What's the point of Escape Velocity?

[Jackissimus]

Hi everyone and thanks for this great forum!

I don't understand how is calculating escape velocity important to space travel. I think I am right to say that you can never really escape the gravity of a body. However far your rocket goes, be it 400km or 14 billion light years and making the distance so much closer to infinity, you will still just send it to an orbit (if we assume that the universe is infinitely large) and it will eventually fall back.

And in space travel the question is not how fast you need to go to reach infinity. The question is how fast you need to go to reach the next body that you want to orbit around. You can never 'escape a body', you just encounter another body which puts you on a different orbit. For example if you want to go to the Moon, you need to give your rocket an elliptical orbit with apogee at moon orbit, and intercept. So your initial velocity depends on how far the Moon is. If you want to do interstellar travel, you would want to calculate the ideal initial speed to leave the solar system. But there is no right answer! It simply depends on which star you want to travel to and put yourself basically on a high elliptical orbit around the Sun. To me this shows that to escape a planetary body calculating escape velocity is useless, because it basically always depends on the situation.

The only use of escape velocity seems to be for comparison reasons. So that you can say that leaving Moon's surface is much easier than leaving Earth's surface. It's just to give you a quick idea of what's hard to do and what's easy. But for planning the route I don't think you need to consider escape velocity, because you simply don't have to reach that velocity.

Am I right?

I hope this post doesn't generate too many tl;dr responses :)

Jacob

 

[Subductionzon]

You are right that you can never escape gravity. But that does not mean you cannot escape away from an object forever. If you are going fast enough away from Earth its force of gravity will never stop you. You will continually slow down, but the rate of deceleration will continually drop.

I am sure that the probes we have sent out to the galaxy had a high enough speed to escape from earth, and I am very sure that they also had enough velocity to escape from the even higher escape velocity of the Sun at Earth's orbit. So it does apply to space travel. I am not sure if the astronauts who went to the Moon reached escape velocity for the Earth, but they probably did. Anyone know for sure? It is not an easy question since escape velocity goes down the further away you are from that object.

 

[mfb]

I think I am right to say that you can never really escape the gravity of a body.

You can. While the body will always attract you, your velocity can be sufficient to overcome this drag and fly away forever without additional thrust.

you will still just send it to an orbit

This (and all repetitions of it) is wrong.

The question is how fast you need to go to reach the next body that you want to orbit around.

Typically, this is extremely close to the escape velocity.

By the way: All other velocity values are related to the escape velocity as well. The speed required for a low, circular orbit, is 1/sqrt(2) times the escape velocity.

 

[Jackissimus]

You can. While the body will always attract you, your velocity can be sufficient to overcome this drag and fly away forever without additional thrust.

If you send a rocket away from the Earth at 11km/s, the rocket will slow down gradually. The speed will get to 10km/s, then 9km/s, due to gravitational pull. The deceleration rate will get lower and lower, but it will stop the vehicle and then turn it around, because there is a constant, neverending attractive force and there are no other forces to counter it.

 

[Jackissimus]

I am sure that the probes we have sent out to the galaxy had a high enough speed to escape from earth, and I am very sure that they also had enough velocity to escape from the even higher escape velocity of the Sun at Earth's orbit. So it does apply to space travel. I am not sure if the astronauts who went to the Moon reached escape velocity for the Earth, but they probably did. Anyone know for sure? It is not an easy question since escape velocity goes down the further away you are from that object.

Yep, the Voyagers are leaving our system right now for example, they had enough speed to reach the Heliopause, but that doesn't necessarily mean they won't come back. They most likely won't, because they will encounter another star that will pull them in, but they didn't need to reach escape velocity then, something lower would suffice.

As for the Apollo, they didn't quite reach escape velocity. Escape velocity for Earth is 11.2 km/s, maximum speed reached by any human was Apollo 10 returning from the Moon, 11.08 km/s.

 

[Jackissimus]

Ok guys, I gave it some more thought and I think I got it. What you are saying is that as my rocket is flying away from Earth, the deceleration rate gets smaller and smaller and it converges to a certain number, which for escape velocity would be 0km/s. If I would fly a rocket at 13.2km/s from the Earth, then I would have a speed of 2km/s at infinite distance.

Is it correct?

Gosh, they really should write the wikipedia article in better terms. To me the phrase ""break free from a gravitational field" seemed like hogwash.

Thank you, guys!

 

[Travis_King]

Circular and Elliptical orbits aren't the only type of orbits. There are parabolic and hyperbolic orbits as well.

