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'escape' velocity

  1. Aug 4, 2005 #1
    Why do people always use the misleading term 'escape velocity'?

    If I had a rocket whose engines provided a thrust of 1N more than its weight (thus yielding a miniscule net force up), i would be able to go to mars with it, given the required amount of time, right? Thrust > Weight, so it should accelerate up...

    The space shuttle, for example, never escapes the Earth's gravitational field -- nor can it ever (strictly) because of the horizontal asymptote in g vs. distance graph.

    The shuttle merely reaches the required velocity to attain a stable orbit around the Earth. So, when in stable orbit, the shuttle is virtually accelerating towards to Earth at whatever g is at that distance from the centre of the Earth but does not fall down because its velocity is always tangential to its orbit (and perpendicular to its acceleration).

    Am i correct? I'm having a little argument with a friend.

    Cheers
     
  2. jcsd
  3. Aug 4, 2005 #2

    arildno

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    Quite so; if you were able to keep up a continuous thrust exceeding the weight, you would accelerate away, and make the distance between you and Earth arbitrarily long.

    However, where would you store all that fuel needed to keep up the continuous thrust?

    The "escape velocity"-concept refers to the situation where the SOLE force acting upon the vehicle is force of gravity.
    Thus, the "escape velocity" is the velocity you need to start out with at Earth in order to come to rest infinitely far away from Earth (i.e, when you've escaped Earth's gravitational field).

    EDIT:
    I just scanned the first few paragraphs you wrote, so I didn't really go into the last parts of your posts.
     
    Last edited: Aug 4, 2005
  4. Aug 4, 2005 #3

    Janus

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    Escape velocity is that velocity needed to enter an orbital path that is parabolic.( open not closed.) Thus the body "escapes" the vicinity of Earth as its trajectory never will bring it back to the Earth.

    If the Earth and Mars were the only two bodies in the Solar system, then yes, you could reach Mars without ever exceeding Earth's Escape Velocity, simply by putting yourself in an orbit around the Earth that has an apogee that far away. In real life, however, we also have to deal with the Sun. In order to reach Mars from the Earth, we must at the very least put the ship in a solar orbit that intersects Mars' orbit. Since the Escape velocity from Earth decreases as you move away from the Earth, you can't help but exceed Earth escape velocity doing so.
     
  5. Aug 4, 2005 #4

    VietDao29

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    Yup, you can go to Mars by that way. 'Escape velocity' is the needed speed that can help you travel from a planet to a very very far place from that planet without requiring any force accelerating it up.
    Nope. Not every object that has the velocity which is tangent to its orbit and perpendicular to its acceleration does not fall down. It must reach the required velocity to move arround and arroung a stable arbit. If the magnitude of the velocity is greater, then the radius of its orbit is bigger than required. If the magnitude of the velocity is smaller, then the radius of its orbit is smaller than required.
    If the magnitude of the velocity is too small, then the object will be pulled towards Earth. (eg : Throw a ball parallelly to the Earth surface. The initial velocity will be parrallel to Earth surface, but it will fall down).
    Viet Dao,
     
  6. Aug 4, 2005 #5

    ZapperZ

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    I am not sure what is so "misleading" about the term escape velocity. I can think of many other terms in physics that are more misleading than that (quantum teleportation comes to mind immediately).

    Remember what is the definition of "escape velocity". It is the MINIMUM launching velocity from typically the surface of the earth until the object completely escapes the earth's gravitational potential. What you seem to be complaining about is the fact that it is a highly idealized case. Well of course it is! It is no different than the idealized case of the Carnot cycle in thermodynamics. You have to assume a spherically symmetric earth, you assume no air resistance, you assume that there are no other gravitational bodies around in the universe, and you assume that you can get to infinitely far away from the earth.

    What this represents then is the LOWER PHYSICAL LIMIT of the launching velocity, the same way the Carnot cycle presents the highest possible physical limit to the efficiency of a heat engine. They are both idealized exercise that we tell students to do in class. This is because the real-life problem is not tractable and would scare students away from physics classes. We'll leave those till when they're in graduate schools!

    Zz.
     
  7. Aug 4, 2005 #6

    rbj

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    i've always been taught that "escape velocity" was the velocity necessary to take an object (with mass [tex] m [/tex]) from the surface of a planet with mass [tex] M [/tex] and of radius [tex] R [/tex] to a point at infinity, assuming no atmosphere. i.e. at the surface of the planet, the object has sufficient kinetic energy to take you infinitely far from the planet (where it is all potential energy). that is a finite amount of energy:

    [tex] E = G \frac{M m}{R} = \frac{1}{2} m v^2 [/tex]

    solving gets you the escape velocity for the atmosphereless planet of mass [tex] M [/tex] and of radius [tex] R [/tex].

