Analyzing Coronal Mass Ejections

Analyzing CMEs
Background
Coronal mass ejections, or CMEs, are the result of collapsing magnetic fields containing immense amounts of hot, ionized gases. These large explosive events tend to last for only a few hours, but eject massive amounts of matter into space at very high speeds. CME events have a bulbous structure and the explosions are often larger than the Sun’s diameter. For this assignment, review the accompanying lecture notes prior to beginning your calculations.
Assignment
Remember to use the Assignment Rubric to review grading requirements.
Part 1: Speed of a CME
The series of images below show a coronal mass ejection that occurred January 4, 2002. The Sun’s surface has been blocked using a coronagraph (dark disc area), and the Sun’s surface is represented as the circular ring in the center of the image.

⦁ Examining the center “clump” of material that can be traced over a series of images, complete the following table. (HINT: Need help getting started? Check out this ⦁ video.)
Time Hours Distance (mm)
10:45 0
11:42 1
12:42 2
13:42 3
14:42 4
15:42 5

⦁ Graph your data from the table above. Then create a best fit line through these data by drawing a straight line that follows the trend in the data. (Insert a chart, then select the dropdown arrow for the Linked Chart Options to edit your data. Open Source to use the data from the table above as your values. Input your data into the source file, then come back to this file. You will then need to click Update on the chart to reflect the changes. The equation of the trend line is displayed in the legend.)

⦁ Find the slope of the line by picking two points and using the slope formula. The result will be in millimeters per hour (mm/h). Recall the slope formula is . (HINT: If you obtained the equation from your trendline in the previous question, simply use the slope from the trendline.)

⦁ Calculate the scaling factor for the images. The small circular ring in the center of the images represents the Sun’s linear surface. If the actual diameter of the Sun is approximately 1.4 x 106 km, then
⦁ Measure the diameter of the Sun in the image series using millimeters.

⦁ Divide the actual diameter of the Sun by this value. (For example, if the small central circle measured 7 mm, then in this image, every 7 mm represents 1.4 x 106 km. The scale would then be calculated as .)

⦁ Find the average speed of the CME over the 5-hour observation period by multiplying your slope (question 3) and the image scale (question 4). The result should be expressed in kilometers per hour (km/h).
This particular CME was not directed towards Earth. However, other solar events occasionally do place Earth in the wave of a CME. The distance between the Sun and the Earth is about 1.5 x 108 km. Dividing this distance by the speed of the CME determines the time it takes for the CME to arrive at Earth.

⦁ Consider a similar CME event directed towards Earth. How many hours would it take for the CME to reach Earth? Days?

Part 2: Accounting for Uncertainty
Your results for determining the speed of the CME in Part 1 depend primarily on an accurate measurement of the Sun’s diameter, locating the same clump in each of the images, and accurately determining the center of the clump as the material spreads out. Consider just the measurement of the Sun’s diameter on the image as that error is systematic – our measurements may be too small or too large, and will affect the results either way.

⦁ What is your original measurement of the Sun’s diameter in millimeters (from question 4a)?

⦁ What would the range of diameters be if your measurement were off by 25%?

⦁ What would the range of scale factors be if measurements were off by 25%?

⦁ Based on the values in questions 8 and 9, continue the calculations to determine the range of speeds of the CME in kilometers per hour (km/h).

⦁ What is the range for how many hours it would take for the CME to reach Earth?