The deceleration rate will get lower and lower, but it will stop the vehicle and then turn it around, because there is a constant, neverending attractive force and there are no other forces to counter it.

Yes, but the point at which the kinetic energy, or velocity, is equal to zero (i.e. the turn-around point) is infinity.

Here's the wiki.

Edit, yep, you've got it.

It helps when you visualize it the other way around. If you start at zero at x=infinity, by the time you reach the object (earth in this case) you will have gained enough kinetic energy (due to the gravitational potential of that body) that your velocity will be ~11.2 km/s. (barring all other sources of gravitational potential energy, like the much larger body, the Sun)

 

[Cthugha]

The deceleration rate will get lower and lower, but it will stop the vehicle and then turn it around, because there is a constant, neverending attractive force and there are no other forces to counter it.

No, this is where your reasoning goes wrong. There is no CONSTANT attractive force. The force will get smaller and smaller as distance increases. For an arbitrary initial velocity you can easily calculate the distance from the starting point where the vehicle "turns around" and will find a general relation on the initial velocity. You will also find that there is a wide range of initial velocities where you will not find such a turning point as the force goes to zero too quickly. This range of velocities is above escape velocity.

edit: Oh, you gave the right response yourself faster than I did.

 

[QuantumPion]

If you send a rocket away from the Earth at 11km/s, the rocket will slow down gradually. The speed will get to 10km/s, then 9km/s, due to gravitational pull. The deceleration rate will get lower and lower, but it will stop the vehicle and then turn it around, because there is a constant, neverending attractive force and there are no other forces to counter it.

No it will not because the force of gravity decreases as the rocket gets further and further away. The definition of escape velocity is the initial speed such that the rocket would eventually be decelerated to zero at infinite distance. Since the rocket can never be infinitely far away from the object, it will never actually be reduced to zero velocity.

 

[mfb]

Ok guys, I gave it some more thought and I think I got it. What you are saying is that as my rocket is flying away from Earth, the deceleration rate gets smaller and smaller and it converges to a certain number, which for escape velocity would be 0km/s. If I would fly a rocket at 13.2km/s from the Earth, then I would have a speed of 2km/s at infinite distance.

Its speed "at infinity" would be $\sqrt{13.2^2-11.2^2}=7.0$ km/s, this can be shown via energy conservation (where the squared speed appears). The difference between 2km/s and 7km/s is related to the Oberth effect.

 

[xlsdx]

it seems to me that "escape acceleration" would be a better term. as long as i can travel upwards continually at any velocity, large or small, i will eventually leave earth. it's the acceleration due to gravity that you need to overcome, really. and the only way to do that is with a counter-acceleration. "escape velocity" can mean anything, as long as you have the acceleration to maintain that velocity...

 

[xAxis]

it seems to me that "escape acceleration" would be a better term. as long as i can travel upwards continually at any velocity, large or small, i will eventually leave earth. it's the acceleration due to gravity that you need to overcome, really. and the only way to do that is with a counter-acceleration. "escape velocity" can mean anything, as long as you have the acceleration to maintain that velocity...

So what is the 'escape acceleration' for the Earth then?

 

[Vorde]

it seems to me that "escape acceleration" would be a better term. as long as i can travel upwards continually at any velocity, large or small, i will eventually leave earth. it's the acceleration due to gravity that you need to overcome, really. and the only way to do that is with a counter-acceleration. "escape velocity" can mean anything, as long as you have the acceleration to maintain that velocity...

No.

Escape velocity is how fast I would have to throw a baseball directly up to make it never come back down. The second it leaves my hand the acceleration on the baseball is instantly -10 m/s/s, yet if the velocity is above escape velocity it will never come down.

 

[sophiecentaur]

As in nearly all cases, it is far, far better to discuss this in terms of Energy. The Potential Energy at the surface of a Planet is the minimum energy needed to move your object to infinity. We are in a Potential Well and the GPE is Negative, wrt Zero at infinity (it's up-hill all the way).
If you can give the object more Kinetic Energy than this - when it's on the surface - then it will escape permanently from the planet, despite the fact that the gravitational force never actually goes to zero. The so-called Escape Velocity corresponds to this value of kinetic energy to leave the surface permanently. You can supply the necessary energy 'gradually' if you like if you had a long ladder, for instance- but rockets are necessary in practice so it is more efficient to go as fast as you can for space travel launches. GPE is proportional to 1/R, where R is the distance from the Earth's centre so, if you go to 2R, the GPE is half the magnitude and so the Escape velocity from there is 1/√2 of what it is at the surface.
Escape velocity depends not just on planetary mass but on its radius. It is a totally artificial term, based on having a gun on the surface, and is of no use except for use in comparing different situations. The OP asks a very good question - what IS the point?