    [tex] v = \sqrt{2 G \frac{M}{R}} [/tex]

    r b-j
     
  8. Aug 4, 2005 #7

    krab

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    The only thing misleading about the term "escape velocity" is that it is not a velocity. Velocity is a vector, but "escape speed" (as it should be called) is a scalar. IOW, the direction does not matter; you could start horizontally and still escape (all this of course assumes no atmosphere to give drag).

    If you had only a vanishingly small net upward force, it would take a very long time to escape, so the amount of energy needed would be very large, this stored energy would be very massive, and to move it requires even more thrust... you see that the model of having just enough thrust to counteract weight results in no practical rocket. A simpler and more realistic model is where you have enough thrust to achieve close to escape speed soon, and then coast into outer space, since this minimizes the amount of fuel needed.
     
  9. Aug 5, 2005 #8

    Janus

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    Same thing. At anything below escape velocity your trajectory will be closed (an ellipse). Your maximum separation from the planet will be finite.
    At escape velocity your trajectory will be open (parabolic) and the maximum separation becomes infinite.
    Above escape velocity, the trajectory becomes hyperbolic.
     
  10. Aug 5, 2005 #9
    Ahh, that's cleared everything up.
     
  11. Aug 5, 2005 #10

    amt

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    But wait...there's more....

    Try throwing a stone high up in the air...it will fall back. Now try even harder- after a while it will fallback. Seems like the speed with which you throw, does not let it to escape the Earth's gravity.

    Now, let's say that you have developed super human strength and this time you throw the stone really really hard. You keep looking for it to fall down, but it seems to have disappeared from sight forever. It's not coming back.

    The force with which you threw it is the escape velocity.
     
  12. Aug 5, 2005 #11

    Art

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    On a related note unmanned balloons have reached heights of 52 km but what prevents them rising any further? Is it because the athmosphere becomes too thin at that height?
     
  13. Aug 5, 2005 #12
    suppose we have a stone with infinite melting temperature, what is (numerically) the escape velocity if we consider the AIR FRICTION?

    has any body seen the figure yet? how do we calculate it?
     
  14. Aug 6, 2005 #13

    Danger

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    The maximum height that an unpowered balloon can reach is when the density of the atmosphere is no longer greater than that of the lifting gas (usually helium). I think that it's more complicated than first appears, because the helium itself will become less dense with increased altitude. (If you notice, a high-altitude balloon is pretty much condom-shaped at launch and becomes nearly spherical at it's operational altitude.) There must be some fancy formulae for figuring it out, but I have no idea what they are.

    The escape speed (hats off to Krab for that observation) doesn't change regardless of the medium being traversed. The force required to produce that speed does.
     
  15. Aug 6, 2005 #14
    really? i thought it's about initial momentum or kinetic energy?
    in fact krab says smthg different about the atmosphere.

    i still think the escape velocity is medium dependent, by looking at the way the integral was performed. :confused:
     
  16. Aug 6, 2005 #15

    Danger

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    I don't want to put words in Krab's mouth, but my interpretation of his statement was that a horizontal impetus would involve passing through more atmosphere than a vertical one would, and would thus require a higher initial speed to overcome more drag. Once the atmosphere is behind, things are equalized. In other words, your escape speed will be the same once you're in near-vacuum. The amount of initial speed required to reach that point will vary dependent upon how much air you have to go through to get there.
    I suspect due to Janus' post that I might be defining 'escape velocity (speed)' wrongly. To me, it has always implied the ballistic speed required to leave a closed orbit. That is, if something were fired from a cannon, how fast it would have to go. If sustained thrust is available, then anything over the mass of the vehicle is sufficient. My only disagreement with Krab's statement is in regard to the fuel mass. Something externally powered such as a solar or laser sail would still take a long time to do anything useful, but wouldn't require any onboard fuel.

    Out of my league, bud. I don't even know what an 'integral' is. My math knowledge ends at some part of elementary geometry.
     
    Last edited: Aug 6, 2005
  17. Aug 6, 2005 #16

    ZapperZ

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    If people look at the definition that was derived by rbj, one would see that it IS a ballistic speed, i.e. no sustained thrust. So this is equivalent to the velocity that Danger talked about, the velocity it needs if it were shot out of a cannon, for instance. If not, you cannot use the energy conservation that was used in the first line to derive that result.

    If one adds sustained thrusts to it, then one needs to call it something else. It is no longer the "standard" escape velocity that we defined in intro physics textbooks.

    Zz.
     
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