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Cannonball shot off of Earth, Escape velocity, max height

  1. Sep 25, 2011 #1
    I'm having some difficulty with a homework problem which states: A cannonball is shot straight up out of a very powerful cannon located at the South Pole and reaches a maximum height of one Earth radius above the surface of the Earth. What was its initial speed? How long did it take to reach its max height?

    The relevant equations I've gathered are:

    1. Radius of Earth Re = 6.4*10^6 meters.
    2. Escape Velocity Ves = (2GM/ro)^(1/2), where ro=Re.
    3. v^2 = Vo^2+2GM(1/r - 1/ro). Is v the final velocity at its max height if I plug in Re for r and ro?
    4. Time to reach max height [itex]\tau[/itex] = (2/GM)^(1/2)*rmax^(3/2)*[([itex]\pi[/itex]/4)-([itex]\theta[/itex]o/2)+(sin2[itex]\theta[/itex]o/4)] What would [itex]\theta[/itex]o represent in this equation? I'm assuming zero since the cannon is shot straight up.

    I would like to know how I would approach the problem using these equations. Would I have to calculate escape velocity and then calculate an additional velocity exerted by the cannonball after it escapes earth and travels an additional Re? Any help would be greatly appreciated. Thanks.
     
  2. jcsd
  3. Sep 25, 2011 #2

    dynamicsolo

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    You won't be using escape velocity at all, since that is the initial speed that would just send the cannonball out to "infinity" after "infinite time"; we know it didn't get that high up...

    You'll need equation (3), with the initial radius set to r0 = RE and the final radius set to r = (?) [the cannonball is at an altitude of RE , so its distance r from the center of the Earth is...?]. At that maximum radius, the cannonball has come to rest (momentarily), so the final velocity is v = ? You are then solving for v0.

    As for equation (4), it should say somewhere in your source what [itex]\theta_{0}[/itex] is. In some contexts concerning projectiles, it is the launch angle measured from the horizontal, but that would be [itex]\frac{\pi}{2}[/itex] here, which would give a time of zero(!). You are likely correct to set the angle to zero; I have not seen the time expressed this way before (I think it is an approximation to save you having to compute an integral).
     
  4. Sep 25, 2011 #3
    Ok, so correct me if I'm wrong, we don't consider escape velocity because it states the cannonball reaches a max height? If it reached escape velocity it would then enter into space outside of Earth's gravitational field and have no "max height."

    Here is how I solved the problem...

    Re = 6.4*10^6m
    Me = 5.9742*10^24 kg
    G = 6.674*10^-11

    Using equation (3) from above [V^2 = Vo^2 +2GM (1/r - 1/ro)]

    0 = Vo^2 + 2*(6.674*10^-11)*(5.9742*10^24) * (1/2Re - 1/Re)

    Vo = 7876.968 m/s

    Then to calculate the time using equation (4) [τ = (2/GM)^(1/2)*2Re^(3/2)*[([itex]\pi[/itex]/4)]]

    τ = 2548.085 seconds.

    Please don't hesitate to correct any mistakes I may have made.

    Thank you.
     
    Last edited: Sep 25, 2011
  5. Sep 25, 2011 #4

    HallsofIvy

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    Yes, you don't use "escape velocity" because the cannon ball does not "escape"!
     
  6. Sep 25, 2011 #5

    dynamicsolo

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    Contrary to the journalistic phrase "outside of Earth's gravitational field" of long standing, there is no "outside". The gravitation of an object is treated as extending for "infinite" distances. The model for gravitational force (or potential) is applied at any distance.

    But, yes, "escape velocity" is the speed at which an object departing along a radius from the body's surface would be just carried to an "infinite distance"; it would "arrive" there at zero speed, so it would require "infinite time" to "complete" the trajectory. A higher speed would allow the object to "reach infinity" with a non-zero speed.

    Yes, in fact it is the same speed you get if you calculate the "circular velocity" of an orbit just about at the surface of the Earth ( r0 = RE ) , that is,

    [tex] v_{c} = \sqrt{\frac{GM}{r_{0}}} = \sqrt{\frac{GM}{R_{E}}} .[/tex]

    Again, I'm taking their word for this equation, but the factor [itex](\frac{R^{3}}{GM})^{1/2}[/itex] certainly belongs there, having units of time and coming from Kepler's Third Law. (I'd have to work this through to see if the "dimensionless constants" are right.)

    The result from the formula simplifies to

    [tex]\tau = (\frac{2}{GM})^{1/2} \cdot (2 · R_{E})^{3/2} \cdot (\frac{\pi}{4}) = (\frac{2^{2} \cdot R_{E}^{3}}{GM})^{1/2} \cdot (\frac{\pi}{4}) = (\frac{\pi}{2}) \cdot (\frac{R_{E}^{3}}{GM})^{1/2} . [/tex]

    The basic time constant for Earth is [itex] (\frac{R_{E}^{3}}{GM})^{1/2} \approx 806.7 [/itex] seconds, so I think you may have had a factor slip away from you somewhere. Keep in mind that since rmax = 2 · RE , the "2" and the "RE" are under that 3/2 power...
     
    Last edited: Sep 25, 2011
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