Apparent Magnitude of Brightness

Algebra Level 5

The apparent magnitude of a celestial object is a measure of its brightness as perceived by an observer on Earth. If the brightness of two celestial bodies is $F_1$ and $F_2$ , then their apparent magnitudes $m_1$ and $m_2$ are related as:

$m_2 - m_1 = 5 \log_{100} \left( \frac{F_1 }{F_2 }\right)$

Spica (the brightest star in Virgo) has apparent magnitude 1.04. If Sirius (the brightest star in the night sky) appears 10 times brighter then Spica, then what is the apparent magnitude of Sirius?

The answer is -1.46.

1 solution

Pranshu Gaba
Apr 21, 2016

Let Sirius be object 1 and Spica be object 2. We know $m_2 = 1.04$ . Since Sirius appears 10 times brighter than Spica, we also know $\dfrac{F_1}{F_2} = 10$ . We can now find $m_1$ .

$\begin{aligned} 1.04 - m_1 & = 5 \log_{100} (10) \\ 1.04 - m_1 &= 5 \times \frac{1}{2} \\ m_1 &= 1.04 - 2.5 \\ m_1 &= -1.46 \end{aligned}$

The apparent magnitude of Sirius is $\boxed{-1.46}$ . Note that the brighter the object, the lower is its apparent magnitude. $_\square$

To understand this scale better, here are apparent magnitudes of some celestial objects: (Source: lcogt.net )

- 26.7	Sun
-12.6	Full Moon
-4.4	Venus (at brightest)
-3.0	Mars (at brightest)
-1.46	Sirius (brightest star)
+3.0	Naked eye limit in an urban neighborhood
+5.5	Uranus (at brightest)
+6.0	Naked eye limit
+9.5	Faintest objects visible with binoculars
+13.7	Pluto (at brightest)
30	Faintest objects observable by the Hubble Space Telescope

Put 1.46 didn't know magnitude could be negative (thank term was swintching on the equation)

damien G - 5 years, 1 month ago

Isn't this a 2016 InPhO question?

Milind Blaze - 5 years, 1 month ago

A problem on apparent magnitudes had appeared in INAO 2016, but it was much more complicated.

Pranshu Gaba - 5 years, 1 month ago

Ah.... yes. It had involved calibration curves...?

Milind Blaze - 5 years, 1 month ago

Apparent Magnitude of Brightness

The answer is -1.46.

1 solution

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