Calculating the distance between a "5 cent coin sized" sun to Proxima Centauri

[shelltree]

After reading this http://www.wired.com/wiredscience/2011/12/universe-size/ posted by Jon in a previous topic in this section, I became interested in trying to calculate the distance between the Sun and Proxima Centauri, using a 5 cent piece with a 19.41mm width as a scale.

Now I know NOTHING about Maths, I failed high school maths so it all gets a bit confusing sometimes but I was determined to see if I could do it anyway.

I will put my very rough calculations below. If anyone could tell me how completely wrong I am, that would be very helpful :P I think I have gone completely off the track with how I tried to calculate it and if someone could please help me solve this, I would be forever greatful :)

40000000000000 km to Proxima Centauri

5 cent piece = 19.41mm

1km = 1000000mm so 40000000000000km = 40000000000000000000mm?

1,391,980 km = width of sun

Width of sun in millimetres = 13919800000000mm

So, 19.41mm = 13919800000000mm

40000000000000000000 divided by 13919800000000 = 2873604.5058118651130044971910516

19.41 x 2873604.5058118651130044971910516 = 55776663.457808301843417290478312

55776663.457808301843417290478312mm = 55.7766634578km

55.7766634578 - 0.0000001941 = 55.7766632637km

So, the distance between the Sun and Proxima Centauri, using the scale of a 5 cent piece as the Sun (19.41mm) = 55.7766634578km

I'm crazy, I know :screwy: And have got this all completely wrong but would appreciate some help :thanx:

[sjastro]

Mark 2

1 light year = 9.46073 x 10^18mm
Distance Sun to Proxima Centauri = 4.2 light years =4.2 X 9.46073 x 10^18mm
=3.97350 X 10^19mm

Scale 5 cent coin to Sun's diameter (in mm)= 19.41/1.391X10^12= 1.395X 10^-11.

Distance at this scale= 3.97350 X 10^19 X 1.395 X 10^-11
= 5.54304 X 10^8 mm
= 5.54304 X 10^2 km
= 554 km.

About 500 km.

Regards

Steven

[CraigS]

Yep Shelley; (Ignore this post if you're already aware …)

Further to Steven's post, the notation he's used is scientific notation:

10^0 = 1
10^1 = 10
10^2 = 100
10^3 = 1000
… etc, etc.
10^-1 = 0.1
10^-2 = 0.01
10^-3 = 0.001
… etc, etc

If you count up the zeros at the end of the big numbers, you'll find its the same count as the number following the '^' above. Similar for the little numbers .. the number of digits after the decimal point is the same as the number following the minus sign. (Just makes it a lot easier to work with the big and small numbers).

Cheers

[Robh]

???
1 light year ~ 9.46073 x 10^12km = 9.46073 x 10^18mm

Regards, Rob

[sjastro]

OK Rob you nailed me.:(

I used a software program used "Convert".
Next time I'll use the Internet for conversions.

Alterations included in my first post Shelley.

Regards

Steven

[ballaratdragons]

That is approx half way from Sydney to Melbourne or Brisbane :)

At the size of a 5c coin imagine how dim our star would be at that distance!

[sjastro]

To confirm these calculations are now correct.

Using the 5c scale the Sun Earth distance is around 2 metres.
Two metres is now our new AU.

In terms of AUs the distance of Proxima Centauri from the Sun is:
(554X1000)/2= 277,000 AU.

According to NASA



So the calculation is fairly accurate.

Regards

Steven

[Robh]

Shelley,

Here is your error. One too many zeros for the Sun in mms.
Should read 1 391 980 000 000mm.

This will increase your calculated distance by a factor of 10 to roughly 558km, which is in pretty good agreement with Steven's result.

Regards, Rob

[renormalised]

Shelley, just to add more to your confusion....if you scaled the distance between Proxima and the Sun to 1 mile, the distance across our Galaxy would be approximately equal to the circumference of the Earth, roughly 25000 miles (if you take the accepted textbook diameter of the Galaxy, which in fact is quite a bit short of the latest measurements).

And, if you then decided to use the same scale for the rest of the Universe, the most distant objects we can see (at their co-moving distance, around 42 billion light years), would be about the same distance away from us as the diameter of the Solar System...about 11.25 billion miles :)

It means, in terms of actual AU's, the Universe is approx' 3.12 x 10^15 times the distance in its radius as the distance between the Earth and the Sun :)

In other words, it's HUGE :)

[ballaratdragons]

So just to clarify (because the answer in here has been changed), is the answer now 500 km?

[CraigS]

554 kms … which is approximately 500kms.

Cheers

[renormalised]

That's an awfully loose approximate!!!:):P

[sjastro]

Still it's a damn side better than the original calculation of 940,000 km.:shrug:

Have absolutely no idea how according to "Convert"
1 light year = 1.68029 X 10^25mm???????

It works now.

Even doing a google search on this conversion factor doesn't provide any clues.

Regards

Steven

[renormalised]

True....how did it end up that big?? Convert must've stuffed up big to end up with that answer. In any case, you could round up the figures and do these calcs in your head. They're not that difficult.

Whoever wrote "Convert" slipped up in their conversions.

[shelltree]

Phewf! Well, that took awhile but I finally managed to understand it all. Within reason...

Here are my calculations the second time around, using Steven's amazing helpful equations! It took me awhile to figure out what some of the figures meant but google is my friend :D Craig, thanks for the help with scientific notations, seems easy enough to understand, for now :P A big thanks to Rob for letting me know I was kind of slightly close but going about it all the wrong way haha and no need for the huge numbers! And Carl, well you never cease to teach and boggle my mind :) One thing at a time, my brain hurts! :lol:

Okay, so here goes!