 

[phinds]

it seems to me that "escape acceleration" would be a better term. as long as i can travel upwards continually at any velocity, large or small, i will eventually leave earth. it's the acceleration due to gravity that you need to overcome, really. and the only way to do that is with a counter-acceleration. "escape velocity" can mean anything, as long as you have the acceleration to maintain that velocity...

No, escape VELOCITY is a ballistic concept, as Vorde said. It is about the initial velocity from a given point. If you bring in acceleration, you have changed the concept.

It's confusing because we always think of rockets achieving escape velocity by accelerating, but that's just so that they can get the the escape velocity. Once they do that, they could just coast.

If you had enough fuel, escape acceleration would be way lower than escape velocity. Basically, you would just accelerate at a low rate until you were so far away that your velocity exceeded the escape velocity for the point you were at.

 

[xlsdx]

right, this is where the confusion comes from i think. people always use the term "escape velocity" when talking about rockets, but rockets accelerate most of the way up. I'm not denying there is such a thing as "escape velocity" or challenging its definition, I'm just saying when talking about rockets it seems more appropriate to talk about it in terms of acceleration since that's ultimately what's causing the rocket to move upwards...

 

[sophiecentaur]

Escape acceleration means nothing - except, perhaps that it could be equal to the field times the mass, at all points on the journey. It doesn't tell you what you would need to know about the exercise - which is how much fuel you need (i.e. Energy) The fact is that an acceleration of g (measured on earth) would only need to be exceeded over the first few metres. By the time you got to the distance of a geostationary satellite, you would only need a fraction of g etc. etc..

Escape Velocity tells you all you need to know about the minimum KE needed. It might be more 'grown up' to specify it in J per kg of load perhaps but old habits die hard. They can stick to their escape velocity if I'm allowed to stick to my Pint.

 

[QuantumPion]

right, this is where the confusion comes from i think. people always use the term "escape velocity" when talking about rockets, but rockets accelerate most of the way up. I'm not denying there is such a thing as "escape velocity" or challenging its definition, I'm just saying when talking about rockets it seems more appropriate to talk about it in terms of acceleration since that's ultimately what's causing the rocket to move upwards...

A bullet accelerates extremely quickly but it does not reach escape velocity. It's the final velocity which counts, not how long it took to reach it.

 

[cjl]

right, this is where the confusion comes from i think. people always use the term "escape velocity" when talking about rockets, but rockets accelerate most of the way up.

No, they really don't (at least not unless you're talking about something like a spacecraft using electric propulsion). Rockets gain all of their velocity within the first couple hundred kilometers of Earth's surface, which is a pretty small distance relative to the size of the earth. Spacecraft sent to Mars (for example) are basically accelerated to escape velocity within a fairly short distance, and then they are basically coasting (at above the Earth's escape velocity) afterwards.

 

[mfb]

To add some numbers: a rocket designed to reach a low Earth orbit might accelerate for about 5 minutes, while an orbit needs 90 minutes. The Apollo rockets went to a low Earth orbit first, performed a quick burn there (again ~5 minutes) and coasted for three days all the way to moon afterwards.

 

[ModusPwnd]

So the claim is that after 5 minutes of accelerating the rocket is at an altitude who's escape velocity is nearly the same as on the surface of the earth? Thus we can effectively apply escape velocity to rockets?

 

[mfb]

The escape velocity in a low Earth orbit (~400km) is just ~3% lower than the escape velocity from sea level.

 

[xAxis]

Well it does represent the minimum energy needed for an object to leave the gravitational system, so it's not completely useless.
Also better term is 'escape speed', it's shorter and direction is not important anyway.

 

[phinds]

Also better term is 'escape speed', it's shorter and direction is not important anyway.

I disagree. THAT "speed" could be pointed towards the ground, which would be quite important.

 

[QuantumPion]

The escape velocity in a low Earth orbit (~400km) is just ~3% lower than the escape velocity from sea level.

The escape velocity from low Earth orbit altitude is only 3% lower than the surface. But if you are in low Earth orbit than you already have ~75% of (surface) escape velocity.

 

[russ_watters]

I disagree. THAT "speed" could be pointed towards the ground, which would be quite important.

We've had that discussion before: if the velocity vector were toward a building, an overflying airplane or any other obstacle, that would be bad too. Otherwise, direction doesn't matter.

