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I have read multiple explanations of escape velocity, including that on Wikipedia, and I don't understand it.

If I launch a rocket from the surface of the Earth towards the sun with just enough force to overcome gravity, then the rocket will slowly move away from the Earth and we see this during conventional rocket launches.

Let's imagine I then use slightly excessive force until the rocket reaches 50 miles per hour and then I cut back thrust to just counterbalance the force of gravity. Then my rocket will continue moving at 50 mph toward the sun. I don't see any reason why I can't just continue running the rocket at the same velocity and keep pointing it towards the sun. The rocket will never orbit earth (by "orbit" I mean go around it). It will just go towards the sun at 50 mph until it eventually reaches the sun. There seems to be no need whatsoever to ever go escape velocity (25,000 mph).

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marked as duplicate by knzhou, John Rennie newtonian-mechanics Feb 11 at 16:46

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    $\begingroup$ With just enough force to overcome gravity - I read that as "with enough force to reach escape velocity". Mind you, gravity does not just stop at one point, it only decreases with the squared distance. If you want to properly escape gravity, ie. move with enough force so that you cannot be pulled back, that is literally escape velocity. $\endgroup$ – ComicSansMS Feb 11 at 13:14
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    $\begingroup$ Escape velocity decreases with altitude. Eventually, as you get far enough from the Earth, it will drop below 50 mph. (Well, it would, if the Earth was floating alone in interstellar space. In practice, the Sun's gravity will start to dominate long before that happens.) $\endgroup$ – Ilmari Karonen Feb 11 at 13:17
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    $\begingroup$ I don't see any reason why I can't just continue running the rocket at the same velocity and keep pointing it towards the sun. Who says you can't do this? If you look at the derivation of escape velocity, it is pretty obvious that you are assuming that you start off with the escape velocity and don't apply any other forces afterwards. But no one is saying that is the only way to "escape Earth" $\endgroup$ – Aaron Stevens Feb 11 at 14:02
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    $\begingroup$ Re, "I don't see any reason why I can't just continue running the rocket at the same velocity and keep pointing it towards the sun." It's because you can't build a rocket that carries enough fuel to do it that way. Also, if it's your intent to send a probe that goes 50 miles per hour all the way to the Sun, it's not even going to get half way there during your lifetime. Space is Big! $\endgroup$ – Solomon Slow Feb 11 at 15:07
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    $\begingroup$ Possible duplicate of Can we escape Earth's gravity slowly? $\endgroup$ – knzhou Feb 11 at 16:19
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Escape velocity is the velocity an object needs to escape the gravitational influence of a body if it is in free fall, i.e. no force other than gravity acts on it. Your rocket is not in free fall since it is using its thruster to maintain a constant velocity so the notion of "escape velocity" does not apply to it.

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    $\begingroup$ I think his rocket is not accelerating, but there is a constant thrust applied by the rocket. Either way, as you point out, what he describes is not escape velocity. $\endgroup$ – garyp Feb 11 at 12:27
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    $\begingroup$ @garyp Well, yes, I meant "accelerating" in the sense of "being under the influence of an acceleration other than that of gravity", not in the sense of "having non-zero net acceleration". It doesn't actually matter for this scenario whether the rocket is maintaining a constant speed or not, as long it's outputting thrust. $\endgroup$ – ACuriousMind Feb 11 at 12:29
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    $\begingroup$ The rocket is not accelerating. It is moving at a constant speed of 50 mph. The acceleration with respect to the earth is 0. $\endgroup$ – Ambrose Swasey Feb 11 at 12:44
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    $\begingroup$ @AmbroseSwasey A body moving at a constant speed in a gravitational field is accelerating. Think about what happens when you throw a ball into the air - without any additional force input, it slows down and reverses direction. The rocket needs to constantly output thrust just to maintain its speed. $\endgroup$ – Nuclear Wang Feb 11 at 13:40
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    $\begingroup$ @NuclearWang accelerating means changing velocity, not "experiencing propulsive force". $\endgroup$ – Ján Lalinský Feb 11 at 14:00
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If I launch a rocket from the surface of the Earth towards the sun with just enough force to overcome gravity, then the rocket will slowly move away from the earth and we see this during conventional rocket launches.

This is not what happens in actual spaceflight. The actual rockets work for a short time and after that, the spacecraft is moving by inertia. And they don't really work against the Earth's gravity - the vertical launch purpose is to shoot the rocket to the high altitude where the atmosphere is thin. Then the rockets turn and accelerate horizontally to hain enough velocity to get on the orbit or the desirable escape trajectory. What you describe would be extremely ineffecient and no rocket exist to actually do that in real life.


