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Pytels Dynamics 12.8: Missile dynamics, acceleration and escape velocity

  1. Jun 10, 2017 #1
    1. The problem statement, all variables and given/known data
    A missile is launched from the surface of a planet with the speed v0 at t=0.
    According to the theory of universal gravitation, the speed v of the missile after launch
    is given by
    v2 = 2gr0(r0 / r -1) + v02

    where g is the gravitational acceleration on the surface of the planet and r0 is the mean
    radius of the planet.
    (a) Determine the acceleration of the missile in terms of r.
    (b) Find the escape velocity that is the minimum value of v0 for which the missile will not return to the planet.
    c)Using the result of part (b) calculate the escape velocity for earth, where g = 9.81m/s2 and
    r0 = 6370km

    2. Relevant equations


    3. The attempt at a solution
    I have tried part (a) . My attempt is found in the attached file.
    Another way to go is to use the v2 = 2a(x-x0) + v02
    and compare with given equation. I have tried that and couldn't get -g(r0/r)2
    How do i solve part (b)? any hints
     

    Attached Files:

  2. jcsd
  3. Jun 10, 2017 #2

    haruspex

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    ##\sqrt{a-b}## is not the same as ##\sqrt a- \sqrt b##.
    Try differentiating without taking the square root.

    For b, if it does not return, how large does r get?
     
  4. Jun 10, 2017 #3
    Check for part a and part b in attached files
    If it doesn't return then r→∞
    Is part b correct?
     

    Attached Files:

  5. Jun 10, 2017 #4

    haruspex

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    Remarkably, you got the right answers to both a and b via flawed algebra.

    In a, you differentiated the left hand side with respect to t and the right with respect to r. You must always do the same thing to both sides of an equation. If you differentiate both sides wrt t then on the right you get a factor ##\frac{dr}{dt}## (chain rule).
    But in the next line, the derivative of v2 with respect to t is ##2v\frac{dv}{dt}=2va##, not just 2a. These two errors cancelled out to give the right result.

    In b, you again have made the blunder of distributing √ across a sum of terms. ##\sqrt{v^2+2gr_0}## is not ##\sqrt{v^2}+\sqrt{2gr_0}##. Is √1+√4=√(1+4)? You got away with that because in the case you are interested in v=0.
     
  6. Jun 11, 2017 #5
    Ok got it!
    On the left side we let z=v2 and take the dz/dt = (dz/dr)(dr/dt) = 2va
    On the right side we do the same : 2gr02 dz/dt where z=1/r => 2gr02(dz/dr)(dr/dt) =
    = 2gr02(-1)r-2v
    the 2 and v cancel out
    This is implicit differentiation isn't it?
     
  7. Jun 11, 2017 #6

    haruspex

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    Yes. But I think you mean dz/dt=dz/dv dv/dt on the left.
     
  8. Jun 11, 2017 #7
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