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Conservation energy to find escape velocity

  1. Mar 15, 2014 #1
    Guy, i have a problem. When we use conservation energy to find the escape velocity, ie the root of potentiality is unlimited, why we don't compute the velocity of earth? We knew the earth rotated its axis, then everything in surface was had linear velocity. So when we launch a thing, it's general velocity is sum of 2 vector velocity, one is linear velocity, other is velocity we supplied. That is
    (0.5 ×mv^2) +(0.5×mV^2)-(GMm÷(r^2))>=0
    Which v,V are linear velocity and velocity of thing we launched respectively
    We obtain,..........
     
  2. jcsd
  3. Mar 15, 2014 #2
    We knew on the surface v= (2π÷T)×R which R is radius of earth and T is period and we have
    V^2>=(GM÷R)-[(4π^2)÷T^2]×R^2 and result is Min V= 7.893961 km/s
     
  4. Mar 16, 2014 #3
    What you're saying is true. That's why it is easier to launch a rocket from the equator than from the poles (at the poles you wouldn't get the extra boost due to Earth's rotation). There is a mistake in you calculation though. You can't just add the two kinetic energies. If V = v + v' than V^2 = v^2 + v'^2 + 2 v.v'. You forgot to include the last term in your calculation. I think you also missed a factor of 2 somewhere in there...
     
  5. Mar 16, 2014 #4

    Dale

    Staff: Mentor

    Note that the rotation of the earth does not alter the escape velocity at all. That is why we don't use it in the computation. What the rotation of the earth does is make it easier to achieve escape velocity, i.e. the delta V to reach escape velocity is smaller if we launch with the rotation near the equator.
     
  6. Mar 17, 2014 #5
    I know. The escape velocity depends on the angle we launched. I forgot tell you i assumming the angle is 90 (perpendicular to surface)
     
  7. Mar 17, 2014 #6
    Thanks for answer
     
  8. Mar 17, 2014 #7

    Bandersnatch

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    But it doesn't. The escape velocity, despite its name, is actually escape speed, as it is completely irrelevant which direction the body is moving - as long as we make sure it doesn't crash/slow down in the atmosphere. If it were a micro black hole, or a neutrino, for example, you could shoot it directly below your feet, and as long as the speed was sufficient, it would pass through and escape.

    EV value is dependent only on the mass producing the gravitational field and the distance from that mass.

    What DaleSpam is saying, is merely that once you've calculated the value of the EV, you can then use a variety of ways to achieve it most economically - like choosing the best launch angle and lattitude. But it's got nothing to do with escape velocity as such.
     
  9. Mar 17, 2014 #8

    D H

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    No, it doesn't. Escape velocity is a bit of a misnomer; a better name would be escape speed. Escape velocity is not a vector. It is a scalar function of distance from the center of the gravitating body (the Earth in this case), the mass of the gravitating body, and nothing else.

    An object launched with a velocity whose magnitude is exactly equal to the escape velocity will just barely escape the gravitational clutches of the gravitating body. Anything less than escape velocity means the object will remain gravitationally bound to the gravitating body. Anything more than escape velocity means velocity remain non-zero as the distance tends to infinity.
     
  10. Mar 17, 2014 #9
    Thank you very much
     
  11. Mar 17, 2014 #10
    I think it depends on the angle and absolutely the mass escapes the gravitational field of earth. If the angle is enough small to escape the field the the velocity we supplied will be smaller. Because Vector(V)= vector(v1)+vector(v2). The module V^2=(v1)^2+(v2)^2+2(v1)(v2)cos(c) where c is angle form vector(v1) and vector(v2). V1 is velocity we supplied and v2 is linear velocity. So v1 will be smaller than. It is depends reference frame we chosen. In my reference frame i chosen is station reference in the universe.
     
    Last edited by a moderator: Mar 17, 2014
  12. Mar 17, 2014 #11

    Dale

    Staff: Mentor

    This is not correct. The escape velocity does not depend on the angle, only the speed.
     
  13. Mar 17, 2014 #12
    You're getting confused because you think the scape velocity is measure with respect to the moving ground in which case it depends on the direction as you said. But the escape velocity is actually defined relative to the center of mass reference frame and is therefore independent of the direction.
     
  14. Mar 17, 2014 #13

    D H

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    Note that your post contains ©, where you obviously meant (c). I find it to be rather hard to convince the autocorrect that, yes, I really did mean "c between two parentheses" rather than the copyright symbol. I fixed your post.


    Regarding the technical details of your post, escape velocity is just a speed. It is independent of direction, independent of current velocity. That said, if an object is already moving at some velocity, achieving escape velocity is most easily accomplished by making the change in velocity, the Δv, be parallel to that initial velocity, and is hardest if the change in velocity is directed against the initial velocity.
     
  15. Mar 18, 2014 #14
    Sorry i mean the velocity we supplied is smaller if we launched it on equator and it depends on the angle we launched. Sorry i forget it is the velocity we supplied and THE ESCAPE SPEED DOES NOT DEPENDS ON REFERENCE FRAME AND THE ANGLE WE LAUNCHED.thanks i am understand now. The escape speed is the value of velocity that everything could escape gravitational field
     
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