To understand why no rocket can reach far this way let's do a quick calculation of the amount of fuel required. Let's assume that going at a constant speed 50 mph (80 kmh) you want to reach a 80 km altitude (the altitude one needs to be awarded by the astronaut wings in US). At that altitude the gravity acceleration $g$ is almost the same as on the ground. That's why we will assume it to be constant. Then you rocket fighting this acceleration for an 1 hour should have so much fuel that if it were in an empty space without any gravitating body it would speed itself to the velocity equal $\Delta v= 1 \mathrm{hour}\cdot g$. The Tsiolkovsky equation relates this speed to the ratio of the mass of the fueled rocket $m_0$ to its final mass $m_f$. \begin{equation} \frac{m_f}{m_0}=\exp\left[\frac{\Delta v}{g I_{sp}}\right]=\exp\left[\frac{1\,\mathrm{hour}}{I_{sp}}\right] \end{equation} where $I_{sp}$ is a so-called specific impulse depending on the type of the rocket. For the idealized LH2-LOX rocket $I_{sp}=450\,\mathrm{sec}$. This means that for such rocket $\frac{m_f}{m_0}=e^{8}\simeq 2980$. I.e. to elevate 1 ton just to this altitude this way you need the same amount of fuel as the mass of the whole Saturn V rocket. And this computation is idealized i.e. all rocket engines, the supporting structure, fuel tanks etc are included into this 1 ton. If we raise the altitude the mass ratio grows exponentially i.e. you need $\simeq 10^{17}$ tons of fuel just to elevate 1 ton to the altitude of the ISS.

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    $\begingroup$ Thank you for going beyond the other answers and explaining the issue with doing what OP described. I've had this question since I was a child; now, at 25, I finally understand why we bother with the notion of an escape velocity. It just seemed so irrelevant if we have rockets... it's not as if we try to reach the velocity on the ground and then take a ramp upwards. $\endgroup$ – Luc Feb 11 at 13:26
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    $\begingroup$ This is discussed further in Why are rockets so big $\endgroup$ – Kyle Kanos Feb 11 at 14:14
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As per other answers, your operational example simply doesn't correspond to the >>definition<< of "escape velocity". An operational example that does correspond to the definition is the cannon in Jules Verne's classic sci fi story https://en.wikipedia.org/wiki/From_the_Earth_to_the_Moon

So what you're suggesting can absolutely be done exactly as you say, and it will work exactly as you say. But it has nothing to do with "escape velocity". That, instead, would be the minimum speed a cannoball has to be fired with to just escape the Earth (ignoring atmospheric resistance), with >>no further forces<< after it's initially fired. That's just the definition of the term.

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Escape velocity is necessary if you turn all your rockets off. Without any rocket thrust, and if you are going below escape velocity, you will go into orbit or crash into the object you are trying to escape. If you keep your rockets on, you don't need escape velocity.

To put this into practical terms, during the Apollo lunar missions, the rockets were off nearly all of the time. The burn called "translunar injection" gave the spacecraft enough velocity to get out of earth's gravitational influence into the moon's. Another burn was needed when the spacecraft arrived in the neighborhood of the moon. This burn was to put it into lunar orbit. There was another burn needed to escape the lunar pull, and put it back on a trajectory to earth. The were a few minor burns for mid course correction. And there was the great big burn, at launch time.

The lunar exploration module had to make a few more burns, to land on the moon, to return to lunar orbit, and to link up with the mother ship.

Other than that, the trajectory of the craft was determined by gravity and inertia (momentum).

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If you're in a car at highway speeds and jump out in any direction, are you still going at highway speeds?

If you shoot a ball at 50mph with a cannon out of the back of a car that's driving 50mph it would stand still: https://www.youtube.com/watch?v=BLuI118nhzc

So imagine Earth is the car where you're jumping out of, and the Earth is traveling around the Sun at about 30km per second. You'd still need to add 19km/s to the "escape velocity" in the opposite way before you come to a standstill and fall down towards the sun. 19km/s is a lot of fuel!

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    $\begingroup$ Perhaps worth showing a picture of how accelerating directly towards the sun generates an ellipse? $\endgroup$ – akozi Feb 11 at 13:59
  • $\begingroup$ While correct, I don't see how this answers the OP's question. $\endgroup$ – Ilmari Karonen Feb 11 at 14:24
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    $\begingroup$ @IlmariKaronen I'm trying to explain why you can't just "point your rocket at the sun" and go there because of your initial orbit when you leave Earth's gravity. $\endgroup$ – DrTrunks Bell Feb 11 at 14:27

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