be causing it to spin around the larger body. Am I missing something? --Paulie Peña 23:33, 1 March 2006 (UTC)

Indeed, you are missing something. You are absolutely right that this result is counterintuitive. But think about the sum of kinetic energy and potential energy. If the projectile's kinetic energy at its present position is more than its potential energy at infinite distance from the planet, then it will escape. Michael Hardy 00:34, 2 March 2006 (UTC)

A large horizontal speed also helps the object escape.--Patrick 02:52, 2 March 2006 (UTC)

Definition

Patrick, I disagree with your reverting to the previous version, claiming that mine was "not an improvement." I have run the two versions by a number of people, who all agree that my version is easier to read without losing any of the content. Of course we're talking about "physics," and of course it's for a given gravitational field. The current version is redundant. - Ztrawhcs.

"In physics" is the usual way to provide context. Escape velocity is a property of a gravitational field and a position, it does not depend on the object. Your version ignores dependence of position, and suggests dependence on the object. "its gravitational field" is also confusing. --Patrick 09:52, 27 January 2006 (UTC)

your version ... suggests dependence on the object ... Touche. Good point. I still think, however, that we need not completely sacrifice readability. The intro paragraph doesn't have to scare people away with its complexity. I can just imagine an entire generation of 10 year old would-be scientists seeing the wikipedia entry on Escape Velocity and turning away in brain-twisting horror, deciding never to become scientists but instead to become garbage men. I hope you're prepared to deal with that. Ztrawhcs 21:05, 2 August 2006 (UTC)

Newer results

Somebody should review the following references, our current data seems inexact:

• Annexed the number that appears in Wikipedia on escape velocity from Solar

System from Earth: 43.6 km/s. In agreement with my paper A note on Solar Escape Revisited * the correct figure is 16.6 km/s. When we get 43.6km/s or 13.6 km/s we neglect the recoil kinetic energy of the earth and that is nonnegligible in Sun's coordinate frame.

• The escape velocity from a position in a field with multiple sources is

derived from the total potential energy per kg at that position, relative to infinity. The potential energies for all sources can simply be added. For the escape velocity this results in the square root of the sum of the squares of the escape velocities of all sources separately.*

• For example, at the Earth's surface the escape velocity for the combination

Earth and Sun is √(11.2² + 42.1²) = 43.6 km/s. As a result, to leave the solar system requires a speed of 13.6 km/s relative to Earth in the direction of the Earth's orbital motion, since the speed is then added to the speed of 30 km/s of that motion.* Diaz-Jimenez_Anthony_P._French-MIT_Professor,_American_Journal_of_Physics,_56,_85,_1988 Diaz-Jimenez_Anthony_P._French-MIT_Professor,_American_Journal_of_Physics,_56,_86,_1988

(from OTRS) David.Monniaux 08:10, 16 February 2006 (UTC)

You mean 13.6 should be 16.6? How is that computed?--Patrick 02:44, 18 February 2006 (UTC)

Pr Diaz-Jimenez sent the Foundation several published papers that revisit the escape velocity. I can forward them to you by email (you can contact me with "send mail to"). David.Monniaux 09:23, 18 February 2006 (UTC)

Well what speed do e.g. rockets typically have to attain? If 16.6 is right then they would have to have attained that speed already. 203.218.86.162 09:28, 31 May 2006 (UTC)

formula in section 2

i have swopped the approx= and = because v is approximately the thing in the root, which exactly equals 11200. 203.218.86.162 09:28, 31 May 2006 (UTC)

Direction of rotation

I have added some clarifications on the direction of rotation of the body. Though rotation is not a must to account for the escape velocity, the direction in which the escaping body is launched is important, if the main body is rotating. --Wikicheng 04:06, 29 June 2006 (UTC)

Escape velocity 50 km above the surface of Venus

I'm working on the Colonization of Venus article right now, and part of it talks about floating cities at 50km above the surface where the atmospheric pressure and temperature are the same as Earth. What would be escape velocity be there? It doesn't seem like it would be that much different from the surface given that even 9000 km from Earth doesn't even halve the number. Venus rotates extremely slowly but these theoretical cities would be up where the wind velocity causes them to whip around the planet every four days. Mithridates 05:58, 9 August 2006 (UTC)

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)