Escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body. However, gravity has infinite range. Object $A$ is always getting pulled by the gravity of object $B$ no matter the distance between object A and B. (Maybe the force becomes extremely small with increasing distance, but still not zero)

So, how can an object possibly “escape the gravitational influence of another object”? Is there a more accurate definition of escape velocity? Won't the orbital path just become infinitely large with increasing initial velocity?

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    $\begingroup$ I've removed some comments which answered the question. Please use comments to improve the post they're attached to, and use answers to answer the question. $\endgroup$ – rob Aug 15 '19 at 19:11
  • $\begingroup$ Not every infinite sum has an infinite result: How can Achilles ever catch up with the tortoise? $\endgroup$ – Peter - Reinstate Monica Aug 16 '19 at 13:47

12 Answers 12


Is there a more accurate definition of escape velocity?

First, consider an object with a radial orbit (zero angular momentum orbit) in a $1/r^2$ central force field. The total energy $E$ of the object (which is constant) is the sum of the (negative) potential energy $U(r)$ due to the force field and the kinetic energy $T(v)$ due to the radial speed.

There are three cases to consider:

  • $E \lt 0$: The orbit is bounded, i.e., there is a maximum, finite distance $r = r_{max}$ where the speed (and thus kinetic energy) is zero, and the object has maximum (least negative) potential energy

  • $E \ge 0$: The orbit is unbounded, i.e., the object never has zero speed. As $r \rightarrow \infty$, the kinetic energy $T$ asymptotically approaches $E$ from above.

  • $E = 0$: A special case of the unbounded orbit in that $T \rightarrow 0$ as $r\rightarrow\infty$.

It is this special case that is relevant to the definition of escape velocity. At any radius $r_0$ in the central force field, there is a speed $v_e$ such that

$$T(v_e) = |U(r_0)|$$

Thus, an object starting at $r = r_0$ with outward radial velocity $\vec{v}_e$ has just enough kinetic energy to, ahem, 'coast to a stop at $r = \infty$'.

More precisely, the object will coast arbitrarily far away with speed arbitrarily close to zero. In this sense, the object 'escapes' the central force field, but just so.


Yes, what you have really discovered is that the phenomenon under consideration has been mislabeled as "escape". At least as far as a layman's likely understanding. Or perhaps the problem is the layman's understanding of "from what". Where you are thinking of "from gravity", what the scientist means is "from a closed orbital path" (elliptical), or in short "from orbit" itself, "orbit" implying a return. As to how an object can possibly evade the persistent force of gravity trying to return it, maybe this will help: an object moving away from a gravitational body is experiencing a decrease in the body's gravitational pull as distance increases. It is possible for an object to have sufficient speed that the decreasing pull always remains below the object's ability to "outrun" that pull. The object's speed never decreases below what would allow the body to pull it back in from whatever distance it has currently managed to obtain. It "wins" the energy race.

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    $\begingroup$ After I read the first sentence, my expectation was that if I continued reading, I'd learn the correct label for the mislabeled velocity in question. But it isn't made clear in your answer what that correct label is. Is it non-return velocity? Is it outrun velocity? Is it winning velocity? Something else? $\endgroup$ – Alfred Centauri Aug 15 '19 at 20:46
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    $\begingroup$ A correct label has not yet been coined to my knowledge. $\endgroup$ – Jeff Y Aug 16 '19 at 2:17

The object doesn't escape the gravitational influence; as you have noted, $1/r^2$ is never equal to $0$. However, what we mean by "escape velocity"is that the object has enough energy (large enough velocity) so that its path would have to "turn around at infinity". In other words, the object's velocity is high enough and decreasing slow enough to where the object will never turn around.

  • $\begingroup$ +1, but I'm a bit confused. How can an object be "at" infinity if infinity isn't a point? $\endgroup$ – user238497 Nov 21 '19 at 14:24
  • $\begingroup$ @Ramanujan_π See my final sentence that gives that idea a more physical backing. $\endgroup$ – BioPhysicist Nov 21 '19 at 14:30
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    $\begingroup$ @Ramanujan_π, Sometimes it's easier to say something like "an object launched at escape velocity turns around at infinity" than it is to say something more formally correct such as, "the distance at which the object turns around increases without bound as the launch velocity approaches escape velocity." $\endgroup$ – Solomon Slow Nov 21 '19 at 14:34

The important part here is the speed of the orbiting object.

