# Couldn't I escape Earth's gravity traveling only 1 mph (0.45 m/s)?

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 (0.45 m/s) - directed away from the surface and you will eventually reach space. So why is the escape velocity 7 miles/s (11 km/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?

• Comments are not for extended discussion; this conversation has been moved to chat. Aug 31, 2016 at 17:22
• Escape velocity is a mathematical definition. What initial speed do you need to reach infinite distance to the planet. It takes infinite time to get there, but the escape velocity has the necessary kinetic energy for infinity. To reach a low or higher orbit, less energy and velocity is needed.
– Uwe
May 30, 2017 at 22:00
• Copying David Hammen's comment here as IMO it serves as perfect answer to this question: 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.
– Atul
Jul 17, 2018 at 16:22

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.

• Comments are not for extended discussion; this conversation has been moved to chat. Aug 31, 2016 at 17:22

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.

• Using a rocket to proceed upwards at a constant speed of 1 mph would be a gigantic waste of propellants. The waste is so gigantic that this is impossible.
– Uwe
May 12, 2017 at 20:00
• It's only a gedankenexperiment to help understand the concept of escape velocity... May 13, 2017 at 1:26
• Technically, Earth's Hill sphere is closer to 10^6 km instead of 10^5. Jan 23, 2020 at 21:04

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.

• Practically speaking, there are other objects in the vicinity that will affect you so this is just an academic exercise. Jun 6, 2014 at 11:58
• Why 'just'? By that standard, isn't every orbital calculation an academic exercise? What application of Newtonian physics would not be? Jun 8, 2014 at 1:01
• This answer is based on there being other celestial objects besides Earth. If there were not (a vacuum experiment, so to speak), eventually the Earth and the object will come back together in the absence of any propulsion.
– Neo
May 31, 2021 at 10:56
• @Neo that is incorrect, an object can escape from a lone primary body (albeit at a higher speed than escaping the Hill Sphere when you do factor in larger bodies). In particular, because the difference in gravitational potential at any point surrounding the primary body and a "point at infinity" is defined and finite, having that much or more initial potential energy ensures that the bodies will never start moving towards each other again. Sep 2, 2022 at 21:09

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.

• 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. Jun 3, 2014 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
Jun 3, 2014 at 22:54
• @TimGostony, you're right it was a typo, but I'm now defiantly not changing it! :D Jun 3, 2014 at 22:55
• @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. Jun 6, 2014 at 16:05
• @GodricSeer, what you wrote is actually incorrect. You specifically said "thrust", so you are assuming a rocket. You have ignored the utter nastiness of the rocket equation. You are also ignoring the Oberth effect. The math changes drastically with a space elevator as opposed to rockets, but you explicitly said "thrust". Jun 6, 2014 at 17:57

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.

• 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, 2015 at 20:02
• The important thing is that you need to be powered stronger than earth's gravity (which is pulling you back). Mar 4, 2018 at 9:54
• I read all more popular answers, but they still were missing something. These two lines can hold their own without the technical stuff and still answer the question. Now I will never forget how simple the distinction is. Jul 20, 2019 at 18:17

Escape velocity is the speed at which you'll leave the Earth and not return if you don't continue to propel your craft. Below that speed, gravity will pull you back down.

If you want to keep propelling your rocket vertically at 1 m/s for 100,000 seconds, you'll need an indescribably vast amount of fuel to do so, because you have to maintain sufficient thrust to cancel out Earth's gravity for that entire time.

Also, merely being in space isn't enough to keep you from falling back to Earth, as has been discussed in many other Q/A here. XKCD's got one of the more accessible explanations.

