A short but spectacular Video of a Solar Eruption

Ramani Aswath

Ramani Aswath

@ramani-VR4O43 • Oct 22, 2024
The Solar Dynamics Observatory (SDO) satellite captured a stunning solar eruption. Tremendous magnetic forces jutted out the plasma. However, the even more powerful gravity of the sun pulled back almost all of the plasma.
Looks best full screen.


#-Link-Snipped-#
Like 0 Replies 5

Replies

Welcome, guest

Join CrazyEngineers to reply, ask questions, and participate in conversations.

Sign up Log in

CrazyEngineers powered by Jatra Community Platform

  • Kaustubh Katdare

    Kaustubh Katdare

    @thebigk • Jan 9, 2013

    Indeed awesome! I bet the eruption is several thousand kilometers in height! The Sun's very BIG!
    0
  • Ramani Aswath

    Ramani Aswath

    @ramani-VR4O43 • Jan 10, 2013

    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!
    0
  • Kaustubh Katdare

    Kaustubh Katdare

    @thebigk • Jan 10, 2013

    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: <a href="https://www.suntrek.org/solar-surface-below/around-solar-surface/giant-eruptions-from-sun.shtml" target="_blank" rel="noopener noreferrer">Giant eruptions from the Sun - On & around the solar surface - Solar surface & below - Sun|trek</a> 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?
    0
  • Ramani Aswath

    Ramani Aswath

    @ramani-VR4O43 • Jan 10, 2013

    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.
    0
  • Ramani Aswath

    Ramani Aswath

    @ramani-VR4O43 • Jan 10, 2013

    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.
    0