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Solid Bismuth
2008-Oct-26, 03:46 AM
You people have helped me before with some tricky questions and I hope you won't fail me now. Don't let me down! ;)

Anyway, I was wondering if someone could work out the apparent magnitude of the Sun at 100 light years away.

I really appreciate this.

Thanks!

Neverfly
2008-Oct-26, 04:07 AM
Launch Window might know (http://www.bautforum.com/off-topic-babbling/17877-luminosity-apparent-magnitude-formula-sun.html);)

Solid Bismuth
2008-Oct-26, 04:21 AM
Thanks a lot but, unfortunately, I am not that great with math.

Anyone willing to help a sweet young boy out?
:lol:

Solid Bismuth
2008-Oct-26, 04:21 AM
Also, I'm changing the distance to 100 light years.

loglo
2008-Oct-26, 04:42 AM
Here is my rough calculation, knowing my math ability it will need checking. :)

Absolute visual magnitude of Sun (from wiki) = 4.83 =
apparent magnitude of Sun at 32.6 light years or 10 parsercs).
So since 150 light years ~ 4.6 x 32.6ly
and brightness is inverse proportional to distance^2 then
brightness of Sun at 150ly ~ 4.6^2 ~ 21 times less and since
a ~21 factor of brightness is about 3.3 magnitudes then
the apparent magnitude of the Sun at 150ly is 4.83 - 3.3 ~ 1.5.

formulaterp
2008-Oct-26, 09:46 AM
then
the apparent magnitude of the Sun at 150ly is 4.83 - 3.3 ~ 1.5.

If an object is dimmer, it's apparent magnitude gets larger in value, not smaller.

Try flipping the (-) into a (+).

Eroica
2008-Oct-26, 10:04 AM
M = m - 5 logd + 5 - A

This is the basic formula. M is the absolute visual magnitude, m is the apparent visual magnitude, d is the distance in parsecs, and A is the interstellar absorption in magnitudes (ignore this term if you're not taking this into account).

100 light-years is about 30.67 pc.

4.83 = m - 5 log(30.67) + 5
=> m = 7.26.

Say 7 to one significant figure.

Solid Bismuth
2008-Oct-26, 02:19 PM
M = m - 5 logd + 5 - A

This is the basic formula. M is the absolute visual magnitude, m is the apparent visual magnitude, d is the distance in parsecs, and A is the interstellar absorption in magnitudes (ignore this term if you're not taking this into account).

100 light-years is about 30.67 pc.

4.83 = m - 5 log(30.67) + 5
=> m = 7.26.

Say 7 to one significant figure.

Thank you for simplifying that for me.

tony873004
2008-Oct-26, 06:22 PM
The 6th calculator on this page does all the math and unit conversions for you.

http://orbitsimulator.com/formulas/

Using the same inputs as Eroica, I also get 7.26. So from 100 light years, the Sun would only be a binocular object.

loglo
2008-Oct-27, 10:44 AM
If an object is dimmer, it's apparent magnitude gets larger in value, not smaller.

Try flipping the (-) into a (+).

Blast!! forgot Loglo's Law:-
If you haven't checked your math at least 5 times then you have a sign wrong!!

Argos
2008-Oct-27, 02:02 PM
With a blistering 32 C of a summer that got early, there´s nothing wonderful about the Sun. In days like these I wish it were a red star [or I lived in Patagonia].