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Escape velocity and gravitational potential energy

  1. Mar 24, 2007 #1
    Hi guys...something does not seem right in physics...
    escape velocity is defined as the initial velocity required by a projectile to rise vertically and just escape the gravitational field of a planet

    and gravitational potential energy = 0 only at infinity displacement from planet
    (i.e. there will only b no gravity on the object if it is at infinity displacement from the planet)

    well escape velocity is given by the velocity required to escape earth's gravitational field

    which also means escape velocity is given by the velocity required to bring an object to infinity displacement from a planet

    which is impossible...... and contradicts laws of physics...
    anyone got a clue?:uhh:
     
  2. jcsd
  3. Mar 24, 2007 #2
    It's something similar to the fact that you assign potential energy as zero at the sea level. You can assign that potential to zero at the altitude of your home town or whatever suitable for your calculation. In the case of escape velocity, they consider gravitational energy of the rocket as zero at infinity, so if it moves any closer to earth, it creates work and gravittational potential envery has a minus value. And vice versa, to make it go further from earth, it must consume some work, or energy.
     
  4. Mar 24, 2007 #3

    Doc Al

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    Why in the world do you think that impossible? Are you bothered by the word "infinity"? Think of it this way: Toss a ball in the air at a certain speed, then calculate the time for it to come to rest and return. As you increase the speed, it takes longer. Given the inverse-squared law of gravity--it rapidly gets weaker as distance increases--it's perfectly possible to launch a rocket with enough speed that the planet's gravity would be too weak to ever stop it.
     
  5. Mar 24, 2007 #4

    Danger

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    Hi, Doc... it's been a while. I just want to add a minor touch here. Russ pointed out quite some time ago that since 'velocity' is a vector measurement, the topic of this thread should be called 'escape speed'. The direction isn't critical. Everyone knows what someone means by the phrase, but it isn't strictly correct. Just nit-picking.
     
  6. Mar 24, 2007 #5

    HallsofIvy

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    And that's impossible why? Don't just say "contradicts laws of physics". Tell us which law of physics it contradicts! You cannot argue that it is impossible to go an infnite distance in a finite time- that's not required. The escape velocity (or "escape speed" as Danger says) is the speed required so that, nothing else interfering, the object would eventually, after infinite time, go to an infinite distance.
     
  7. Mar 24, 2007 #6
    As originally stated, the gravitational potential energy is zero only infinitely far from the planet.
    Also, one would need to impart to an object a velocity sufficient to break out of the potential energy well experienced on the earth's surface. Stated another way, you need to give your rocket enough kinetic energy to break away. So, call the earth's gravitation potential Phi and Ms the mass of the space ship.

    PE = KE
    Ms * Phi(on earth's surface) = (Ms * v**2)/2

    v = sqrt(2*Phi(on earth's surface))
    Phi has a 1/R dependence.

    Notice that the velocity is independent of the mass of the spaceship.

    If you point your spaceship down toward the earth, you won't escape. Moreover, pointing it up is imprecise. If you are taking off from a rotating object, you need to figure this contribution into the calculation. This correction isn't large for the earth's gravity even at the equator, but it does depend upon latitude. So, your escape velocity, figured with a little more detail (but still ignoring winds, drag etc.), should include the latitude, as well as the direction of travel relative to the rotation of the earth. Hence, I prefer velocity.

    Nonetheless, ignoring rotation and using G=6.67e-11 m/(kg*s**2) R(earth) = 6.36e6 m, M(earth) = 5.94e24 kg, you can figure the escape velocity from a planet with the earth's mass, but half its radius as ca. 15.8 km/s.
    -Jim
     
  8. Mar 24, 2007 #7

    Danger

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    Sorry, but that's incorrect. The rotation of the involved body means only that you need a different ground speed. Your speed relative to the planet's centre of mass, from which gravity is measured, remains constant.
     
  9. Mar 24, 2007 #8
    What I said is correct for an observer on the planet. I also assumed a spherical planet and some other details.
    -Jim
     
  10. Mar 24, 2007 #9
    Escape velocity, as measured from the Earth, certainly depends on the location and the direction of the rocket.
    After all, the Earth is not a perfect sphere and it is spinning!
     
  11. Mar 24, 2007 #10

    Danger

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    Yes, I realize that. But escape speed is not calculated relative to a ground-based observer. Your speed is based upon the centre of mass, which isn't anywhere near the surface.
     
  12. Mar 24, 2007 #11
    Velocity can be measured relative to any observer. Einstein developed special relativity to show how it remains true even as one approaches the speed of light.
    -Jim
     
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