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What's the point of Escape Velocity?

  1. Feb 19, 2013 #1
    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
     
  2. jcsd
  3. Feb 19, 2013 #2
    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.
     
  4. Feb 19, 2013 #3

    mfb

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    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.
    This (and all repetitions of it) is wrong.
    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.
     
  5. Feb 19, 2013 #4
    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.
     
  6. Feb 19, 2013 #5
    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.
     
  7. Feb 19, 2013 #6
    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!
     
  8. Feb 19, 2013 #7
    Circular and Elliptical orbits aren't the only type of orbits. There are parabolic and hyperbolic orbits as well.

    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)
     
  9. Feb 19, 2013 #8

    Cthugha

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    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.
     
  10. Feb 19, 2013 #9

    QuantumPion

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    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.
     
  11. Feb 19, 2013 #10

    mfb

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    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.
     
  12. Feb 25, 2013 #11
    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...
     
  13. Feb 25, 2013 #12
    So what is the 'escape acceleration' for the earth then?
     
  14. Feb 25, 2013 #13
    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.
     
  15. Feb 26, 2013 #14

    sophiecentaur

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    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?
     
  16. Feb 26, 2013 #15

    phinds

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    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.
     
  17. Feb 26, 2013 #16
    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...
     
  18. Feb 26, 2013 #17

    sophiecentaur

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    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.
     
  19. Feb 26, 2013 #18

    QuantumPion

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    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.
     
  20. Feb 26, 2013 #19

    cjl

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    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.
     
  21. Feb 26, 2013 #20

    mfb

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    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.
     
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