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Calculating Escape Velocity

  1. Apr 20, 2009 #1

    Stratosphere

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    How do you Calculating the escape velocity? I googled it but they give bad instructions on how to do it. What is the formula?
     
    Last edited: Apr 20, 2009
    Stratosphere, Apr 20, 2009
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  3. Apr 20, 2009 #2

    Stratosphere

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    Every time I try using the formula i have i have to solve for V^2 and I get it to equal 35
    3541. Is this right? I plug in the numbers like it says it I still dont get the right answer.
     
    Last edited: Apr 20, 2009
    Stratosphere, Apr 20, 2009
  4. Apr 20, 2009 #3

    mgb_phys

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    Gravitational potential energy is = GMm/r (where M is the mass of the earth)
    Kinetic energy = 1/2 m v^2

    You start at r=radius of earth and end up at r=infinity.
    A bit of rearranging and a subtraction and it shoudl be easy
     
    mgb_phys, Apr 20, 2009
  5. Apr 20, 2009 #4

    Stratosphere

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    I set KE=GPE i get a velocity of 465,561 for a 10kg object. This doesnt seem right.
     
    Stratosphere, Apr 20, 2009
  6. Apr 20, 2009 #5

    Nabeshin

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    Well escape velocity is a function of radius so this question, as you stated, makes no sense...
     
    Nabeshin, Apr 20, 2009
  7. Apr 20, 2009 #6

    Janus

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    What formula are you using?
     
    Janus, Apr 20, 2009
  8. Apr 20, 2009 #7

    Stratosphere

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    V[tex]^{2}[/tex]=[tex]\frac{2GM}{R}[/tex] to find the escape velocity of the earth and i get, 465561.808.
     
    Stratosphere, Apr 20, 2009
  9. Apr 20, 2009 #8

    sylas

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    You've got "tex" tags inside your formula. I've tried to fix tags in the quoted extract; but it is still not displaying for some reason. But the forumla looks fine to me. Just make sure all your variables are in consistent units. You give a number, but no units. Are you meaning km/hr, or m/s, or something else?

    The gravitational constant G is usually given in SI units. If you do everything in meters and seconds and kilograms, it will work.
     
    Last edited: Apr 20, 2009
    sylas, Apr 20, 2009
  10. Apr 20, 2009 #9

    Janus

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    Okay, from this answer, I'd say that you are using kilometers instead of meters for the radius of the Earth.
     
    Janus, Apr 20, 2009
  11. Apr 20, 2009 #10

    mgb_phys

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    Why would the mass of the object matter ?

    GM for the Earth is almost exactly 400,000 km^3/s^2 (worth remembering)
    The radius of the Earth is around 6400km

    so v^2 = 2GM/r = 2*400,000/6400 = 225km^2/s^2
    v = 11.2km/s (roughly)

    edit - sorry i posted this straight after you replied, but it didn't show up (got a database error message)
     
    mgb_phys, Apr 20, 2009
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