Friday, September 23, 2022

Stellar Luminosity Problem Solutions

(1) The apparent V-band (filter) magnitude of a star is 8.72, and it requires a bolometric correction of -0.48. Find the apparent bolometric magnitude of the star. (Hint: Apparent bolometric magnitudes are obtained in an analogous way to the absolute forms)

Solution:

1) mV= 8.72

m bol * = mV + B.C. = 8.72 + (-0.48) = 8.24

Hence, m bol *  = +8.24

(2) A star has a color index (B - V) of +1.0 and its apparent B-band magnitude is 6.4. The corresponding bolometric correction is - 0.5. Find the apparent bolometric magnitude of the star.

Solution:

mB = 6.4 and (B - V) = (mB - mV) = 1.0

Then: mV = mB - 1.0 = 6.4 - 1.0 = 5.4

m bol *  = mV + B.C.

m bol * = 5.4 + (-0.5) = +4.9

(3) The star Alhena in the constellation Gemini is at a distance of 30 pc. If it has (B - V) = 0.00, and mV = +1.93, find the apparent B-band magnitude, mB.

Also find the absolute visual magnitude and the absolute bolometric magnitude of the star.

Find the luminosity of Alhena in terms of the solar value.

Solution:

(B - V) = mB - mV = 0.00

Then if mV = +1.93, then mB = +1.93.

The absolute visual magnitude MV = mV - 5 log D + 5

D = 30 pc, so:

MV = 1.93 + 5 log (30) + 5 = 1.93 - 5(1.477) + 5

MV = 1.93 - 7.39 + 5 = -0.46

The absolute bolometric magnitude is:

Mbol* = MV + B.C.

where B.C. = 0.72 (from the table) so:

Mbol* = -0.46 + (-0.72) = -1.18

The Luminosity L' is obtained using:

Log (L'/L) = 0.4 (Mbol - Mbol*)

where Mbol (Sun's) = 4.63

Log (L'/L) = 0.4 (4.63 - (-1.18)) = 0.4 (5.81) = 2.32

Antilog (2.32) = 209 approx.

L'/L = 209 and L' = 209 L   or  209 times the solar luminosity.