Every distance of the orbiting object has its own orbiting speed (orbiting along the circle). If you increase the speed, the orbiting path changes to an ellipse; if you increase the speed more, the ellipse becomes more and more prolonged.

If you continue to increase speed, there will be the moment in which the path becomes parabolic — and parabola is not closed curve anymore, so the object escapesnot from the gravitational force, but from the orbiting around the object.

  • $\begingroup$ How about throwing an object up to the sky vertically? Escape velocity still applies to this scenario and the orbiting speed is zero. $\endgroup$ – doge99 Aug 13 '19 at 15:25
  • $\begingroup$ @doge2048719 The other answers do not rely on orbits. $\endgroup$ – BioPhysicist Aug 13 '19 at 15:36
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    $\begingroup$ If the initial speed is high enough, the object will still move away, albeit with still decreasing speed. The limit of speed will be zero, but the speed never reach this limit, so it will be still positive. So object escapes — again, not from the gravitational influence, but from closed path; it's distance will never be shorter. $\endgroup$ – MarianD Aug 13 '19 at 15:40
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    $\begingroup$ @doge2048719, Every ballistic trajectory in the vicinity of a planet is a conic section with one focus at the planet's center of mass. If you throw a baseball, it follows an extremely skinny elliptical path--an orbit, literally--until that path intersects the ground. If you throw the baseball absolutely straight up, it becomes a degenerate ellipse. Or, if you could throw it hard enough, it would be a degenerate parabola or a degenerate hyperbola, and it would never fall back. $\endgroup$ – Solomon Slow Aug 13 '19 at 17:42
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    $\begingroup$ @NuclearWang, while your point is worth noting, I don't understand your pushback here. It seems to me that the OP likely has no problem with the idea of a transfer orbit from LEO to the vicinity of the moon but, rather, is concerned with the notion of escape velocity in the idealized central force case. $\endgroup$ – Alfred Centauri Aug 15 '19 at 21:32

You are right. The notion of "escape" from a gravitational field is something that only exists in an idealized mathematical model of the situation. In that idealized model, we have only one gravitator, in a space-time of infinite extent. With that, we can state the idea of the escape velocity as thus:

It is the minimum speed that an object must be thrown away from the gravitator, in order that it will reach infinite distance, after an infinite lapse of time.

If you throw the object with less speed than this, it will come back: either it will strike the gravitator's surface, or it will go into a repeating orbit around it that nonetheless stays within some maximum distance from the object. But if you throw it outward with speed at least the escape velocity or higher, then the orbital path "becomes infinitely large" - it goes out to, and intersects (in a sense) infinity.

In formal language, if $\mathbf{r}(t)$ is a vector pointing from the gravitator toward the thrown object and which traces its trajectory, and at $t = 0$ we throw the object, the escape velocity is the minimum speed $v_\mathrm{esc}$ such that if

$$|\mathbf{r}'(0)| \ge v_\mathrm{esc}$$

i.e. the speeed at time of throw is larger, then

$$\lim_{t \rightarrow \infty} |\mathbf{r}(t)| = \infty$$

, that is, the distance from the planet grows forever. In reality, of course, there are two limitations: for one, we can never have an infinite time go by, which means that the object will, at all times, still be theoretically under the influence of the gravitator, even if that influence is extremely small. The other limitation is that in reality, there will be other gravity sources present at distance, which will then change the trajectory in other ways once their influence begins to dominate control over the motion of the thrown object and this, in turn, will also prevent reaching that infinite distance. For example, if we are talking about the Solar System, and we are "escaping" the Earth, if we launch a craft away with no more than Earth's escape velocity, then it will only "escape" until the Sun's gravity begins to dominate it, at which point it will now follow a more complicated trajectory based on mutual influence between the Earth and Sun (mostly, it will orbit the Sun, since the Earth will move away in the interim, but it may encounter the Earth again), and also, to some extent, the other planets.

