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Escape Velocity at an angle

  1. Mar 20, 2013 #1
    The escape velocity of an object on the surface of the Earth is the minimum speed required for it to "break free" from the Earth's gravitational field without requiring further propulsion.

    Its value is mathematically determined to be [itex] v = \sqrt{\frac{2GM}{R}} [/itex] with M being the mass of the Earth and R being the Earth's radius.

    If the object is fired pointing radially away from the Earth, then it will travel in a straight line and certainly reach "infinity".

    If the object is fired pointing radially towards the Earth, it will travel in a straight line and simply accelerate towards the Earth.

    My question is what happens if the object is fired at some angle (with the escape speed of course). Most sources say that if it is fired with some upward angle, it will still reach out to "infinity" because of energy considerations, and the fact that energy is scalar.

    But if the object is fired with some downward angle, it will not reach out to "infinity" and simply spiral towards the Earth.

    What if the object is fired perpendicular to the radial line?

    And how can we prove these results using a rigorous approach, rather than a merely observation that energy is a scalar? I am more of a mathematician, so would prefer a very rigorous attempt to address this question.

    All help is appreciated. Thanks!

    BiP
     
  2. jcsd
  3. Mar 20, 2013 #2

    tiny-tim

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    Hi Bipolarity! :smile:
    No, it will follow a parabola.

    If the parabola intersects the earth, of course the object will stop!

    If the parabola doesn't intersect the earth, the object will "reach infinity" exactly as if it had been fired outward.
    That is rigorous. :redface:

    Energy shows perfectly rigorously that an object launched at escape velocity will "reach infinity at zero speed", no matter what the angle of launch.
     
  4. Mar 20, 2013 #3
    Thanks tim, but I don't understand how these claims are rigorous. How can you prove them? Suppose I wanted to prove it without using energy...

    BiP
     
  5. Mar 20, 2013 #4

    russ_watters

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    As you get further from the earth, you get closer and closer to flying directly away from it regardless of what angle you started from.
     
  6. Mar 20, 2013 #5
    Energy Proof (simple)

    This is basically the proof that's given on Wikipedia, very simple and very easy. I' can't prove the parabola thing that tiny-tim wrote about, so you'll have to figure that one out some other way. However, since energy is scalar, as long as you don't run into the planet (like tiny tim said) the energy is conserved as shown.

    We start with conservation of energy.

    [tex] (KE + PE)_1 = (KE + PE)_2 [/tex]

    Initial kinetic energy is some velocity, we'll call it ve for escape velocity. Likewise, its potential energy is whatever it is on the surface, at radius r from the planet center. As we go to infinity, the potential energy goes to zero as r goes to infinity, and we'll pretend the kinetic energy goes to zero at the point where r goes to infinity.

    Additional notes: m is mass of escaping object, M mass of planet, G gravitational constant.

    [tex] (1/2)mv_e^2 + (-GMm/r) = 0 + 0 [/tex]

    So, we're then left with:

    [tex] v_e = \sqrt{2GM/r} [/tex]

    Another additional note: I'm not sure why they defined the potential energy the way they did, but that's how they showed it.
     
    Last edited: Mar 20, 2013
  7. Mar 21, 2013 #6

    tiny-tim

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    the "parabola thing" is easy

    a small body near an isolated mass must follow an ellipse (eccentricity < 1) a parabola (eccentricity = 1) or a hyperbola (eccentricity > 1)

    as you increase the speed (for a fixed direction at a fixed position), you increase the eccentricity

    if it's an ellipse, the body will never "reach infinity"

    the smallest value of eccentricity, and therefore the smallest speed, for which the body can "reach infinity" is when the path is a parabola :smile:
     
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