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Escape velocity for object on Earth

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose the gravitational force of the Earth on a body was [itex]F = \frac{KMm}{r^3}[/itex]. What escape velocity would a body need to escape the gravitational field of the Earth?


    2. Relevant equations

    [itex]v_e = \sqrt{\frac{2GM}{R}}[/itex]
    [itex]F_g = G*\frac{m_1*m_2}{r^2}[/itex]

    3. The attempt at a solution

    [itex]G*\frac{m_1*m_2}{r^2}=\frac{KM_em}{r^3}[/itex]
    m_1 and m_2 cross out with M and m. r^3 and r^2 can be reduced to give:
    [itex]G=K/R[/itex]

    Re-plug back into v_e equation:

    [itex]\sqrt{\frac{2KM}{R^2}}[/itex]

    But the given solution says:

    [itex]\sqrt{\frac{KM}{R^2}}[/itex]

    Where did I go wrong? Thanks!
     
  2. jcsd
  3. Apr 9, 2013 #2

    cepheid

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    No, you're misunderstanding the problem. You shouldn't use Newton's Universal Law of Graviation, because the whole point of the problem is to ask, "what if we lived in a universe where Newton's universal law of gravitation was different and was an inverse-cube law instead of an inverse-square law?"

    Hint: for a conservative force, we have:[tex]F = -\frac{dU}{dr}[/tex]which means that[tex]U = -\int F(r)\,dr[/tex]where this integral will give you an arbitrary constant, but we have the freedom to set this arbitrary constant to whatever we want, and we typically set it to 0 so that the potential energy goes to 0 at infinity.
     
  4. Apr 9, 2013 #3
    Ok thank you!

    I can take the integral of (KMm/r^3) and get:

    [itex]\frac{KMm}{2r^2}[/itex]

    then:

    [itex] \frac{1}{2}mv^2 - \frac{KMm}{2r^2} = \frac{-KMm}{r_{max}}[/itex]
    [itex] v^2 = 2KMm\times (\frac{1}{2Rr^2})[/itex]
    [itex] v = \sqrt{\frac{KM}{R^2}}[/itex]

    Thanks again!
     
  5. Apr 9, 2013 #4

    cepheid

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    I don't understand your second equation, particularly the KmM/r_max term. All you have to do is equate the potential energy to the kinetic energy. (1/2)mv^2 = KMm/(2r^2). This gives you the right answer.
     
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