Nonetheless, it is useful because we can take it as a ballpark for the minimum speed, and hence minimum effort which our engine must expend, to "get away" from the gravity source and get to a more remote destination like, say, Mars. Of course, actually travelling there will require more speed change (so-called "delta vee"), because we will also need to navigate to intercept that destination and then settle onto its surface safely.

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    $\begingroup$ "The other limitation is that in reality, there will be other gravity sources present at distance, which will then change the trajectory in other ways once their influence begins to dominate control over the motion" - given the significance of this year, would you consider mentioning the crossing, by the various Apollo spacecraft, of the equigravisphere? $\endgroup$ – Alfred Centauri Aug 14 '19 at 0:36

Throw a stone up, and it will fall back on Earth. But if you give it very, very large speed, it will go away, into space. Surely, it's always under the gravitational influence of the planet but it has speed great enough that it never returns back.

The minimum speed needed for this is called escape velocity. And at that speed (or greater) the path of the object is an open curve; this means the curve extends to infinity and the object, hence, has no bounds on how far it can go.


How can an object possibly “escape the gravitational influence of another object”? Is there a more accurate definition of escape velocity? Won't the orbital path just become infinitely large with increasing initial velocity?

The apparent paradox can be resolved by considering the kinetic and potential energy of an 'object' projected from a planet's surface at some initial velocity.

Imagine an object travelling outwards from the centre of gravity of a planet at uniform speed and calculating the potential energy being gained. If the force on the object due to gravity was uniform with distance then the potential energy would increase linearly with distance. If the object was released from a given height it would fall and impact with a velocity V (assuming no atmosphere).

Energy = mgh = 1/2mv^2
So V from a given height = (2gh)^0.5

However, with increasing distance from the 'planet's" center of gravity the inverse square gravitational effects means that the energy needed to reach that altitude increases more slowly with distance. The integration of required energy with increasing distance has a finite value at infinity. |

Conversely, an object "falling" from an infinite distance has non-infinite potential energy and so non-infinite impact velocity. This is the escape velocity.

An object projected from the planet at escape velocity will lose kinetic energy as it "rises" but the rate of loss will decrease with distance due to the inverse square gravitational effect. It will ultimately come to a stop at infinite distance.
An object projected at lower than escape velocity will "stop rising" at some distance less than infinity.

An object projected at above escape velocity will still have net kinetic energy and "outward" motion both at infinity and at all lesser distances.


You know the old saying, "What goes up, must come down"? If you go up at escape velocity or faster, it's not true.

You'll keep slowing down, but never come down, because the gravitational influence of the body you're "escaping" falls off (by the inverse square rule) as you get further away. It never goes away, but it's never strong enough to reverse your velocity.


This notion confused me too, but a simple way to explain it would be to compare the gravitational pull on the object at each given moment to the sum of an infinite series of numbers, say the series (1/2)+(1/4)+(1/8)+(1/16)… It's easy to see that this series goes on forever, as you can simply cut each new term in half to create the next term. However, while this series is always getting bigger, it will never approach infinity. If you could sum up all the infinite terms of this series, it would ultimately add up to 1.

Think of a empty box with a volume of 1 cubic meter. Fill in half the volume with water, then fill in half the remaining volume with water, then half the remaining volume again... you will get closer and closer to filling the box exactly, which is a finite volume, despite the fact that you are adding water an infinite number of times. If you place the box on a scale opposite an identical box that is completely filled with water, the box you are adding water to will never tip the scale in its favor.

In the same way, if you were to add up the pull of gravity on an object traveling at escape velocity over an infinite time frame, you would find that the pull of gravity on that object is ultimately adds up to a specific number - escape velocity. Therefore, going at or faster than escape velocity means that the object will never be pulled back, similar to how the partially-filled water box in the previous paragraph will never outweigh the completely-filled one.

  • $\begingroup$ OK but gravity is an inverse square law, not inverse exponential, as in your example. And your final paragraph appears to be adding up a sequence of forces to produce a velocity, which doesn't make sense in terms of the units. $\endgroup$ – David Richerby Aug 16 '19 at 10:00

You are right. To talk of escaping the Earth’s gravitational field is nonsense, and confusing nonsense, and therefore immoral.

