# Couldn't I escape Earth's gravity traveling only 1 mph?

It is said that in order for an object or a projectile to leave Earth's gravitational pull, it must reach Earth's escape velocity, meaning reach a speed of 7 miles per second (~11 km per second). Well, as far as I understand, you could easily escape Earth's gravity even at 1 mph - directed away from the surface and you will eventually reach space. So why is the escape velocity 7 miles/s?

Is it because the object has to gain a certain speed once it reaches orbit in order to maintain that altitude? Or is it because practically an object can't carry infinite amount of fuel, and so it has to reach a certain speed to maintain its orbit before all fuel is gone?

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One reason this is a great question is that it is often explained that a black hole's event horizon cannot be escaped 'because escape velocity is greater than the speed of light'. However it is explained here that gravitational bodies can be escaped at any velocity, including 1mph, as long as there is constant thrust. The 'escape velocity' explanation appears to not sufficiently describe the real reason for why a black hole's event horizon cannot be escaped. – user1445967 Jun 3 '14 at 20:40
@user1445967 - It most certainly does. When talking about black holes you are well into the realm of general relativity. Newtonian concepts don't work in that realm. You can't escape because every direction is down inside the event horizon. – David Hammen Jun 3 '14 at 20:47
Your confusion is due to a misunderstanding about precisely what is meant by "escape velocity". Wikipedia's definition will help: the speed needed to "break free" from the gravitational attraction of a massive body, without further propulsion -- the bit that I highlighted is the bit you are missing: without further propulsion. If you are attaining a steady rate of 1 MPH straight up against gravity then something is propelling you, and that is "further propulsion". The escape velocity is the velocity you must attain so that you don't need that propulsion. Make sense? – Eric Lippert Jun 3 '14 at 22:46
If you have a second question then please post a new question. Don't edit an existing question to ask a new question; one question per question! – Eric Lippert Jun 3 '14 at 22:54
For reference: an analogous question on Physics: physics.stackexchange.com/questions/29779/… – David Z Jun 4 '14 at 1:32

The force of gravity decreases with distance. It follows an inverse-square relationship... essential to know when you're grinding out the math, but not essential to a conceptual understanding.

The fact that gravity decreases with distance means that at some distance, it will be negligible; an object sufficiently distant from Earth may be considered to have "escaped" Earth's gravity. In reality, the force of gravity has no distance limit; two objects would have to be at infinite distance from each other to have no gravitational interaction, but for practical purposes, one can think of finite distances where gravitational forces become small enough to ignore.

Consider an object some large distance from Earth... right at the edge of what we would consider the Earth's gravitational "sphere of influence". Some tiny movement toward Earth will increase the gravitational attraction, accelerating the object toward Earth. The process will escalate with the object's velocity and acceleration increasing. If we ignore the effects of Earth's atmosphere, the object will continue its acceleration until it strikes the Earth's surface at some velocity.

Now, let's reverse everything. The object magically launches up from Earth's surface at exactly the same speed as our falling object had at the instant of impact. As it rises up, gravity tugs on it and it slows down. As it gets further away, gravity diminishes so it decelerates more slowly. Eventually, it gets to some distance where it has come to a stop, but Earth's gravity no longer has any effect on it.

The velocity our object had at Earth's surface is Earth's escape velocity. In precise terms, a body's escape velocity is the velocity an object in "free fall" must have in order to escape the gravitational influence of that body - no more and no less. Technically, escape velocity can be specified for any distance from the center of a body, and the value will decrease with distance, but when a planet's escape velocity is stated, it is usually for the planet's surface. Mathematically, it is calculated as an integral of the body's gravitational acceleration from some specified distance to infinity.

An object does not have to travel at escape velocity to escape a planet's gravity, but the same amount of energy needed to accelerate an object to escape velocity must be applied to an object (giving it potential energy) to lift it out of the planet's gravitational sphere of influence. The difference is that at escape velocity, the object needs no external influence to escape; at anything less than escape velocity, some external force must be applied.

