2019-02-11 12:12:53 8 Comments

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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## 5 comments

## @ACuriousMind 2019-02-11 12:14:42

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.## @garyp 2019-02-11 12:27:17

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.

## @ACuriousMind 2019-02-11 12:29:54

@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.

## @Ambrose Swasey 2019-02-11 12:44:01

The rocket is not accelerating. It is moving at a constant speed of 50 mph. The acceleration with respect to the earth is 0.

## @OON 2019-02-11 13:14:54

@AmbroseSwasey ACuriousMind means that it is not in free fall. To compensate gravity acceleration the rocket would have to constantly fire an engine that would spent its fuel. The amount of fuel required will become insane even to reach the near space this way

## @Nuclear Wang 2019-02-11 13:40:33

@AmbroseSwasey A body moving at a constant speed in a gravitational field

isaccelerating. 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.## @OON 2019-02-11 13:41:40

@AmbroseSwasey I've added a calculation of the amount of the fuel required to reach just near space your way to my answer.

## @Ján Lalinský 2019-02-11 14:00:13

@NuclearWang accelerating means changing velocity, not "experiencing propulsive force".

## @OON 2019-02-11 14:10:35

@Ján Lalinský Free fall is in many senses closer to the inertial motion without gravitational fields than keeping the constant velocity by some other force in the presence of gravity. In fact in the relativistic physics you can't really give an universally objective definition of moving at constant velocity in the presence of gravity whereas the objective definition of free falling can be given. That's why using "accelerating" in this sense is actually quite common in this context.

## @Ján Lalinský 2019-02-11 14:20:21

@OON that is true in general relativity, but not true in Newtonian mechanics or orbital mechanics, where the gravity force is just another external force acting on the rocket. The reference frame centered at the Earth and with axes maintaining fixed orientation w.r.t. stars is inertial frame in this setting.

## @Green Grasso Holm 2019-02-11 14:30:12

You should revise your response to say

if it applies no further propulsion". That propulsion is only one factor determining whether the rocket accelerates or decelerates, gravity being the other (assuming no *otherforces are acting on the rocket). If the nose of your car is right up against a large brick building, you can step on the accelerator all you want, applying propulsion to it, but you won't be accelerating.## @Shufflepants 2019-02-11 15:43:48

@OON As a corollary, if we stipulate the OP's conditions that the rocket is traveling with a constant speed of 50 mph relative to the Earth and without continuous propulsion, this would mean the rocket is actually in orbit of the Earth rather than traveling away from it. (granted, in this case his stipulated speed of 50mph gives an orbital distance of 5.3 AU which wouldn't really be possible unless earth was off in interstellar space by itself as the sun would be the dominating gravitational force at that distance.) Which would still mean it hasn't escaped.

## @Shufflepants 2019-02-11 15:46:22

Though, at that distance, it would only need another 21 mph impulse to achieve escape velocity of the Earth.

## @DJClayworth 2019-02-11 15:57:58

@ACuriousMind I too disagreed with your answer, but understood it once I read your first comment. I think the answer would be improved by putting what you say in the comment into the answer.

## @garyp 2019-02-11 16:08:01

Good edit. That's an answer I can live with. I find it interesting that there was a discussion about the nature of acceleration. Who knew there were two points of view.

## @Walter Mitty 2019-02-11 14:49:46

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).

## @OON 2019-02-11 12:26:15

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

exponentiallyi.e. you need $\simeq 10^{17}$ tons of fuel just to elevate 1 ton to the altitude of the ISS.## @Luc 2019-02-11 13:26:09

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.

## @Kyle Kanos 2019-02-11 14:14:27

This is discussed further in

Why are rockets so big## @DrTrunks Bell 2019-02-11 13:50:30

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!

## @akozi 2019-02-11 13:59:30

Perhaps worth showing a picture of how accelerating directly towards the sun generates an ellipse?

## @Ilmari Karonen 2019-02-11 14:24:05

While correct, I don't see how this answers the OP's question.

## @DrTrunks Bell 2019-02-11 14:27:55

@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.

## @John Forkosh 2019-02-11 12:38:37

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_MoonSo 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.