# Please help with Astronomy questions **URL:** \ **Category:** Factual Questions\ **Created:** [January 31, 2004, 12:15am UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855 "2004-01-31T00:15:10Z")\ **Posts on this page:** 9\ **Page:** 2
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [January 31, 2004, 3:09am UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/21 "2004-01-31T03:09:07Z")
I am not saying you need to, but if you want to read it, I have written [a one-page summary of trig used in basic astronomy (PDF!)](http://cfa-www.harvard.edu/~cpilman/docs/trig.pdf). You may find it helpful. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [January 31, 2004, 3:12am UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/22 "2004-01-31T03:12:31Z")
> [@JohanDane](#): > > **Q.E.D**., Got it! > > So I was on the right track at first. 360-171=189 (Angle1). If I multiply that times 26000 (years), then divide the answer by 360 (Angle 2, being the whole circle), and get 13650. Correct? Bingo! Now you should be able to do the same problem using the more precise time of 25,800 years. ---
**Author:** ![JohanDane](https://avatars.discourse-cdn.com/v4/letter/j/e0b2c6/32.png) [@JohanDane](https://boards.straightdope.com/u/JohanDane)\ **Post date:** [January 31, 2004, 3:19am UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/23 "2004-01-31T03:19:31Z")
> [@Q.E.D.](#): > > Bingo! Now you should be able to do the same problem using the more precise time of 25,800 years. 13,545! Thanks alot! ---
**Author:** ![JohanDane](https://avatars.discourse-cdn.com/v4/letter/j/e0b2c6/32.png) [@JohanDane](https://boards.straightdope.com/u/JohanDane)\ **Post date:** [January 31, 2004, 4:04am UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/24 "2004-01-31T04:04:50Z")
> [@Q.E.D.](#): > > Bingo! Now you should be able to do the same problem using the more precise time of 25,800 years. QED One last thing. I have one last question to solve and i’ve been working at it for a while now: What is the distance to the star whose parallax angle is 5.585e-7. I’m using formula alpha=D/R or (after converted) R= 206264.8’’ times D/alpha. D = 1 AU, αlpa is the angle in radians, which is 5.585e-7, and R is the distance to the star in AU. 1 AU is equal to about 149,597,870 kilometers. Do I have everything I need to figure out the problem? ---
**Author:** ![so\_da\_ne](https://avatars.discourse-cdn.com/v4/letter/s/d9b06d/32.png) [@so\_da\_ne](https://boards.straightdope.com/u/so_da_ne)\ **Post date:** [January 31, 2004, 4:35am UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/25 "2004-01-31T04:35:16Z")
Let’s see if this helps. I converted the angle to arcsec because that’s what I’m more familiar with (thanks **wolf\_meister** ) If I understand correctly the angle you have been given is 5.585 ee -7 radians or 0.1152 arcsec. The diameter of the object is 1 AU, so the distance will also be given in AU. Using the small angle formula I get 206265(1)/0.1152" = 1.7905 ee 6 AU. I rounded to 206265 and rounded my answer but if this makes sense to you you can do it yourself a little more accurately. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [January 31, 2004, 5:26am UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/26 "2004-01-31T05:26:28Z")
There are two things going on here. First, there’s the equation that relates the angular size (theta) of an object to its actual size (D) and distance ®: D = R × theta[sub]radian[/sub] That’s the formula you want for the second question in the OP. There’s also the formula for parallax, which is what you want for this third question. This formula relates an object’s parallactic angle (alpha) to its distance ® and the size of the baseline (D): D = R × alpha[sub]radian[/sub] (Equation 1) You can see that it’s similar, and indeed both formulas are true for the same reason. However, theta and alpha do not measure the same thing. Usually, for either of the above equations, D and R have to be in the same units, and alpha has to be in radians. However, because it’s useful, astronomers have defined a certain unit called a parsec which makes the following equation true: D[sub]AU[/sub] = R[sub]parsec[/sub] × alpha[sub]arcsec[/sub] If you’re measuring the distance to stars, it’s assumed that the baseline is D = 1AU. That simplifies the formula to this: R[sub]parsec[/sub] = 1 / alpha[sub]arcsec[/sub] (Equation 2) Now, there are two ways to solve this problem. First, you could simply use Equation 1, which will give you R in AUs. Second, you could convert alpha into arcseconds, and use Equation 2 to get R in parsecs. ---
**Author:** ![Find\_Friends](https://avatars.discourse-cdn.com/v4/letter/f/c37758/32.png) [@Find\_Friends](https://boards.straightdope.com/u/Find_Friends)\ **Post date:** [January 31, 2004, 6:14pm UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/27 "2004-01-31T18:14:31Z")
> [@wolf\_meister](#): > > … So let’s convert that 1.04 yads into millimeters. Here’s a great converter for that: > [http://www.1728.com/convleng.htm](http://www.1728.com/convleng.htm) Yes, it is My site 😃 > … Great site, wm. I’ll bet you could convert c into furlongs per fortnight in your head! * * * True Blue Jack ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [January 31, 2004, 8:45pm UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/28 "2004-01-31T20:45:08Z")
Not to disrespect **wolf\_meister** ’s well-put-together site, but I like the interface of [Google calculator](http://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=c+in+furlongs%2Ffortnight&btnG=Google+Search) myself. 🙂 ---
**Author:** ![wolf\_meister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wolf_meister/32/15202_2.png) [@wolf\_meister](https://boards.straightdope.com/u/wolf_meister)\ **Post date:** [February 1, 2004, 2:01am UTC](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855/29 "2004-02-01T02:01:48Z")
**Achernar** I must admit Google’s calculator is impressive. Of course my calculators require much less typing and don’t require that “1 unit to 1 unit” “from to” conversion style. Then again, I don’t really consider Google a rival or anything. Heck, I’m proud to say a lot of my calculators finish #1 in Google searches. They are obviously doing something right. LOL 😃 [Previous page](https://boards.straightdope.com/t/please-help-with-astronomy-questions/226855.md?page=1)