A particle at rest at the distance of Sirius and free of all other gravitational influences will fall towards the Earth. The acceleration is 56 microns per century per century.

If you imagine an object projected up from the Earth at high speed, the Earth’s gravity will slow it down and carry on slowing it down for ever. Because gravity is governed by an inverse square law, the total slowing down adds up to a finite figure. (If gravity had been $1/r$ rather than $1/r^2$ the sum would have been infinite, not finite).

“Escape velocity” is thus defined as “the speed at which you have to start if you want to get all the way to infinity without falling back”. Or rather, since that is impossible, it is the speed greater than all those that do fall back eventually.

Talking about escaping the Earth’s gravitational field is a lie, but at least an understandable one, since if you are next to Sirius you will be subject to gravitational accelerations much greater than 56 microns per century per century.

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    $\begingroup$ "To talk of escaping the Earth’s gravitational field is nonsense, and confusing nonsense, and therefore immoral." There seems to be a logical disjunction between your first 2 points and your third. Your argument appears to be missing a few steps or key definitions, and is thus nonsense (and confusing nonsense). After all: An apple is a fruit, and a tree fruit, and therefore from Asia. $\endgroup$ – Chronocidal Aug 14 '19 at 15:49
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    $\begingroup$ Therefore immoral? Science education is built on teaching more and more refined models knowing all the while they are inaccurate. We can't just jump to field theory. $\endgroup$ – Schwern Aug 14 '19 at 19:05
  • $\begingroup$ I took what he was saying to mean "If intentionally taught to confuse, it's immoral." I suppose that my interpretation of what he was saying might also be hasty, but it doesn't seem helpful to beat that statement up too awfully much. $\endgroup$ – tgm1024--Monica was mistreated Aug 15 '19 at 1:40
  • $\begingroup$ Regarding the 4th paragraph: there are an infinity of speeds that meet the definition of escape velocity that you give. Using the 2nd sentence of that paragraph as a start, isn't it more the case that an object with any speed less than $v_e$ will return eventually? $\endgroup$ – Alfred Centauri Aug 15 '19 at 21:16

If you fell from infinity to the body you would go splat on it with a given amount of energy, energy you picked up gradually as you fell. However if instead you set off from the body with slightly more kinetic energy than that which you went splat with, then you would reach infinity and beond! The speed you need to have that amount of kinetic energy is the escape velocity.


Won't the orbital path just become infinitely large with increasing initial velocity?

No, is the short answer (more on that in a moment).

But firstly, your basic confusion appears to arise from mixing up two separate concepts:

a. For the orbital path to be infinitely large implies an infinite initial velocity (which obviously is not possible in any realistic scenario: nothing in the universe travels with infinite velocity).

b. For an orbital path to be possible at all implies a finite initial velocity -- and we're back to "no is the short answer" -- because:

i) If it was possible to have infinite initial velocity, by definition that must exceed the escape velocity, which is a defined, finite, velocity. The moment it does exceed that (finite) velocity, which would be long before reaching infinite velocity, any type of orbit becomes impossible: the rocket must go shooting off into space.

Orbit is an equilibrium state, where the force pushing the orbiting object away from the planet (basically, its momentum) is exactly equal to the force pulling it toward the planet (basically, gravity). If the outward impetus is in exact balance with the inward impetus, the equilibrium thus achieved causes the orbiting object to maintain a constant distance from the planet: an orbit.

ii) A slight imbalance in those forces will cause the orbiting object to orbit out to a greater radius distance, or to orbit in to a lesser radius distance; but a large (positive) imbalance will result in the object leaving orbit: it will cease to follow a curved path about the planet: the larger imbalance will cause it to follow, increasingly, a less curved path, until the path has largely ceased to curve at all: the object is now going in a straight line (approximately), so has ceased to orbit.

iii) Having once attained escape velocity, whether or not this follows upon an orbiting stage, if the rocket does no more than maintain its constant velocity it can never be recaptured by the planet's gravity, because the gravitational strength falls off with distance: under the inverse-square law, when the distance from the planet doubles the gravitational strength is not constant, but reduces to one-quarter. With merely constant speed, the rocket can never be captured by the falling gravitational force (because the latter is falling).

