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Just a quick question of the escape velocity equation

  1. Nov 29, 2009 #1
    Okay, I just found this new equation (new to me) for the escape velocity of a body.

    escape velocity = √(2GM)/R

    So say for instance I plug this in to the escape velocity of the earth

    I am going to say the radius is 4001 miles so I it would take a mile above the earth to escape (4001 * 1600) = 6401600 meters
    escape velocity = √((2)(6.674*10^-11)(6.0*10^24)/(6401600)
    = 11185.08934 m/s

    This obviously isn't the speed you need to escape the earths gravity a mile off of its surface because we have left earth already with much lower speeds. This is so fast we would get a little time dilation, can someone tell me what I am doing wrong?

    Thanks in advanced.
  2. jcsd
  3. Nov 29, 2009 #2


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    The Apollo missions acheived speeds around 11000 m/s or more, so you're number seems OK, other than the atmosphere would be an issue if only 1 mile up.
  4. Nov 29, 2009 #3
    Wait I am confused, the escape velocity it the speed you need to leave the earths system, Is this data I collected implying I must go that speed to go above the earth for one mile?

    That doesn't make sense, I wouldn't have to go that speed to go just a mile above the surface.
  5. Nov 29, 2009 #4


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    What does this even mean? I can't make heads or tails of this business you're talking about "a mile above to escape." Also, time dilation (indeed, all relativistic effects) are negligible for the vast majority of travel up until about .1c (for example, at this speed gamma is 1.00000000067).
  6. Nov 29, 2009 #5


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    11 000 m/s is the speed you need to escape from Earth if you start a mile up. It's not the speed needed to travel 1 mile up from the ground.

    BTW, the escape velocity is the speed needed for a FREEFALLING object to escape. That means it doesn't apply to rockets, as rockets have propulsion. To illustrate, you could fire your rockets and inch upwards at 1 mm/s and still escape Earth completely.
  7. Nov 30, 2009 #6
    A helicopter can go a mile above earth's surface, so obvosly escape velocity is not how fast you need to go to go to a certain height.

    Escape velocity is the speed at which you must go to escape earth's gravity. For instance, let's say you shoot a rocket straight up at 10000m/s. It will fly up and up and up, but will constantly be slowed down by earth's gravity, until it slows all the way to 0 and starts falling back down....and it will hit the ground. Below escape velocity, "what goes up must come down" (kind of).

    If however, you shoot the thing at 12000m/s, it will also fly up and up and up. Earth's gravity will still slow it, but keep in mind earth's gravity diminishes as you go further away. If you're going at escape velocity, your rocket will basically never be slowed down to 0. It might approach 0, but it will keep getting further away from earth, and earth's gravity will keep diminishing, with the result that it will keep on going into infinity and never fall back down.

    Also, if you were hanging out at a far-far-far distance from earth at a speed of 0, and you let earth's gravity pull you in, you will end up traveling at escape velocity when you finally hit the ground (disregarding the atmosphere).
  8. Nov 30, 2009 #7
    Keep in mind that escape velocity is the speed (it's actually a scalar) that a body would need to have to escape the gravitational field it is in, without the benefit of any propulsion.

    So the reason why the number is so large, is because that is the speed you would need to initially have, to be able to then "coast" (no propulsion!) out of the gravitational field. The reason rockets seem to escape at lower speeds is because they are continually applying force.
    Last edited: Nov 30, 2009
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