# I require help with understanding beginner astronomy questions.

1. Oct 18, 2014

### SigmaOctans

First, I'd like to make clear that this isn't exactly homework. I've been reading books about astronomy out of pure interest and I've bought myself an introductory astronomy textbook and I've been reading through it and answering all the questions. I've been doing fine... that is until I reached a "difficulty spike" or my brain decided to quit on me. For the past 2 days, I've been trying to find out how to do this:

"From the information in the figure that accompanies the Making Connections box in this chapter, estimate the speed with which the particles in the CME in the final two frames are moving away from the Sun".

The pictures they show are precisely these ones, same order and all: http://sohowww.nascom.nasa.gov/gallery/images/large/nov00cme.jpg

This question is under a section called "Figuring For Yourself" meaning they really don't indicate anywhere in the book how to do this. I imagine this was intended so you could have an instructor to clue you along. I've searched online and have come across this (and many other sites that seem to copy & paste these): http://soho.nascom.nasa.gov/classroom/cme_activity.html

There's just a few things that I don't seem to understand (I've been self-teaching astronomy and mathematics. My math wasn't exactly stellar in high school, but I'm working hard to fix it). In the SOHO site, it states "Select a feature that you can see in all five images, for instance the outermost extent of the bright structure or the inner edge of the dark loop shape. Measure its position in each image." My problem is...from where do I start my measurement? Am I supposed to measure only the feature itself, or am I supposed to start from the "far" end of the Sun's diameter and measure up to my chosen feature? I imagine its the feature only, but I am really unsure.

I kept reading and saw this:
v = (s2 - s 1)/(t2 - t1)
where:

s2 is the position at time, t2.
s1 is the position at time, t1.

Alright, looks simple (this is where my lack of math experience comes into play), but how exactly do I appropriately fit the time into that equation? Am I supposed to put it in minutes or hours? From 08:06 until 11:42 is 216 minutes. So I would be doing 216 - 0 (because my start time, t1, must be zero, right?).
There's another question, but I don't exactly want this to be lengthy and I want to be able to solve this first. I'm not looking for direct answers, I just need a clue. Am I on the right track by going to the SOHO site? Or is it much simpler? To be sure, the book does say how fast a CME travles. It says it "travels outward at about 300 km/s" (though I've read that they can be anywhere from 200-1000km/s), which of course would make this question easy. But I assume they want it only judging from those pictures.

I sincerely apologize if I've violated any forums rules, though I don't think I have.

2. Oct 18, 2014

### Matterwave

These questions are basically rate questions as applied to astronomy. I think you will have to recall back to your high school physics days and remember what you did there. Basically, you want a (average) velocity which is defined as displacement/time. So you need to know 2 things in order to solve this problem. You need to know how far the coronal mass ejection has moved, and how long it took for it to move this far.

To measure how far something has moved, you will need to measure its initial position and its final position. Relative to what point does not matter because in the end you will be making a subtraction. Let's consider only the 1-dimensional case for right now, so that the coronal mass ejection just moves say "to the right" of the page and not "up or down" or "towards or away". Say (just for the sake of an example, these are just random numbers I made up) you took the initial position to be "0" and the final position you found was "250000km", then your s2-s1 will come out to be "250000km". If instead, you decided to measure the "initial position" from the left side of the sun, you might find that the initial position is "400000km", and then your final position you measure also from the left side of the sun will turn out to be "650000km", in this case "s2-s1" will be "650000km-400000km=250000km" which will definitely be the same as before. Try this for yourself and see that it is so.

Now, as far as the time goes, you need a time stamp for the "initial position" and a time stamp for the "final position" and subtract the former from the latter. If you want your answer to come out in km/s (kilometers per second) then you should measure your time in seconds and your positions in kilometers. But physics doesn't tell you that using any other units are wrong. If you wanted to measure your positions in miles and times in hours, that's perfectly fine, according to physics, you will just get an answer that is in miles/hour instead.

3. Oct 19, 2014

### vociferous

There are a lot of ways to approach this problem. The above poster gave you a good rundown. Maybe I can offer how I would approach it.

1. Figure out what the radius of the sun is in meters (you can just look it up).

2. Figure out what the radius of the sun is in pixels for each photo. I would use software, since Astronomy is heavily reliant on doing things that way and it is good to get used to. You can use any old software. You don't need to download a professional astronomy program like SAODS9 or the likes. If you just want to do a rough estimate, instead of using pixels, just put a ruler up to the screen and use millimeters.

3. Likewise, figure out how far the prominence is from the surface in pixels or mm.

4. Using ratios, convert from pixels/mm to meters. Rember, ratios are a1/a2=b1/b2. You can figure out the proper ratio based on the information you looked up for $${R_ \odot }$$. This symbol is just a fancy astronomical way of saying "radius of the sun"

5. Once you have converted everything from pixels/mm on the page to meters in reality, use the formula you looked up to calculate average speed.

4. Oct 20, 2014

### SigmaOctans

Alright, I've attempted this problem many times again after reading the advice (thank you both very much, I appreciate it). Let's see if I understood this correctly.

What I did was find the radius of the Sun which I found was 6.960X10^8 m (according to my book) or 1.392x10^9 m diameter.I then measured the pictures on the computer using MSpaint. In the first frame (well,actually the 3rd frame, but I call it first because it's the first of the relevant ones). In the first frame, the Sun's diameter is 180 pixels and the feature (measuring from the "far" side of the sun's diameter on the page) is 620 pixels. For frame 2, the Sun's diameter was 40 pixels and the feature was 450 pixels. I multiplied them by 4.5 so I can scale it up to the size of the first frame.

Then I looked at the formula on the SOHO website:

dscreen/dactual = sscreen/sactual

Then I simply plugged in the appropriate numbers:

Frame 1:

180 pixels/1.392x10^9m = 620 pixels/4.79x10^9m

Frame 2:

180 pixels/1.392x10^9m = 2025 pixels/1.56x10^10m

Then I applied my new numbers to this equation:

v = (s2 - s 1)/(t2 - t1)

v = (1.56x10^10 m - 4.79x10^9 m) / (216 minutes - 0 minutes)

v = (1.081X10^10 m) / (216) = 50,046,296 kilometers per minute or 834,104 m/s or 834.104 km/s

I'm not sure if I somehow screwed something up. Was I correct to scale up the numbers in for the second frame? I also measured from the picture in the book (rather than on the computer screen) in centimeters and I ended up 967 km/s...

5. Oct 20, 2014

### Matterwave

Did you make sure to measure starting from the same spot in both frames? Also, did you makes sure that you only measured the distance only in one direction? In other words, you didn't tilt the ruler up or down for the second measurement? Provided that you did this, it looks like your answer is roughly right.