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Quick question about Escape velocity

  1. Nov 29, 2009 #1
    Okay this equation is giving me some trouble, I just have a question

    vescape = √(2GM)/R

    Say I wanted to find the escape velocity of myself 3 meters away
    So I would use
    vescape = √(2)(6.674*10^-11)((145*4.44)/9.8)/(3)
    = 5.419003598 *10^-5

    So does this mean something that moves that slow cannot escape me farther than 3 meters away?

    It doesn't seem right. My gravitational field feels so ephemeral. Is this correct?
    Last edited: Nov 30, 2009
  2. jcsd
  3. Nov 29, 2009 #2


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    Homework Helper

    I don't know why your mass is 145/9.8 (you weigh only 15 kg???), but yes, your escape velocity is tiny, and yes, something moving slower than your escape velocity can't escape from you in an ideal world. In the real world, turbulence, Brownian motion, and the gravity of other objects make that conclusion moot.
  4. Nov 30, 2009 #3
    Weight is a force.
    F = MA
    M = F/A or F/G

    So I could take my weight in ibs and divide it by 9.8 to get my mass.
  5. Nov 30, 2009 #4

    Doc Al

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    Staff: Mentor

    Careful! If you knew your weight in Newtons (not pounds), then you could divide by g = 9.8 m/s2 to get your mass in kg. When using equations such as W = mg, you need to use consistent units: in this case, Newtons, kilograms, and m/s2.

    If you weigh 145 pounds, your mass is about 66 kg.
  6. Nov 30, 2009 #5
    Okay, thanks a lot. That really messed me up. I changed it, but back to my point. Was my calculations and logic correct?
  7. Nov 30, 2009 #6
    In theory, if you were isolated in a big region of empty space, and tried to toss say a marble away, it would return to you if the initial velocity were less than escape velocity, or it would go into orbit around you if it had a correct satellite velocity. I think ...


    The proper unit for mass in British enegineering is slugs, that is, your weight in pounds force divided by standard gravity 32.174 ft/s/s. Always work in some consistent set of units.
  8. Nov 30, 2009 #7

    Doc Al

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    Staff: Mentor

    Using your numbers, I get an escape speed of about 5.4 x 10-5 m/s. (So double check your arithmetic.) Which basically says that it doesn't take all that much energy to escape your gravitational field. So if you were in outer space--away from any other masses--and some small object was 3 meters away and moving at less than that speed it could not escape your gravitational field. (But give it the merest push and it's gone!)

    Of course, this calculation assumes that you are a spherical ball! (That's how the escape velocity formula is derived.) So I wouldn't take it too seriously.
  9. Nov 30, 2009 #8

    Thanks, the weight thing in newtons messed me up. Thank you so much anyways. I understand now.
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