 

[QuantumPion]

So the claim is that after 5 minutes of accelerating the rocket is at an altitude who's escape velocity is nearly the same as on the surface of the earth? Thus we can effectively apply escape velocity to rockets?

The escape velocity is just the instantaneous speed you would need starting from the Earth's surface to escape Earth's gravity. For example, for a cannon to launch a bullet out of Earth's gravity, it's muzzle velocity would have to equal escape velocity (actually more, due to drag).

A rocket does not need to actually reach escape velocity at any point to escape the Earth, it just needs enough energy to do so. A rocket accelerates much more slowly than a cannon and is fighting against drag and gravity as it is accelerating. The amount of energy you need is equivalent to if you burned all of the rocket's fuel instantly on the launch pad it would be moving at escape velocity. But real rockets don't do that for obvious reasons.

 

[russ_watters]

Another thing that helps me think about it (not sure if actual rocket scientists think this way), but total energy is really what determines if you will escape. In that way, gaining a lot of altitude can substitute for gaining some of the speed, but the total energy is the same.

But since kinetic and potential energy are both mass dependent, the mass cancels out and you are just left with a velocity that works for any object.

 

[cjl]

Another thing that helps me think about it (not sure if actual rocket scientists think this way), but total energy is really what determines if you will escape. In that way, gaining a lot of altitude can substitute for gaining some of the speed, but the total energy is the same.

Often, yes, they do. Frequently, rocket performance for objects on an escape trajectory is specified based on the achievable C3 for a given payload, which is basically the excess energy per unit mass above the escape energy (and it is often specified in units of km²/s²). An object at exactly escape velocity would have a C3 of 0, and any higher indicates the excess energy the object will have after escape (per unit mass).

(Oh, and the square root of the C3 value gives the velocity the object will have after escape)

 

[mfb]

The escape velocity from low Earth orbit altitude is only 3% lower than the surface. But if you are in low Earth orbit than you already have ~75% of (surface) escape velocity.

Escape velocity is a property of position only, I think. A rocket in LEO still needs ~11km/s as escape velocity - it is just easier to reach it.

A rocket does not need to actually reach escape velocity at any point to escape the Earth, it just needs enough energy to do so. A rocket accelerates much more slowly than a cannon and is fighting against drag and gravity as it is accelerating. The amount of energy you need is equivalent to if you burned all of the rocket's fuel instantly on the launch pad it would be moving at escape velocity. But real rockets don't do that for obvious reasons.

Actually, a rocket needs more energy to escape - if you could burn all the fuel in a second (and neglecting drag), the rocket would be more efficient. This is easy to see in the limit of no rocket acceleration: You have to keep burning fuel to counter gravity, but you don't increase the velocity of the rocket.

 

[QuantumPion]

Escape velocity is a property of position only, I think. A rocket in LEO still needs ~11km/s as escape velocity - it is just easier to reach it.

A rocket in LEO still needs 11 km/s to escape but it is already going 7 km/s due to its orbital velocity. Otherwise it wouldn't be in orbit.

Actually, a rocket needs more energy to escape - if you could burn all the fuel in a second (and neglecting drag), the rocket would be more efficient. This is easy to see in the limit of no rocket acceleration: You have to keep burning fuel to counter gravity, but you don't increase the velocity of the rocket.

Yes, gravity is considered as a type of drag in this sense.

 

[sophiecentaur]

Is it not true to say that no spacecraft has ever been given the full 'escape velocity' with its motors? I think that the craft that have enough energy to escape from the Solar System have all relied on pinching energy, using a Slingshot orbit.

 

[QuantumPion]

Is it not true to say that no spacecraft has ever been given the full 'escape velocity' with its motors? I think that the craft that have enough energy to escape from the Solar System have all relied on pinching energy, using a Slingshot orbit.

New Horizons was launched directly into solar escape trajectory. As far as Earth escape velocity is concerned, any probe leaving the Earth-moon system would need the full escape velocity (unless it possibly received a gravity assist from the moon, although I don't know if that is typically used or not).

 

[rcgldr]

A rocket accelerates much more slowly than a cannon and is fighting against drag and gravity as it is accelerating.

To avoid fighting against gravity, a rockets thrust can be directed so that it's always perpendicular to gravity so that all of the thrust goes into increasing velocity of the rocket, but this can't be done until outside the atmosphere and close to low Earth orbit.

 

[sophiecentaur]

New Horizons was launched directly into solar escape trajectory. As far as Earth escape velocity is concerned, any probe leaving the Earth-moon system would need the full escape velocity (unless it possibly received a gravity assist from the moon, although I don't know if that is typically used or not).

