How do i figure it out: how far is a star with an appearent magnitude of +2 and an absolute magnitude of +4?

Is there a formula for that?





I think the answer is 12.7 light years but i don't know how to get that answer.


thanks!How do i figure it out: how far is a star with an appearent magnitude of +2 and an absolute magnitude of +4?
The absolute Magnitude (we always write it with a capital M to remind us not to get mixed up), is the magnitude the star would appear to be if it were placed at 10 parsecs (32.6163 light-years) from us.





The apparent magnitude indicates how brilliant the star appears at its true distance.





The intensity of light decreases as the square of the distance (push a star three times further, it will appear nine times fainter).





The magnitude scale is geometric. One step in the magnitude sacel corresponds to a ratio of approx. 2.512 (it is the fifth root of 100).


A star of m= 2 is 2.512 times brighter than a star of m=3.


A star of m= 2 is 6.310 times brighter than a star of m=4


...


A star of m= 2 is 100 times brighter than a star of m=7.





Here, we have M=+4 at 10 parsecs and m=+2 at distance X.


+2 is brighter than +4 (the magnitude scale runs backwards with bigger number meaning fainter stars).


+2 is 6.310 times brighter than +4.


Distance X must be 10 parsecs, divided by the square root of 6.310.


X = 10/SQRT(6.310) = 10/2.512 = 3.98 parsecs.


3.98 parsecs = 13 light years.





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The formula to calculate distance based on magnitudes is called the ';distance modulus'; and it involves logarithm. This is good because powers and roots in normal calculations become simple multiplications and divisions in logarithmic calculations.





m 鈭?M = 5*log_10(d / 10pc)





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pc means parsec, a distance unit based on the distance at which the parallax (based on the orbital radius of Earth's orbit around the Sun) is exactly 1 second of angle.





log_10 means; logarithm in base 10 (a.k.a. common logs, not ';natural'; logs).





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M = +4, m = +2





m 鈭?M = 5*log_10(d / 10pc)





2 - 4 = 5*log_10(d / 10 pc)


-2/5 = log_10(d / 10 pc)





to get rid of the log_10, make both sides the power of 10.


(This operation is called: taking the ';antilog';) On many calculators, this is done by pressing ';inverse'; then ';log';. It is also the same as the key marked 10^x.





If a = b, then 10^a = 10^b





Takint the antilog of a log cancels the log and leaves only what is in the bracket:





10^[log_10(x)] = x


antilog[log_10(x)] = x





so, we now have:


-2/5 = log_10(d / 10 pc)





10^(-2/5) = d/10 pc


1/10^(2/5) = d/10





10^(2/5) is the same as (10^2)^(1/5) which is the say to write: the fifth root of 100. That is 2.512 (approximately)





1/2.512 = d/10


10/2.512 = d = 3.981 parsecs





1 parsec = 3.261631 light-years


therefore


d = 3.981 * 3.261631 = 12.9848 light-years.





13 light years


It would be bold to pretend to a lot of accuracy, unless you had determined that m really is +2.0000 and M = +4.0000 (and not +4.0027). Magnitude tables are rarely that precise.How do i figure it out: how far is a star with an appearent magnitude of +2 and an absolute magnitude of +4?
Abs mag is how bright if it were 10 parsecs or 32.6 ly from Earth. Yours is +4 abs, but +2 app, so it is 2 mags brighter which means it is closer than 32.6 ly. 2 mag difference=6.2 times brighter. Brightness increases by inverse square of distance.

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