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Great answer, when I was done reading I went like "aah!" totally got it now! :D – olli Jun 4 '14 at 14:30
The inverse-square law is of fundamental importance, as it makes the gravitational potential tend to a finite limit when going off to infinity. With a simple inverse relationship for instance, there would be no finite escape velocity (or energy) at all. – Marc van Leeuwen Jun 4 '14 at 22:33
@MarcvanLeeuwen: Same principle as Riemann series, I suppose? – Laurent LA RIZZA Jun 5 '14 at 10:58
There's a more specific point at which you normally reach the "edge" of the Earth's gravity: when its gravity balances with some other body (Sun, Moon, etc.) This gives a true zero point (and it's much closer to the Earth than you'd think looking at the Earth in isolation). – Jerry Coffin Jun 5 '14 at 18:49
@JerryCoffin: The escape velocity is defined as the velocity that would be needed to overcome the Earth's gravitational potential in the hypothetic situation that the Earth were the only gravitating body around. If you take into account other bodies, it is not really an escape velocity: the stated velocity does not allow you to overcome the graviational potential of the Sun (let alone that of our galaxy), and therefore will not get you arbitrarily far from the Earth. – Marc van Leeuwen Jun 6 '14 at 7:57

Escape velocity reduces as you get further away from the Earth. If you proceed upwards at a constant speed of 1 mph (which as noted will require continuous thrust to counteract gravity), you will eventually reach a distance where the escape velocity is equal to 1 mph. Then, you will have reached escape velocity and are no longer gravitationally bound to the Earth.

This distance is extremely great; around 4×1012 km or 26000 AU. In practice, third-body effects (moon, sun, other planets) will dominate when you get beyond 105 km away from the Earth.

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So escape velocity is distance related. Thanks – user3339411 Jun 3 '14 at 23:08
By far the best, most direct answer to the question. – Vedant Chandra Aug 18 '15 at 14:41

To sum up the answers: the escape velocity is the velocity that, at a given distance, is sufficient to escape the gravitational field so that no additional energy (= acceleration) is needed.

That is, if you are 26000 AU from Earth, you don't need any more fuel to counteract Earth's gravity, you just float away. However, when at Earth's surface, you will need additional acceleration to sustain the 1mph velocity - otherwise you just fall back down like the tossed ball.

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Great answer - all the others are either flawed or far too complicated. – jwg Jun 5 '14 at 10:47
Practically speaking, there are other objects in the vicinity that will affect you so this is just an academic exercise. – Brian Jun 6 '14 at 11:58
Why 'just'? By that standard, isn't every orbital calculation an academic exercise? What application of Newtonian physics would not be? – Spike0xff Jun 8 '14 at 1:01

You are confusing velocity and acceleration. If you were to jump standing on the surface of the Earth you might experience 8 m/s which is 17 mph velocity upward, but the acceleration of gravity would act to retard your motion, slowing your velocity down. If you have a high enough velocity, the effect of (de) acceleration can not slow you down before you get far enough away from the gravitational source.

So if you could keep a constant velocity of 1 mph, you would defiantly be able to escape the earth. The problem is that would require constant thrust. If you're going 11 km/s then you can just relax and watch the world shrink in your rear view mirror.

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Also note that the amount of energy needed to maintain that upward velocity of 1 mph to the point where escape velocity is 1 mph vastly exceeds the amount of energy needed at attain a velocity of 11 km/s right off the bat. – David Hammen Jun 3 '14 at 20:50
I'm pretty sure defiantly is a typo for definitely, but I really like it in this case. Defiantly escaping the Earth! Who needs it anyway? – Tim Gostony Jun 3 '14 at 22:54
@TimGostony, you're right it was a typo, but I'm now defiantly not changing it! :D – ThePlanMan Jun 3 '14 at 22:55
+1 just for defiantly accepting the DYAC. – David Hammen Jun 4 '14 at 3:58
@DavidHammen that is actually incorrect. The amounts of energy would be equal. The thrust to maintain 1mph would only have to counter the potential energy gained by the increased height. The KE of a 1kg object at 11.115 km/s is 62.217 MJ. The potential energy of a 1kg object gained rising from 6.4x10^6 m to 4x10^15 m is 62.218 MJ, and I am going to count the difference from rounding. – Godric Seer Jun 6 '14 at 16:05

I think if you are powered (rocket/motor ) you can go at any speed and escape the gravity. The escape velocity is only for objects thrown (projected into space), with the initial velocity and they are not powered.

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Thanks. Yes, this makes sense. – user3339411 Jun 3 '14 at 23:03
One of the best answers to the question, pointing out that the escape speed makes sense only for an unpowered spacecraft which must counter the gravity until it is insignificant. – mins Feb 7 '15 at 20:02

The key difference is that "escape velocity" is how fast you would have to throw a stone straight up from the Earth's surface (ignoring air drag), for it to escape from Earth's gravitational influence. It would be coasting the whole way, always losing speed due to Earth's gravitational pull.

