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Finding the Radius of Venus' Orbit


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Hello Folks!

Lots of people seemed to like the activities I posted for download - fun things to do with your telescope and binoculars. :)

I've had several comments and some PM's asking for something a bit meatier, so to speak. This exercise is based on a nice essay in John Clark's book: Measure Solar System Objects and Their Movements for Yourself, and is designed to be done with Stellarium, or similar planetarium software. I particularly did this one for the computer because I've read so many posts about "the clouds.... Oh, the clouds!", so you can do this one indoors whatever the weather!

You could easily do this exercise with a small scope, of course, but the next maximum elongation of Venus won't be for several months yet!

Parents - this activity would make a suitable science fair activity for Jr., and if your kid is able to do this (with or without your help! :( ), have them take it in to school - I'm betting that many science teachers would be happy to give some credit for solid science work done on the side at home.

Teachers - I do this activity over about 3 class days for high school kids. Day 1 is reading the lab, paying attention to the vocabulary words (always bold/italic in the text), and the concepts involved. We also take some time to review the basic geometry involved. Day 2 is Computer Lab day (hooray!), kids use Stellarium and gather their data, make sketches. Day 3 is calculation and analysis day. We finish the activity together, kids write a short "What I learned from this lesson..." essay and turn it in.

C'mon folks - they say that the love of learning is dead, and that we don't care to explore. They say that ignorance is King, and there is a sucker born every minute. I say different! I say there are people who wish to explore and learn and figure out how the Universe works. We want to understand the powerful ideas of our intellectual ancestors - not just admire their statues and portraits in a museum. We are astronomers! We participate in the science!

I'd like to see how many people are willing to dive off the deep end with a little science and give this a whirl. I'll be here to answer questions and help out if you need it - and I'd love to hear how you do! And if that doesn't motivate you, think of the poor slobs who are paying upwards of $500 a semester to do this sort of thing with me, and you lot are getting it for Free! ;)

I Love this stuff!

Dan

Lab 13 - Venus' Orbit - Stellarium.pdf

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I have been imaging Venus at intervals since last July, and it had occured to me that it would be quite easy to get the radius of Venus' orbit from the apparent size of the planet. Just haven't got round to doing it yet. This may encourage me to get on with it.

My images can be seen at Planets

Dave

Edited by Dave Smith
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Wow Dave, your shots are FANTASTIC! I would encourage everyone to follow the link in Dave's post and take a look at these photos! :)

I hope you find my little activity useful. If you should decide to do the activity in the live sky, you simply substitute the observation of Venus with a scope for those off Stellarium and there you are!

Let me know how you get on!

Dan

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Here is what I get. Using your method from the angle between Venus and the Sun, which I didn't measure but recorded from SkyMapPro, 46deg 10 sec gives a value of 0.72 AU (cf average value 0.73) Good agreement but did not involve any measurements by myself.

I have attempted to estimate the radius of the orbit from the apparent change in size.

I printed off my images (reversed to save ink) and measured the diameters of the Venus images when the phase was 50% and have assumed that the distance to Venus is then 1.0 AU (in fact not quite right) and the largest image when I have made the assumption that Venus is at it's closest distance to the Earth (again not quite right).

The measurements were 66mm and 24mm. As the apparent size is inversely proportional to the distance it follows that

66/24 = R/(R-r) where R is the radius of the Earth's orbit (= 1.0 AU) and r = radius of Venus' orbit.

This gives r = 0.64 AU

Considering the assumptions I have made, I'm quite pleased with this result.

Dave

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This doesn't work too good on an iPhone, I think both of your pictures where over lapping, Dan... Oh well. I'm gonna print these out at work tomorrow and read there :)

Dazz

Oh Dave, your pictures are just epic. :•D hope I can get good pictures like that on my SPC900 when I get it all sorted out :(

Edited by Dazzio
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My publisher (Funderstanding.com) has asked me to do a video presentation for this, so it looks like I'll be making a video for this one way or the other.

I'll let you folks know when it's ready.

Dan

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That sounds cool Dan.

Right I've downloaded the "lab" and printed it off, reading it now in my lunch break. However just to make sure we are on the right page. When you say "right triangle" you actually mean a "right angled triangle"?? I know it sounds pretty simple in all honesty, I just didn't want to assume the wrong idea about it. I've never heard it put that way.

Dazz

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That sounds cool Dan.

Right I've downloaded the "lab" and printed it off, reading it now in my lunch break. However just to make sure we are on the right page. When you say "right triangle" you actually mean a "right angled triangle"?? I know it sounds pretty simple in all honesty, I just didn't want to assume the wrong idea about it. I've never heard it put that way.

Dazz

Yes, the classic 90-degree triangle from geometry class. In the States, we call it a "right triangle" in all the text books.

Dan

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Yes, the classic 90-degree triangle from geometry class. In the States, we call it a "right triangle" in all the text books.

Good good, just wanted to make sure it was the same thing. We always called them a "right angle" or the full "right angled triangle". :)

Dazz

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