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Vertical projectile mathematics

  1. Mar 27, 2014 #1
    Hello
    I have been struggling with this problem for a whole day, and finally admitting that i can't solve it, so i ask for help :smile:
    I am familiar with the escape velocity , and the speed needed to brake into interplanetary space - the newton cannonball thought experiment. But i was wonder about launching a cannonball perpendicular to the gravitational field - vertical lunch. Ignoring air resistance, and other celestial bodies , what is the initial speed the cannonball must have in order to break trough gravitational influence of the earth and head to infinity?
    I tried , but i got just mathematical understanding of calculus, i never learned it's applications in physics.
    I know only that this values change over time like this:
    v(t) = Vo-g*t g= M*G/r^2 , and R=ro+V(t) H= Vo*t-(g*t^2)/2 or after brief calculation, H=(Vo^2)/2*g
    where M is the mass of earth
    r the radius
    t-time
    ro initial semi major axis of earth
    vo- initial velocity
    H-max height in homogeneous g-field
    because speed tends to reduce for g each second , and g tends to reduce by inverse square of r
    there should , in my opinion be a sufficiently large initial velocity with which a verticaly launched projectile would leave the earth to infinity, as R-> infinity g ->0 , LimV= Vmin and LimH=infinity
    How to mathematically express that kind of solution, if there is one?
     
    Last edited: Mar 27, 2014
  2. jcsd
  3. Mar 27, 2014 #2

    AlephZero

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    Science Advisor
    Homework Helper

    Your formula v(t) = Vo-g*t is only true when g is constant, so it is wrong for this question.

    Probably the easiest way to do the physics is to use conservation of energy. The work required to raise the ball to height h above the center of the earth is the integral of (force x distance). And force = mass x acceleration, so
    work = ##\int_{r_0}^h \frac{mMG}{r^2}dr## where ##r_0## is the earth's radius.

    So the work to reach height h is ##mMG(\frac{1}{r_0} - \frac{1}{h})##.

    If the initial kinetic energy ##\frac{mv^2}{2}## is greater than ##\frac{mMG}{r_0}## there is enough energy to reach any height h, so the ball can escape from the earth's gravitational field.

    You can also solve this problem by integrating the "force = mass x acceleration" equation, but this is tricky because the force is a given as a function of the distance, but acceleration is a derivative with respect to time not distance.
     
  4. Mar 27, 2014 #3
    [itex]\sqrt{2GM/r}[/itex] -and it is the escape velocity :surprised
     
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