If, on the other hand, you have a rocket engine with sufficient fuel, you can just keep rising slowly (1 mph), which is almost a hover, until you've gotten way out into space and Earth's gravity is overwhelmed by the Sun, Jupiter, etc. You could keep throttling back to maintain the same upward speed (gravity decreases with distance, and the rocket carries less fuel) if you wanted to, or let the rocket speed up.

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escape velocity is a scalar and not a vector. – Sebastian Wozny Nov 30 '15 at 13:33

Unless you are very far away from Earth, if you are only moving away at 1 mph the gravity of Earth will pull you back to Earth (assuming you do not have an infinite fuel supply to maintain a 1mph thrust). So you are correct when you say

Is it because the object has to gain a certain speed once it reaches orbit in order to maintain that altitude.

Think of a ball tossed in the air, it starts by moving quickly, but as it rises higher it goes slower, than stops and falls back down. At some point it is moving away from Earth at 1mph, but gravity overcomes that momentum. Air Resistance has some impact on the ball, but you can throw horizontally much farther than you can up.

Gravity works pretty much them same on the surface of the Earth as it does a 1000 miles up. When you throw something horizontally it falls towards the earth in an arc, attracted by the gravity of the Earth. If it is moving fast enough the curvature of the Earth will match the arc of the falling object, this is called Orbital speed and the object will not hit the earth.

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Looking at this in another way, consider the concept of gravity wells. The gravity well of course is not a "real", physical well, but it is a commonly used metaphor to describe how much energy is required to escape from the gravitational effect of a body, and it provides a reasonably straight-forward way of answering your question. (Space buffs, bear with me below; this is meant as an explanation, not a university-level physics and astronomy lecture.)

If you are at or near the bottom of a gravity well (say, at the surface of the Earth) and want to climb out of that well, you basically have two options. Either climb very fast for a short distance (this is the approach taken for getting off the surface of the Earth, for reasons stated in other answers), or climb slowly for a much longer distance (this works once you are far enough away from the body forming the gravity well that the predominant gravitational forces acting on you are small or negligible). Each way of looking at it represents the same thing: you provide some sort of energy input, usually in terms of fuel of some kind, which is used to climb the "side" of the gravity well. The energy provided as input becomes potential energy as you climb farther from the surface, and at some point, your potential energy exceeds the gravitational pull at that point of the body that forms the gravity well; you "continue on a tangent" and move straight on from that point forward rather than following the curve of the gravity well. Once that happens, you have reached escape velocity from that body.

If you don't climb far enough for your rate of climb at the time you stop actively climbing, then when you stop climbing (let's assume you cannot grab hold of anything, because in space there is nothing to hold on to) you will fall back toward the body that forms the gravity well you are trying to climb out of; you did not attain escape velocity.

Of course, there are usually multiple gravitational forces to contend with at any one point. However, one of them will project a stronger force on you than the others; that's the concept behind the sphere of influence. Near Earth (yes, that most definitely includes low Earth orbit), it's Earth's gravity that dominates; take a trip to Luna and its gravity will exert the greater force once you pass the Earth-Moon system L1 Lagrangian point.

The "depth" of a gravity well is often given as its escape velocity, in km/s or some other convenient measurement of velocity, taken at the bottom of the well. Hence, the depth of Earth's gravity well is approximately 11.2 km/s, which is the escape velocity at the Earth's surface. Wikipedia gives the escape velocity at 9,000 km above the Earth's surface as 7.1 km/s, but as we have seen in other answers, getting to 9,000 km above the surface itself takes a lot of energy, negating the gain from the lower "absolute" speed necessary to break free of the Earth's gravity.

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to maintain a speed of 1 mph long enough to escape, one is accelerating about 34 feet per second per second (1.46 feet per second above gravity) straight up. To get outside the hill sphere (and into "solar space" rather than being in "Earth Space"), you're looking at 107 years of continuous 1.05G acceleration.

So, in theory, yes, but in practice, the delta-V makes it insanely expensive.

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Erm, you're forgetting about the inverse-square law. ;) – TildalWave Jul 14 '14 at 23:35
No, basically, just ignoring it, since the delta-V is about 1/4 what it would be for the whole time. 1G thrust at reasonable exhaust speeds hits 90% mass in a matter of hours. And it's 100 miles to LEO... – aramis Jul 15 '14 at 5:44

## protected by TildalWaveJun 3 '14 at 22:17

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