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Escape velocity and leaving earth

  1. Dec 20, 2012 #1
    Need some help with solving bet
    Do spacecraft (so we're talking about powered object) have to reach escape velocity in any point of travel to leave earth and if so, can you prove it by any scientific definition ?
     
  2. jcsd
  3. Dec 20, 2012 #2

    mfb

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    No. They could slowly increase their altitude and burn fuel forever - quite inefficient, but it would be possible.
     
  4. Dec 20, 2012 #3
    so yes or no ? We're talking about strictly theoretical situation and laws of physics
     
  5. Dec 20, 2012 #4

    mfb

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    => No, they do not have to reach escape velocity anywhere.
     
  6. Dec 20, 2012 #5
    To 'escape' from the Earth requires a certain amount of ENERGY (about 63MJ per kg)
    If this is supplied as KE then the velocity is about 11km/s. This is what is known as escape velocity.
    If you wanted to climb up a ladder then 63MJ still need to be supplied but you can take your time.
     
  7. Dec 20, 2012 #6

    Janus

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    But to be fair, escape velocity decreases with altitude. So if you leave Earth at just 1 km/sec, when you get to about twice as far away as the Moon, the escape velocity from there will have dropped to 1 km/sec.
     
  8. Dec 20, 2012 #7
    To be totally fair
    The original question related to escape velocity from the Earth.
    It is a mistake to see this as a discussion about VELOCITY.
    To escape from the Earth requires ENERGY. If all of this energy (63MJ/kg) is supplied in one go then the velocity required is 11km/sec.
    If, lets say 3MJ can be supplied by climbing to the top of a ladder, then 60Mj need to be provided as KE ie a velocity of 10.9km/sec
     
    Last edited: Dec 20, 2012
  9. Dec 20, 2012 #8

    rcgldr

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    So when you state "leave earth", you mean to "leave earth" so that the object never returns or end up orbiting the earth? (Otherwise, you can just jump up and you've left the earth for a brief moment). Also I assume you're ignoring the effect of other objects in space such as orbiting or landing on the moon as method to "leave earth".

    Assuming a finite exhaust velocity and a finite ratio of (mass of fuel) / (mass of spacecraft), eventually you'll run out of fuel (update based on mfb's next post - unless you can use increasingly less fuel so that total fuel usage approaches some finite amount as time approaches infinity). At this point the rocket will need to have achieved the escape velocity required for that distance from the earth, but at very large distances, the escape velocity will be very small.

    Energy wise, this is very inefficient. It's better to focus on increasing velocity by methods that don't directly oppose gravity (although the increase in velocity will result in a curved path that moves away from the earth).
     
    Last edited: Dec 20, 2012
  10. Dec 20, 2012 #9

    mfb

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    If you can control your thrust with arbitrary precision, a finite amount of fuel is sufficient. Get a bit below escape velocity, cruise for a long time (increasing your distance) to a point where the escape velocity is 1/4 of the previous value, repeat (using 1/4, 1/16, 1/64, ... of the fuel burnt in the first step, with a finite sum).

    The most efficient way is to reach escape velocity as quickly as you can, I agree.
     
  11. Dec 20, 2012 #10

    rcgldr

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    I didn't consider this possibility since I didn't do the math to see if a finite amount of fuel could keep a rocket below escape velocity but away from the earth for an infinite time. I'm not sure if a radial path (straight out) versus a curved path makes a difference. The math for this may be what the OP is asking for.
     
  12. Dec 21, 2012 #11

    jbriggs444

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    For a pure rocket, the optimal path is one with an impulsive acceleration at launch in the direction of the object's pre-existing motion due to the rotation of the earth. That's where a given delta-V will buy the most kinetic energy. The resulting path will curve under the influence of gravity.

    Of course, this ignores the problem of air resistance (which motivates a slower and more vertical launch), the practical limitations on how fast you can burn fuel (the faster you burn it, the bigger your engines and your engines are not payload) and the possibility of a gravitational slingshot.

    If you are operating under the constraint that you are not allowed to actually reach escape velocity then an infinitesimal nudge to this thrust profile with a massive impulse at the start and a decreasing trickle of thrust later on will work. Clearly, for every such thrust profile there is another one that is just a bit better because it decreases that trickle just a tad more.

    It follows that there no _optimal_ trajectory that stays below escape velocity.
     
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