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Maximum height of ball with changing acceleration

  1. Oct 7, 2015 #1

    Ryan_m_b

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    I put basic on this because my maths skills are exactly that and I suspect the answer to this question will be fairly maths heavy.

    I understand that the equation to calculate the maximum height of a projectile (ignoring wind resistance) is

    Hmax = (Vf2 - Vi2)/2a

    But this only works over small distances and doesn't take into account the change in acceleration due to being further away from the center of the mass the object is being fired from. I've tried to google this but I must be using silly search terms because the results aren't quite what I'm asking.

    Any help would be appreciated :)
     
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  3. Oct 7, 2015 #2

    DaveC426913

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    How the heck high is this ball being fired??

    Gravity does not change appreciably at altitude from Earth until you get well past low orbit.
     
  4. Oct 7, 2015 #3

    Ryan_m_b

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    Yeah I know, for the sake of argument assume really, really high. The initial velocity could be a good fraction of escape velocity.
     
    Last edited: Oct 7, 2015
  5. Oct 7, 2015 #4
    The force acting between to masses can be written as

    F = G ⋅ m1 ⋅ m2 / r2

    G ... constant (6.674 ⋅ 10-11 m3/(kg ⋅ s2)
    m1 ... mass 1 (e.g. projectile)
    m2 ... mass 2 (e.g. earth) (5.972 ⋅ 1024 kg)
    r ... distance between the centres of mass of m1 & m2

    Due to energy consistency (with v0 = initial velocity) →

    m1 ⋅ v02 / 2 = ∫ [(G ⋅ m1 ⋅ m2 / r2) ⋅ dr]

    integrating from r1 to r2 → v02 / 2 = G ⋅ m2 ⋅ (1 / r1 - 1 / r2)

    If you start from the surface of the earth r1 ≈ 6.37 ⋅ 106 m, plugging that in, you get r2
     
  6. Oct 7, 2015 #5

    Ryan_m_b

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    Thanks, I'm a little confused with the way it's laid out (basic maths knowledge). I don't get how to fill out the equation without knowing r2, do you mind swapping out the symbols for real figures? Lets say:

    Cannonball weight = 1kg
    Initial velocity = 10km/s
     
  7. Oct 7, 2015 #6
    The weight of the cannon ball doesn't matter, but if the initial velocity v0 = 10 ⋅ 103 m / s

    r2 = 1 / { (1 / r1) - [v02 / ( 2 ⋅ G ⋅ m2 ) ] }

    r2 = 1 / { (1 / (6.37 ⋅ 106)) - [108 / ( 2 ⋅ 6.674 ⋅ 10-11 ⋅ 5.972 ⋅ 1024 ) ] } = 31,708 ⋅ 103 m

    That's the distance from the earth's centre of mass, so you have to substract the radius: 31,708 ⋅ 103 - 6,370 ⋅ 103 = 25,338 ⋅ 103 m → about 25300 km above earth surface.
     
  8. Oct 7, 2015 #7

    Ryan_m_b

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    Awesome :) thanks
     
  9. Oct 7, 2015 #8

    PeroK

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    For large distances and velocities, you could equate total energy,, kinetic and potential.
     
  10. Oct 7, 2015 #9

    Ryan_m_b

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    One more thing :) I've had a go at rearranging the equation but again, maths skills. How would this example work if instead of the initial velocity you had everything else? In other words you knew the and the radius of Earth and were told that an object reached a height of X, how would you work out he initial velocity?
     
  11. Oct 7, 2015 #10
    v0 = SQRT[2 ⋅ G ⋅ m2 ⋅ (1 / r1 - 1 / r2)]

    But you must not forget to add the radius of the earth to the heights above the surface,

    so r1 = 6.37 ⋅ 106 m (if started from the earth surface)
    and r2 = h + 6.37 ⋅ 106 m (if h = height above surface in m)
     
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