Only by decelerating could the rocket reduce its momentum such as to be travelling with less velocity than the critical threshold needed for equilibrium, i.e. an orbit, at its new radius distance from the planet (a very large deceleration would be involved, once the rocket has travelled even a short distance, because in real life, at a very short distance from the planet its gravity gives way to the Sun's far greater gravity, with the planet becoming thereafter a negligible influence).

So a further misunderstanding in your question now emerges: by treating the planet as if it was the only source of gravity in near-space, you have misled yourself into neglecting nearby stronger gravitational forces! Those forces overwhelm the planetary effects at quite short distances, such that your orbital-path assumptions break down, once the rocket has moved only quite a short distance from the planet.



The very small differences in orbital altitude associated with an eliptical orbit are irrelevent to my answer. The only relevent issue is whether the orbit is stable: if it is, minor altitude differences won't clarify the position any. I am addressing only the main issue: is the satellite-cum-rocket able to accelerate (taking it out of orbit), and is it able to decelerate (taking it out of orbit).

Like it or not, in reality all rockets burn fuel at a constant rate, and the only reason they go faster is that gravitational resistance to their motion decreases as altitude increases, obeying the inverse-square law.

Burning fuel at a constant rate, the amount of fuel it is possible to carry sets a real limit on its initial velocity (if it is going further than Earth orbit), and there are practical restrictions on the weight of fuel it is possible to lift (every extra ton of fuel in the tanks increases the mass which must be lifted into space). So increasing the payload of fuel has negative consequences also, not purely positive ones.

And a high enough initial velocity will prevent the rocket going into orbit: it will do what planetary probes do, and depart for Mars or beyond. If it never completes even a single orbit, it is not reasonable to describe it as having an orbit. If it will also depart the solar system, as some do, it will not even be orbiting the Sun, in any meaningful sense.

The gravitational influence of a single quark is tiny, for an individual quark, so it is only logical to conclude that its radius is also tiny. It is therefore not logical to conclude that its influence is infinite. The inverse-square law demonstrates that even with objects of stellar mass, the influence declines rapidly. Doubling the distance, the strength declines to a quarter, i.e. 25 percent (1 over 2 squared, i.e. 1/4). At 8 times the distance, the strength is only one percent (1 over 8 squared, i.e. 1/64th).

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    $\begingroup$ I recommend editing your answer to at least (1) clarify point a which doesn't seem correct (hyperbolic orbits do not require infinite initial velocity), and (2) clarify the 6th paragraph which seems to imply that an object in orbit maintains a constant distance from the planet (but the distance isn't constant for elliptical orbits). $\endgroup$ – Alfred Centauri Aug 15 '19 at 21:07
  • $\begingroup$ Do you truly believe that an object which has an orbital eccentricity greater than 1 is genuinely in orbit, in any meaningful sense? It is self-evident that neither a circular, elliptical, parabolic nor hyperbolic orbit requires infinite velocity, in an engineering sense; and no orbit can be infinitely large, since the universe has a finite size; but in a mathematical sense the existence of an orbit which is infinitely large necessarily implies infinite velocity (how does an object with finite velocity circumnavigate an infinite path?), which is a proof that no orbit can be infinitely large. $\endgroup$ – Ed999 Nov 21 '19 at 21:55
  • $\begingroup$ Ed999, "hyperbolic orbit" is a term of art regardless of whether you or anyone else believes such a path is a genuine orbit or not. $\endgroup$ – Alfred Centauri Nov 22 '19 at 3:18
  • $\begingroup$ You are evading the issue: how can an object which has an orbital eccentricity greater than 1 genuinely be in orbit, in any meaningful sense? $\endgroup$ – Ed999 Dec 15 '19 at 4:50
  • $\begingroup$ Ed999, I've noted that you believe I'm evading "the issue". $\endgroup$ – Alfred Centauri Dec 15 '19 at 20:24

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