1 light year = 9.4605284 x 10^12km
= 9.4605284 x 10^18mm

Distance to Proxima Centauri = 4.2 light years

= 4.2 x 9.4605284 x 10^18
= 3.973421928 x 10^19
= 19.41/1.391^12 (diametre of the Sun in mm I found out, thanks google!)
= 1.391^-11 (so does that mean if the answer were 0.03... instead of 0.3... then it would be 1.391^-10??? Wasn't sure how this works)

3.9734 x 10^19 x 1.391^-11 =
5.5269994e x 10^8 =
5.54 x 10^2 =
554km ;)

I thought I'd also try it for another star, just to see I've got the hang of it so bear with me :)

I'm going to use Aldebaran! I'm working on it now, though not sure if my answer will be correct...

[shelltree]

Here goes, I have a feeling this is totally and utterly wrong, oh the shame!

1 light year = 9.4605284 x 10^18mm
Aldebaran = 65.1 million light years away

65.1 x 9.4605284 x 10^18
= 6.159 x 10^11

Diametre of Aldebaran is approx. 61 million km

19.41/6.1 x 10^12mm
= 3.18 x 10^12

6.159 x 10^11 x 3.18 x 10^12
= 1.958 x 10^10mm
= 1.958 x 10^4 km
= 19580km

:question:

[shelltree]

:lol: that is terrible, oh my goodness!!! Hahaha! I will recalculate tomorrow, what a silly mistake :P

[renormalised]

You're a factor of 2 out in your answer...

1 light year = 9.4605284 x 10^18mm
Aldebaran = 65.1 light years away

65.1 x 9.4605284 x 10^18
= 6.159 x 10^20 (not 11)

Diametre of Aldebaran is approx. 61 million km

19.41/6.1 x 10^12mm
= 3.18 x 10^-13 (not 12)

6.159 x 10^20 x 3.18 x 10^-13
= 1.958 x 10^8mm (195,856,200mm)
= 1.9585 x 10^2 km
= 195.85km

[renormalised]

Try...

Rigel.....810ly and 85 solar diameters
Betelgeuse.....640ly and 900 diameters
Sirius....8.64ly and 1.71 diameters
Gliese 581....20ly and 0.36 diameters
Alpha Cen A.....4.395ly and 1.23 diameters

That should keep you busy :)

[shelltree]

It kept me very busy in my lunch break :P

These are the results I got, though again they are very likely a bit off :screwy:

Rigel

810 light years and 85 solar diametres

810 x 9.4605284 x 10^18
= 7.66 x 10^12

85 x 1.391 x 10^12
= 1.182 x 10^14



19.41/1.182 x 10^14
= 1.642 x 10^-17

7.66 x 10^12 x 1.642 x 10^-17
= 1.257 x 10^14
= 1.257 x 10^8
= 1.25 x 10^2
= 125km

Betelguese

640 light years and 900 solar masses

640 x 9.4605284 x 10^18
= 6.05 x 10^12

900 x 1.391 x 10^12
= 1.251 x 10^7

19.41/1.251 x 10^7
= 1.55 x 10^-10

6.05 x 10^12 x 1.55 x 10^-10
= 9.377 x 10^13
= 9.377 x 10^8
= 9.377 x 10^2
= 937.7km

Sirius

8.64 light years and 1.71 solar diametres

8.64 x 9.4605284 x 10^18
= 8.173 x 10^13

1.71 x 1.391 x 10^12
= 2.378 x 10^12

19.41/2.378 x 10^12
= 8.162 x 10^-13

8.173 x 10^13 x 8.162 x 10^-13
= 6.67 x 10^17
= 6.67 x 10^11
= 6.67 x 10^2
= 667km

Gliese 581

20 light years and 0.36 solar diametres

20 x 9.4605284 x 10^18
= 1.892 x 10^11

0.36 x 1.391 x 10^12
= 5.007 x 10^11

19.41/1.892 x 10^11
= 1.025 x 10^-13

5.007 x 10^11 x 1.025 x 10^-13
= 5.132 x 10^12
= 5.132 x 10^6
= 5.132 x 10^2
= 513.2km

Alpha Cen A

4.395 light years and 1.23 solar diametres

4.395 x 9.4605284 x 10^18
= 4.157 x 10^13

1.23 x 1.391 x 10^12
= 1.710 x 10^12

19.41/1.710 x 10^12
= 1.135 x 10^-14

4.157 x 10^13 x 1.135 x 10^-14
= 4.718 x 10^14
= 4.718 x 10^8
= 4.718 x 10^2
= 471.8km

:shrug::help: ...maybe? I think I'm on the right track but so far I seem to get something wrong :lol:

[renormalised]

You're way off again...check all your figures.

How do you get 810 x 9.4605284 x 10^18 = 7.66 x 10^12, when it should be 7.66 x 10^21????. Unless you're transposing the exponent for some weird reason!!!! :):P

Check all your figures carefully.

[renormalised]

Ah...I think I see what you're doing. You're converting that first figure to kilometres in your head and then writing it down as the answer. You're only confusing yourself and others by doing this....and not getting the right answers.

For instance, you got this for Alpha Cen A...

4.395 x 9.4605284 x 10^18 (assuming your answer was 4.158 x 10^19mm)
= 4.158 x 10^13 (km, not mm)

That answer is correct, but leave it at 4.158 x 10^19 (mm), which will give you the correct final answer.

But even still, when you get the calc' wrong in the first place...

(Rigel)

810 x 9.4605284 x 10^18
= 7.66 x 10^12 (should be 7.66 x 10^21mm)

You get 7.66 x 10^15km (not 7.66 x 10^12km)

What you've done here is multiplied it out but for some reason used 18 as the exponent instead of the correct answer, 21, and then did your conversion (which is unnecessary).