Saturday, September 24, 2011

Solutions to Intermediate Astronomy (8)

The Problems again:

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:

m_V= 8.72

m_bol* = m_V + 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:

m_B = 6.4 and (B - V) = (m_B - m_V) = 1.0

Then: m_V = m_B - 1.0 = 6.4 - 1.0 = 5.4

m_bol* = m_V + 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 m_V = +1.93, find the apparent B-band magnitude. 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) = m_B - m_V = 0.00

Then if m_V = +1.93, then m_B = +1.93.

The absolute visual magnitude M_V = m_V - 5 log D + 5

D = 30 pc, so:

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

M_V = 1.93 - 7.39 + 5 = -0.46

The absolute bolometric magnitude is:

M_bol* = M_V + B.C.

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

M_bol = -0.46 + (-0.72) = -1.18

The Luminosity L' is obtained using:

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

where M_bol (Sun's) = 4.63

So:

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

Antilog (2.32) = 209 approx.

So:

L'/L = 209 and L' = 209 L

or 209 times the solar luminosity.

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