A short but spectacular Video of a Solar Eruption

Replies

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  Kaustubh Katdare

  Indeed awesome! I bet the eruption is several thousand kilometers in height! The Sun's very BIG!
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  Ramani Aswath

  The_Big_K

  Indeed awesome! I bet the eruption is several thousand kilometers in height! The Sun's very BIG!

  You really kindled my curiosity. I thought that the video must have enough data for one to calculate the height of eruption. As I always maintain, any engineer worth his salt (especially CEans) should do an order of magnitude calculation at the drop of the (hard) hat
  I keep a cheap plastic vernier calipers at home for various projects. I traced the frozen image on the monitor on a OHP sheet (I keep some of these for doodling) and measured the chord of the sun's profile and the height of the eruption at maximum point (where I froze the video). Later I measured the depth of the chord from the edge of the sun from the tracing.
  
  Chord = 83 mm depth of chord = 10 mm Height of eruption = 45 mm
  
  Good old ever present Perry and the mensuration formulae were consulted.
  Sun's Radius (average) = 695000 km (from Wolfram Alpha)
  After some calculations (aided by Wolfram Alpha),
  Height of eruption = 343,577 kilometers
  You won the bet, K!
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  Kaustubh Katdare

  That sir! I say! Is awesome way of calculating! I thought the eruption would at the most go to a thousand or two kilometers heigh; and my guess was based on looking at the overall picture - photo frame shows big part of the Sun's surface and the overall height of the eruption occupies most part of the screen. But I'd have thought the eruption would be that tall!
  
  I found this website: Giant eruptions from the Sun - On & around the solar surface - Solar surface & below - Sun|trek that says one of the giant eruption is 200,000 km tall! Holy kaw!
  
  The next question in my mind is - what was the velocity with which the fireball shot up in the skies! Is there any way to measure it?
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  Ramani Aswath

  The_Big_K

  The next question in my mind is - what was the velocity with which the fireball shot up in the skies! Is there any way to measure it?

  Now you have me on a different spin!
  The description of the video mentions something about the time factor. The video does show the falling back of the plasma. We know the gravitation at the sun's surface.
  Back to paper, pencil and Perry! Tomorrow perhaps, Rather sleepy after a long day at the factory.
• 

  Ramani Aswath

  Here goes. Some assumptions have to be made:
  1. The eruption does not meet any solar atmospheric resistance (that is, there is vacuum above the sun's surface.
  2. A geometric mean gravitational field is applicable between the highest point reached by the eruption and the sun's surface.
  
  Given the above the rest is fairly routine arithmetic.
  Sun's acceleration due to gravity at its surface = 273.95 meters/sec/sec (Wolfram Alpha)
  Sun's acceleration due to gravity at 343,577 km above surface = 183.3 meters/sec/sec (this arises out of the gravitation law), which translates to:
  g(h) = g(s)(r/(r+343577))
  where g(h) is the accelration due to gravity at height h and g(s) that at the surface and r the radius of sun.
  The geometric mean acceleration = 224.1 m/sec/sec again to take into account the inverse square law.
  The third equation of motion for free fall in gravity is V^2 = 2g s
  
  V^2 = 2 x 0.2241 x 343,557 (all in consistent units)
  from which, V = 392.42 km/sec
  
  That is the whopping velocity (greater than 0.1% of speed of light) at which the jet should have taken off from the surface.
  
  The value may not be dead on the nose. However, the order of magnitude should be correct.