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Escape velocity & Black-Body Radiation

  1. Jul 18, 2007 #1
    I'm being confused between a Physics past exam paper book and a Physics study guide.

    Is the escape velocity derived from EK + EP = 0 (this makes the escape velocity depend on the mass of the body to escape from i.e. Earth or a planet) or EK = EP (this makes the escape velocity depend on the mass of the escaping object) ?

    (Where EK = 0.5mv2 and EP = - GmM/r)

    Also, the final step I must take in conquering the necessary Quantum Physics is the Black Body Radiation curve.

    Why is there a very large maximum point (or peak), say at approximately 210nm then a sudden plummet in the curve after this ?
    Where is this extra large intensity emitted from ?
  2. jcsd
  3. Jul 18, 2007 #2


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    ?? Except for a sign change, the two equations are identical. Certainly if one depends "on the mass of the body to escape from" the other must also. In either form, you can divide the entire equation by m and eliminate m.

  4. Jul 18, 2007 #3
    EK + EP = 0

    That is the equation for conservation of energy. The equation you wrote above is just the same but you missed a - sign in the right hand side. Also, just use your equation and fill in the values for kinetic and potential energy. You will see that your value for the velocity is a complex number !

    Also, the second conclusion on mass dependence is incorrect since m will be devided out of the two energy equations ! So, no dependence on the object's mass.

  5. Jul 19, 2007 #4
    High School Physics textbooks say the escape velocity is v = √(2GM/r), the highest level of Maths in High School Physics is simple trigonometry (sinθ & cosθ only), fractions (Fc=mv2/r), substitution and quadratics and definitely NO calculus (a=Δv/Δt instead e.g Final Velocity - Initial Velocity/Final Time - Initial Time)
    As for Complex Numbers, that is studied in the highest level of Mathematics offered at my school (which I don't take).
  6. Jul 19, 2007 #5
    If you would have done that calculation, you would have found out that you will need to take the squareroot of a negative number. This does not work unless you use complex numbers. Such numbers are unphysical in classical mechanics. That's all. It is all about the squareroot of the negative number ! Whether you know complex numbers or not is irrelevant in this case.

  7. Jul 20, 2007 #6
    The derivations from both my study-guide and my teacher say:

    0.5mv2 + (-GMm/r) = 0
    0.5mv2 - GMm/r = 0
    0.5mv2 = GMm/r
    mv2 = 2GMm/r
    v2 = 2GM/r
    v = √(2GM/r)

    Otherwise I can't calculate the escape velocity any other way.
    The Physics curriculum from which we learn from has been extremely simplified.
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