- 3522 - ENERGY - of a solar flare, Math. - On November 4, 2003 a Solar Flare was observed by satellite detectors that measured the intensity of radiation in watts per square meter per second. This light intensity was recorded over several minutes. The light intensity rose very quickly, in 3 to 5 seconds. How much energy was released?
--------------------- 3522 - ENERGY - of a solar flare, Math
- The intensity decayed more slowly, in 85 seconds. This review explains how astronomers calculated the energy of such a solar flare. Actually the detectors in the satellites are very sensitive and the peak intensity of the flare was so intense it saturated the detectors so no record could be made of the peak. Calculus to the rescue.
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- During maximum solar activity 1,100 solar flares occur each year. The flares follow the lines of the magnetic fields exiting the Sun’s surface at one point ( the north pole) and entering the Sun’s surface at another point ( the south magnetic pole for each flare).
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- A solar flare can carry gas particles out to 100,000 mile for the Sun’s surface. Temperatures in the flares can measure 5,000,000 degrees Kelvin. This particular flare only lasted a couple minutes but some can last days, even weeks.
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- The satellite detectors saturated at 1.8 thousands of a watt creating a flat line for the peak of the intensity. However, we can plot the growth and decay of the intensity over time and free hand sketch the peak to be somewhere around 0.00200 watts per meter^2. The data for the growth of the intensity for the first 5 minutes was:
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Growth Curve:
------------- Time ------------- Intensity in watts /m^2 / sec
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------------- 0 time start ------------- 0.0010
------------- 1 ------------- 0.0060
------------- 2 ------------- 0.0180
------------- 5 ------------- 0.0200 (estimated)
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Decay Curve:
------------- 10 ------------- 0.0200 (estimated)
------------- 15 ------------- 0.0180
------------- 20 ------------- 0.0140
------------- 25 ------------- 0.0090
------------- 30 ------------- 0.0060
------------- 35 ------------- 0.0040
------------- 40 ------------- 0.0030
------------- 45 ------------- 0.0025
------------- 50 ------------- 0.0020
------------- 55 ------------- 0.0019
------------- 60 ------------- 0.0017
------------- 65 ------------- 0.0016
------------- 70 ------------- 0.0015
------------- 75 ------------- 0.0014
------------- 80 ------------- 0.0012
------------- 85 ------------- 0.0010
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- To calculate the peak of the intensity we need to determine the growth function and the decay function. If this were a continuous function then we could determine the peak using calculus to take the derivative of the function and set it equal to zero. The derivative of a function is the slope and the slope passes through zero when it goes over the peak.
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- In this case we can not do this because the growth function and the decay function for the solar flare are two different processes and the functions are discontinuous. Therefore we will treat each function separately, set them equal to each other, and determine the intersection of the two functions in order to determine the peak intensity.
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- To get the function for the growth curve we put the data in to an Excel Spreadsheet, plot the data, do a least squares regression analysis to fit the best exponential function to the curve and the program gives us the function, y = function (x).
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--------- The exponential curve to fit the growth curve: y = 0.0001 e^1.4452 x
where y is the intensity in watts /m^2/sec and x is the time in minutes.
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----------- The exponential curve to best fit the decay curve: y = 0.002 e^-0.0386 x
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- To find the intersection of these two curves we set them equal to each other and solve for “x”, the time the peak occurred:
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---------- y = 0.0001 e^1.4452 x = y = 0.002 e^-0.0386 x
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---------- Take the natural log of both sides of the equation.
