Properties of the Sun
Goals of the Lab
- To find the Solar rotation rate from images of the Sun.
- To measure the velocity of matter leaving the Sun through a coronal mass ejection.
Requirements: a calculator, a ruler.
Background and Theory
The rotation period of an object is the time it takes to spin on its axis once. For example, the rotation period of the Earth is one sidereal day. In order to observe the rotation period of any rotating sphere, we can measure how many degrees of longitude a surface feature moves in a given time interval. We then solve for the period by taking the ratio of one full rotation (360 degrees) to the observed angular motion, and multiplying by the time interval. For example, if we observe that a given feature moves 30 degrees of longitude in 2 hours, then the rotation period is:
period=(360°)×(2 hours/30°)=24 hours.
This method can be applied to the rotation of the Sun by observing active regions as they are carried across the disk by the Sun's rotation.
Active regions of the Sun come in a number of varieties, such as sunspots, flares, and prominences. All active regions have large magnetic fields. Sunspots are relatively cool areas of the Sun which stand out from the rest of the surface because they are relatively dark. The temperature of a sunspot is about 4200 K compared to a temperature of about 5500 K for the rest of the Sun. If we could see the spots in isolation, they would appear bright red. Flares occur in regions of the Sun where the magnetic field is twisted very tightly by the differential rotation of the Sun. This magnetic twisting then releases energy. These active regions emit very high energy light, such as X-rays. Prominences are regions of cool gas in the corona of the Sun. When seen against the disk, these appear as dark filaments, but when seen above the Sun's limb, they can be very bright.
Coronal mass ejections (CMEs) are caused by the eruption of prominences, and are blasts of gas which move out through the Sun's corona into the rest of the solar system. You can see an image from the SOHO spacecraft of a CME from October 2003 in this movie. The "fuzz" you see in the movie after the CME is ejected results from charged particles hitting the spacecraft detector. The movie uses a coronagraph, where the bright disk of the Sun is blocked out (similar to an artificial eclipes) so that you can see the thin, dim Solar cornoa. The circle is the size the Sun would be in the movie if you could see it.
Part I: The Rotation Period of the Sun
Examine the two images of the Sun below. They were taken 1 day apart.
![[Sun Image 1]](sol104.gif)
![[Sun Image 2]](sol105.gif)
Notice the bright active regions and the dark sunspots in each image. In which compass direction do the regions move over one day? (Hint: the thin white line at the top of the image indicates North.)
Make a diagram of the Sun-Earth system, showing the rotation of both the Sun and the Earth, as well as the revolution of the Earth about the Sun. What can you say about all of these motions? Comment on a possible reason for any similarities or differences.
Find the image scale for the image. To do this, measure the diameter of the Sun in pixels. Use the:
Click on the very left side of the Sun, and note down the coordinates (x1, y1). Then click on the very right side of the Sun, and note down the coordinates (x2, y2). You can figure out the total distance in pixels using the Pythagorean Theorem:
d2 = (x2-x1)2 + (y2-y1)2Divide the actual diameter of the Sun in km (1.4 million km) by the diameter in pixels to get the image scale. This is similar to finding a scale on a map...
Choose one of the active regions near the Sun's equator. Make sure that it is recognizable in both images. Use the measurement tools to find the pixel positions of the active regions in the first image and the second image:
Note down the coordinates of the active region on the first image as (x1,y1) and the coordinates of the active region on the second image as (x2,y2). Figure out the pixel distance the active region moved using the Pythagorean Theorem:
d2 = (x2-x1)2 + (y2-y1)2Multiply by the image scale from step 3 to find the actual distance, D, in km, moved by the active region.
Since the spots are always moving at the same velocity, the period of the rotation of the Sun can be found using a ratio. A point on the surface of the Sun near the equator rotates all the way around the circumference of the Sun, 2*π*R, over the course of one orbital period. The images you are using were taken one day apart. The ratio of the times must equal the ratio of the distances as follows:
P (days) = 2πR / DCalculate the period, P, of the Sun in days. (D is the actual distance calculated above, and R is the radius of the Sun, which is half of the diameter of the Sun.)
Estimate the uncertainty of your measurements (in pixels). For example, do you think your measurement was accurate to one pixel? two pixels? eight pixels? This is your uncertainty. Use the following formula to estimate your fractional error:
fractional error = (uncertainty) / dwhere d is the distance moved by the sunspot (in pixels).
To find the corresponding error in days, multiply your fractional error by the period. Report the period as the number you calculated +/- your error. For example: 28+/-3 days.
Part II: Solar Prominences
Examine the image below from SOHO/EIT, which shows the development of a solar prominence over time.
![[Solar Prominence]](promine.gif)
Now, find the image scale for the top set of frames as in step 3 in Part I above. (All of the subimages in the top row have the same image scale.) Use the:
Use the Sun Prominence Measuring Tool to measure the minimum height (in pixels) of the prominence above the surface of the Sun, as shown in the first upper frame. Use the (x1,y1) and (x2,y2) method with the Pythagorean Theorem as you did above. Multiply by the image scale to get the minimum height in kilometers.
Use the Sun Prominence Measuring Tool to measure the maximum height (in pixels) of the prominence, as shown in the fourth upper frame. Multiply by the image scale to get the maximum height in kilometers.
Find the growth of the prominence by subtracting the minimum height from the maximum height.
Find the time between the frames (in hours) by subtracting the times, which are given in hours and minutes of Universal Time.
Find the velocity of the material as it leaves the Sun by dividing the growth of the prominence by the time interval between frames 1 and 4. Convert this velocity from km/hr to mi/hr, a scale which may be more familiar. (1.6 km = 1 mile). Is this velocity fast or slow compared to the speeds of objects you are familiar with (for example, cars or jet planes)?
This lab is based strongly on the "Properties of the Sun" lab at the University of Washington.