# How to calc absolute magnitudes by comparing to the sun's luminosity

• wyse

#### wyse

hi guys,

(i) By comparing with the luminosity of the Sun, calculate the absolute magnitude of a 60W light bulb.
[The absolute magnitude of the Sun is +4:75; you may ignore colour differences
between the bulb and the Sun.]

(ii) The Hubble Space Telescope (HST) can detect objects as faint as 30th magnitude.
Calculate the maximum distance in kilometres at which the light bulb could be
detected by HST.

(i). can anyone give me any hints for this. i know the formula for two absolute magnitude and the formula for the luminsoity, but not sure how to use them, as we aren't told the luminosity of the bulb.

(ii). we know the absolute magnitude (1/30), and so this is measured at 10 pcs. so do they just want use to calculate 10 pc in metres?

thanks for any help.

(i) Hint: The unit of luminosity is Watt.

(ii) I am sure you have a formula relating apparent magnitude with distance and absolute magnitude. You know the absolute magnitude of the light bulb from (i) and you know that when the HST can just observe the light bulb, the light bulb has an apparent magnitude of 30. Calculate d.

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(i) Hint: The unit of luminosity is Watt.

(ii) I am sure you have a formula relating apparent magnitude with distance and absolute magnitude. You know the absolute magnitude of the light bulb from (i) and you know that when the HST can just observe the light bulb, the light bulb has an apparent magnitude of 30. Calculate d.

(i) so since luminosity is related to absolute magnitude, we have $\frac{L_{sun}}{L_{bulb}} = \frac{M_{sun}}{M_{bulb} }$
so we have $\frac{3.8 e 10^{30}}{80}= \frac{4.75}{M_bulb}$
which we reaarange to get $7.5 e 10^{-29}$

(ii) so we need to use the forumla $m - M = 5 log (d/10pcs)$. it's just that i get the answer 10.1pcs and i expected something a bit smaller.

thanks again

The equation you use for (ii) is correct, however I am not familiar with the one you're using for (i). The formula you should use is $M_1-M_2=-2.5 \log_{10} \frac{L_1}{L_2}$. Yes spotting an 80W light bulb at 10pc seems a bit extreme.

Also you know that the magnitude scale is a bit weird, the lower the number the brighter the source. The absolute magnitude of the sun is +4.75. The absolute magnitude you calculated for the light bulb is of the order 10^-30. This is smaller than +4.75, so according to your calculations the bulb is a lot brighter than the sun! This cannot be right of course, so that should be your first hint to why the distance seems a bit off.

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You may find it easier to use the 1/R^2 relationship to obtain the correct distance. It is definitely less than 10 parsecs. Start by calculating the distance at which the luminosity of an 80W light bulb equals that of a magnitude 1 star.

This assumption in the problem:

you may ignore colour differences
between the bulb and the Sun

is very strange as a typical 60 W lightbulb radiates most of its energy in the infrared. I think you now have 15 Watt lighbulbs that radiate most off the energy in the visible range that give off more light than a 60 Watt traditional lightbulb.

is very strange as a typical 60 W lightbulb radiates most of its energy in the infrared.

So does the sun. Well, almost. A light bulb is about the same color as an M-class star.

Hi All

Seems someone has forgotten the inverse square law. Absolute magnitudes are based on seeing at 10 pc (as you know) and you know how far that is in metres. The 60 W from the light spreads over an area of 4π(10 pc)² as does the Sun's. To get the relative magnitude of the two is just 2.5*log(Lo/L) then add 4.75 to make it absolute. The Sun puts out about ~48% of its energy as visible light and a lightbulb about ~5% but that's quibbling. Their bolometric luminosity is more useful thus the ratio is (3.847E+26/60) = 6.412E+24 (units cancel) and 2.5*log(that) is 62 and so the absolute magnitude of the bulb is 66.75.

If the Hubble can see the bulb at magnitude 30 then that means it has to be 10^[(66.75-30)/5] times closer than 10 pc. But I'll let you figure out why.

It is more fun to compute this for a 5 milliwatt green laser pointer.

It is more fun to compute this for a 5 milliwatt green laser pointer.

Sorry. I only deal in terawatt Death-Beams myself. If it can't slice Ceres in two, then it's not worth powering up in the morning.