• It is important to note that escape velocity is a horrible term as it is a scalar value -- escape speed would be better. If you have escape speed in any direction (including down) you will escape the body -- as long as you don't hit anything.
– Erik
May 30, 2017 at 22:08
• So it's a practicality requirement, and note an absolute? Theoretically you could just go straight up at any speed and just use a few megatons of fuel? May 30, 2017 at 22:17
• @AlexMann Far more than just "a few megatons" of fuel (more like $~5 × 10^{135}$ tons of fuel per ton of payload, more fuel than there is mass on Earth), and not an escape: if you're 100km up at 1 m/s when you run out of fuel, you fall back to Earth. May 30, 2017 at 22:57
• Escape velocity does not have to be along the vector you describe @AnthonyX.
– Erik
May 31, 2017 at 3:09
• This is the most correct answer. If you were to throw a ball in any direction at escape velocity, a planet's gravity would not be enough to overcome the ball's motion. However, if the ball is under thrust, then you need to factor in the delta-v (the total velocity change of the object) to determine whether the object will reach escape velocity with the available thrust and fuel. Oct 27, 2019 at 22:27

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.

• escape velocity is a scalar and not a vector. Nov 30, 2015 at 13:33
• As @SebastianWozny points out, escape velocity is a scalar. You can be moving any direction.
– Erik
May 16, 2017 at 16:13

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.

Edit 4 years later to consider a solar sail

If you had a nearly infinite fuel supply, and you kept moving away from Earth at 1 mph, yes you could escape. You could do this with a solar sail there are a couple of issues using the sail near Earth but assuming you start in a high stable orbit, you could easily expand until your escape. Of note, using a solar sail, as you move farther from Earth your speed would increase unless you lowered the efficiency of the sail. In other words, if you started with a solar sail to get 1 mph thrust, you would need to work to maintain that speed, otherwise you would soon be going faster.

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.

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.

• Erm, you're forgetting about the inverse-square law. ;) Jul 14, 2014 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... Jul 15, 2014 at 5:44
• You do not need to accelerate above gravity continuously. Once you reach 1 mph, it is enough for the acceleration to match the gravity. Other than that, yes delta-V of straight up is insanely expensive. Kerbal Space Program (or Orbiter for more scientific minds) is excellent for such experiments.
– Suma
Aug 5, 2018 at 22:47
• No, @Suma, it's not. You have to account for friction. And there is measurable friction well past 200 km. (Enough to require adjustments even on geostationary orbits. Very small adjustments, but adjustments none the less. Sep 14, 2018 at 4:55
• @aramis The friction above 200 km is measurable for satellites / stations moving at orbital speed (8 km/s). If you will be moving at 1 mph (0.5 m/s), your friction will be about 100 000 000 times lower. Below 20 km there will be some friction, but it will still be order of magnitude weaker than gravity.
– Suma
Sep 14, 2018 at 11:11

What everybody who is used to thinking about rockets takes for granted but which may not be intuitively clear is: Every gram of fuel burned to counteract gravity is a gram of fuel wasted. Think about the worst case: Hovering on top of a running rocket engine which is just strong enough to keep you afloat. You are burning fuel without going anywhere. Your suggestion to go somewhere at 1 m/s is a bit better, but just a bit.

This leads directly to the main criteria for an efficient start trajectory: Minimize the time accelerating against gravity, because you could achieve the same acceleration plus 9.81m/s2 with the same fuel accelerating sideways! As soon as a starting rocket has any speed to speak of, it will begin to tilt sideways as much as possible. Ideally it would accelerate perpendicular to gravity right from the start, e.g. after having been lifted by a carrier plane, or on a magnetic levitation track on a celestial body without atmosphere.

Now if you don't want to counteract gravity for the reasons outlined above, you need speed to leave a gravity well. The direction is irrelevant, unless it gets you on a collision course. Theoretically, once you are in space (no drag) you could spiral out with a small but continuous tangential acceleration and slowly escape Earth's gravity well without ever reaching 11.2 km/s. I suppose this would be a viable option for a photon drive equipped probe already in space, if it is initially slower than escape velocity. (A photon drive can run for a long time because energy can be transported or transmitted easier than reaction mass.)

Yes you could. That's how the a space elevator would work -- provided you're using a earth rotational frame of reference.

By crawling up the tether to the geosynchronous altitude, then continuing to travel away from the Earth (but now down the tether) one reaches the point where one can let go, and centrifugal force will do the rest.