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How far away from earth before the pull is negligible?

  1. May 29, 2007 #1

    My girlfriend asked me this question - how far from earth do you have to be before you don't really feel any gravity from it anymore?

    I suppose the question to ask is - how negligible does 'don't really feel' actually mean ...

    After a bit of thought I think one way of defining this might be ...

    If you could build a static platform at any distance you like from the center of the earth (that is not in orbit, but hangs there as the earth spins beneath), how far away from the center of the earth would it have to be before you could jump, using purely the power of your own legs to reach escape velocity?

    Some further definitions.
    The power of the jump is sufficient to allow the jumper to jump to a height of 1 meter when on the earths surface.

    I think that should be enough to work it out ... if not please feel free to include other assumptions you had to make in your replies.

    I'm also very interested to know how much the strength of the jump effects how far away you have to be to reach escape velocity - for example, if the jump is only good enough to reach a height of half a meter at the earths surface - what does this change the required distance from the center of the earth to? What is the equation for this in terms of

    d = f(x)

    where d = distance from the center of the earth before you can reach escape velocity and x = height jumped to on earths surface

    If the value of x within a reasonable range radically changes the distance d then this is not a very good way to answer the original question which was ...
    Is there a better way to define this question? - if so, what would that be, and more importantly - what is the ANSWER!!? :smile:

    Last edited: May 29, 2007
  2. jcsd
  3. May 29, 2007 #2
    I think I have a simpler way to ask this ...

    What is the ratio between my percieved weight on the earths surface and my percieved weight when standing on my platform at a distance of 2r from the center of the earth, and at 3r, 4r etc - what does the graph of this look like?

    (where r is the radius of the earth)

    ie, what is the equation for ... I think...

    y = f(x)

    as x varies as my distance in radius's of the earth (r) from the center of the earth
    where y is the ratio m/n
    where m is my perceived weight (if I had that platform to react against me) at distance x
    where n is my perceived weight at the earths surface (1r from the center of the earth).

    I'm sorry if I have just managed to suggest an impossible equation with too many other undefined variables here - please feel free to point out my stupidity!

  4. May 29, 2007 #3
    hehe, that sounds like one of those fermi question.

    in everyday experience, it seems reasonable that if g is reduced by a factor of 16, it would be negligible, so gravity falls off like r^2, so at 4 times the radius of the earth, then the gravity would be pretty negligible, and the radius of the earth is a little more than 6,000,000km, so i would say around 30,000,000km, or roughly the distance light travel in 100 seconds.
  5. May 30, 2007 #4
    Might I suggest when the gravitional pull of the next body (sun, planet, whatever) exceeds the gravitional pull of Earth. This should mean that eventually you would be pulled into that other body rather than back to Earth. To explain it to your girlfriend this: when part way between the Moon and the Earth gravity will pull you towards the Moon rather than the Earth.
  6. May 30, 2007 #5

    D H

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    This is not quite the definition you want. The gravitational pull of the Sun exceeds that of the Earth once a body is more than 259,300 km from the Earth. In contrast, the Moon orbits the Earth at 384,400 km.

    The "Hill sphere" (http://en.wikipedia.org/wiki/Hill_sphere" [Broken]) is much closer to what you are asking for.
    Last edited by a moderator: May 2, 2017
  7. Jun 1, 2007 #6
    Thankyou - all great answers. Mission accomplished :biggrin:
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