Interesting link. This was the highest launch speed so far, apparently and well above escape velocity - but Wiki says it still made a gravity-assist pass of Jupiter. Well, I suppose they didn't want to hang about. You know - places to go, people to see. . . .

 

[mfb]

Escape velocity is a property of position only, I think. A rocket in LEO still needs ~11km/s as escape velocity - it is just easier to reach it.

A rocket in LEO still needs 11 km/s to escape but it is already going 7 km/s due to its orbital velocity. Otherwise it wouldn't be in orbit.

That is what I said?

Yes, gravity is considered as a type of drag in this sense.

Well, it is completely different from air drag or similar types of drag.

Is it not true to say that no spacecraft has ever been given the full 'escape velocity' with its motors?

Starting from what? Every interplanetary mission uses the velocity of Earth (~30km/s), and most rockets use the rotation of Earth (up to ~500m/s) to begin the mission.

 

[QuantumPion]

Well, it is completely different from air drag or similar types of drag.

Actually gravity drag it is very analogous to air drag in the context of rockets. Both are forces which oppose the acceleration of the rocket during its launch. The difference is air drag is proportional to speed, while gravity drag is constant but only in the vertical direction.

 

[sophiecentaur]

Actually gravity drag it is very analogous to air drag in the context of rockets. Both are forces which oppose the acceleration of the rocket during its launch. The difference is air drag is proportional to speed, while gravity drag is constant but only in the vertical direction.

That is a strange way of looking at it. When you increase your GPE, that energy is of some use to you (it's half of the energy involved in your final orbit). Energy lost because of drag is of no use at all.
I understand the message in that link but it strikes me that it's not much more than an alternative explanation for the process of carrying fuel and extra rocket stages with you in a launch. It is taking two values of 'wasted energy' and lumping them together. There is no 'gravity drag' on the bit of the vehicle that is finally up there and in orbit. It still possesses the same amounts of GPE and KE it was given.

 

[QuantumPion]

That is a strange way of looking at it. When you increase your GPE, that energy is of some use to you (it's half of the energy involved in your final orbit). Energy lost because of drag is of no use at all.
I understand the message in that link but it strikes me that it's not much more than an alternative explanation for the process of carrying fuel and extra rocket stages with you in a launch. It is taking two values of 'wasted energy' and lumping them together. There is no 'gravity drag' on the bit of the vehicle that is finally up there and in orbit. It still possesses the same amounts of GPE and KE it was given.

Correct, energy lost to gravity drag is of no use at all. As MFP said, if you had a rocket burning at full thrust just to hover, you have 100% of your energy lost to gravity drag. There is no air drag on the vehicle that finally gets to orbit either. Gravity drag and air drag are terms used for determining how much extra delta-V over the theoretical minimum you need to reach orbit. If you had a spaceplane that could fly to the very upper reaches of the atmosphere using jet engines, and then ignite rocket engines to accelerate horizontally to orbital velocity, losses due to gravity drag would be very low for example.

 

[mfb]

Actually gravity drag it is very analogous to air drag in the context of rockets. Both are forces which oppose the acceleration of the rocket during its launch. The difference is air drag is proportional to speed, while gravity drag is constant but only in the vertical direction.

In the limit of burning all the fuel at once, gravity still acts on the rocket, but there is no energy loss due to gravity. Air drag, on the other hand, is always there if the rocket is in motion, just its amount is variable.

Who is MFP?

 

[QuantumPion]

In the limit of burning all the fuel at once, gravity still acts on the rocket, but there is no energy loss due to gravity. Air drag, on the other hand, is always there if the rocket is in motion, just its amount is variable.

Who is MFP?

I meant mfb not P, sorry. But yes you are correct.

 

[sophiecentaur]

Who is MFP?

Your long lost dyslexic brother, perhaps.

I really don't like the term Gravity Drag but, if it's used by people in the know, I think we have to go along with it. After all, it is Rocket Science.

 

[Travis_King]

I just want to chime in and say that drag should should be used for forces acting in the opposite direction as the velocity. Typically drag is defined as:
From Merriam-Webster

the retarding force acting on a body (as an airplane) moving through a fluid (as air) parallel and opposite to the direction of motion

So while a component of gravity may be acting as a "drag" force, certainly when the craft is pointed directly opposite the direction of Earth's gravitational acceleration, I don't think it's at all useful to call it gravity drag...

In the upper atmosphere, the craft must still overcome gravitational acceleration to continue climbing, but due to it's thrust vector and velocity, it's using very little energy overcoming the "drag" component of that force.