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Escape Velocity, with varying acceleration as a(y)

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Find: The minimum initial velocity at which a projectile should be shot vertically from the surface of the Earth so that it does not fall back to Earth. Note: This requires that v=0 as y approaches infinity.

    2. Relevant equations
    a = -g[(R^2)/((R+y)^2)]
    g = 9.81 m/s^2
    R = 6356 km (6.356E6 m)

    3. The attempt at a solution
    My thought process is to substitute dv/dt for a and integrate to get to velocity, but it seems like I want to find velocity as a function of position? I am stuck, it is the acceleration as a function of position that is confusing me.
     
  2. jcsd
  3. Apr 9, 2015 #2
    acceleration due to gravity is a function of position
     
  4. Apr 9, 2015 #3

    vela

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    You can use the chain rule to write
    $$a = \frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v \frac{dv}{dy}.$$
     
  5. Apr 9, 2015 #4

    Dick

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    Using conservation of energy is the easiest way to approach this if you know how that works.
     
  6. Apr 9, 2015 #5
    So using the substitution a = vdv/dy and the given equation for acceleration, then separating variables, I should be able to integrate both sides and solve for velocity with respect to y. Let me try that.
     
  7. Apr 9, 2015 #6
    After integration, I find v^2 = 2g(R^2)/(R+y) + C. Is it valid to say that C is zero because v is zero as y goes to infinity?
     
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