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Initial Velocity of a volcano

  1. Jun 9, 2009 #1
    Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500km (or even higher) above the surface. Io has a mass of 8.94×10^22 [kg] and a radius of 1815km . Ignore any variation in gravity over the 500km range of the debris.

    How high (in km) would this material go on earth if it were ejected with the same speed as on Io?

    I've calculated g from g = GM/R^2. What equation would I use to calculate the launch speed, v_0, for y= 500,000 m?
     
  2. jcsd
  3. Jun 9, 2009 #2
    Perhaps consider conservation of energy:

    [tex]mg_{Io}h = \frac{mv^2}{2} \implies v^2 = 2g_{Io}h [/tex]

    Where [tex]g_{Io}[/tex] is the acceleration due to gravity on Io.
     
  4. Jun 9, 2009 #3
    I don't think we've covered that equation yet. Anything else that you can think of? This chapter covers Newton's Law of Gravitation.
     
  5. Jun 9, 2009 #4
    You can derive this equation under constant acceleration considerations, but the result is the same. Specifically, the equation is:

    [tex]2a\Delta y = v_f^2 - v_0^2[/tex]

    In your situation you should see that you know what a is and you know what v_f is.
     
  6. Jun 9, 2009 #5
    V_f = 0 since it's falling back down?
    I plugged my values, and got 55.4223 km. Wasn't correct though. Is my v_f value correct?
     
  7. Jun 9, 2009 #6
    At the maximum height right before it starts to fall down the velocity should be zero, so yes [tex]v_f = 0[/tex].

    Please write out what you used for your equation replacing [tex]a[/tex] and [tex]v_f[/tex] with what they should be.

    What do you find for [tex]g_{Io}[/tex]?

    What do you find the initial velocity to be on Io?

    What equation do you use to solve for how high it goes on the Earth?
     
  8. Jun 9, 2009 #7
    g= G_m/R^2 which came out to 1.81*10^6 m/s^2

    From there I plugged my variables into [tex]2a\Delta y = v_f^2 - v_0^2[/tex]

    2(-1.81*10^6) 500,000 = 0 - v_f^2 = 1.08*10^6 m/s

    Then I replaced a with g of earth, 9.81, since I calculated v_0.

    2(-9.81) (delta y) = 0 - (1.08*10^6) = 55.4223 km
     
  9. Jun 9, 2009 #8
    It looks good dREAPER. Please check over your units (does your answer for [tex]g_{Io}[/tex] make sense?) and please check over your calculations one more time (what is [tex]v_0?[/tex])

    You should end up with ~92km.
     
  10. Jun 9, 2009 #9
    Ahh thank you. I missed the units for gravity (it was in km, not m). Calculated it and got 92.3469 m/s^2. Was correct.
     
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