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---------- ln 0.0001 + 1.4452 x = ln 0.002 - 0.0386 x
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---------- -9.21 + 1.4452 x = -6.21 - 0.0386 x
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---------- 1.4835 x = 3
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---------- x = 2.02
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- So the peak occurred at 2.02 minutes. To find out the energy intensity at that point substitute this time back into the first equation for the growth of intensity:
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------------ y = 0.0001 e^1.4452 x
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------------ y = 0.0001 e^1.4452 * 2.02
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------------ y = 0.0001 e^2.92
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------------ y = 0.0001 * 18.53
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------------ y = 0.00185
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------------ The peak intensity was 0.00185 watts / m^2/sec
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- To double check lets substitute the same time into the decay curve to get the peak intensity:
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------------ y = 0.002 e^-0.0386 x
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------------ y = 0.002 e^-0.0386 *2.02
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------------ y = 0.002 e^-.078
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------------ y = 0.002 / 1.08
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------------ y = 0.00185
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------------ The peak intensity was 0.00185 watts / m^2/sec
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- To calculate the total energy we need to determine the area under the two curves, the growth curve up to 2.02 minutes and the decay curve from that point , 2.02 minutes out to 85 minutes.
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- ( Calculus: Integration is used to determine the area under the light curve. Integration is the summation of little rectangles “y” high by “dx” wide. The area of each rectangle being “y*dx”. The summation of these tiny rectangle areas being the total area, or the total energy intensity under the curve and over the time that we measured it.)
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-------- Integral from 0 to 2.02 minutes of y = 0.0001 e^1.4452 x
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-------- The general integral for e^ax * dx = 1/a * e^ax
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--------------------------- Integral from 0 to 2.02 minutes :
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-------- Integral y*dx = 0.0001 / 1.4452 * ( e^1.4452 * 2.02 - e^0 )
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------- Integral y*dx = 0.0000692 * ( e^2.919 - 1 )
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------- Integral y*dx = 0.0000692 * ( 18.53 - 1 )
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------- Integral y*dx = 0.0000692 * (17.53 )
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------- Energy intensity under the growth curve = 0.0023 watts / m^2
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---------------------------- Integral from 2.02 to 85 minutes of y = 0.0020 e^-0.0386 x
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--------- Integral y*dx = 0.0020 / -0.0386 * ( e^-0.0386 * 85 - e^-0.0386 * 2.02 )
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--------- Integral y *dx = -0.0518 * ( 1 / e^3.281 + 1 / e^ .07797 )
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--------- Integral y*dx = -0.0518 * ( 1 / 26.60 + 1 / 1.081)
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--------- Integral y*dx = -0.0518 * ( .0376 + .925)
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--------- Integral y*dx = -0.0518 * .96267
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--------- Integral y*dx = -.0499 watts / m^2
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------- Energy intensity under the decay curve = .0499 watts / m^2
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- The total energy under both the growth and the decay curve = 0.0023 watts /m^2 + 0.0499 watts / m^2 = 0.0522 watts / m^2.
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- Our detector was 93 million miles away from the Sun ( 147*10^6 kilometers , or 147*10^9 meters). The total surface area of this spherical radiation is 4*pi*radius^2.
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- (Calculus: The summation of the surface areas from the center out to the radius is the volume of the sphere. The integral of 4*pi*r^2 is ¾ * pi *r^3. The general integral for this function is a*x^n = a/(n+1) * x^(n+1).)
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- The surface area of the sphere at 147*10^9 meter radius = 4*pi*(147*10^9)^2 = 12.567 ( 21,609*10^18) = 27.16 * 10^22 meters^2.
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- Therefore, if one square meter is 0.0522 watts then the total energy intensity across the entire surface area of the sphere at that radius is (0.0522 watts/m^2) * (27.16 * 10^22 m^2) = 1.42*10^22 watts for the solar flare.
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- The total luminosity, or intensity of energy radiation from the Sun is calculated the same way. We get 1,400 watts / square meter of the Sun’s energy on the Earth’s atmosphere. 1,400 * 27*10^22 = 3.827 * 10^26 watts for the total intensity of the Sun.
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- For each second of time this is 3.827*10^26 watt * seconds, or 3.827 * 10^26 joules of energy.
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- I hope you enjoyed the math.
March 28, 2022 ENERGY - of a solar flare, Math 3522
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--------------------- --- Monday, March 28, 2022 ---------------------------
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