Apparent magnitude of the Moon from Mercury?


by rtfirefly
Tags: apparent, magnitude, mercury, moon
rtfirefly
rtfirefly is offline
#1
Sep12-13, 01:38 AM
P: 2
I know there there has to be a simple answer to this, but I can't understand where it's at. Here is the question.

Find the apparent magnitude of the Moon [Earth's] as seen from Mercury. Assume Mercury is 0.52 AU from the Moon and that Mercury sees the Moon fully [it's a full moon].

Okay, here is where my thinking lays. I'm going to be using M = m + 5 - 5log(d), where M = absolute magnitude of the Moon, m = apparent magnitude of Moon, and d = distance from Moon to Mercury in parsecs = 2.52110^(-5) pc.

But the problem I run into is M - I don't know it. I know that the absolute magnitude is the apparent magnitude measured at 10pc, but then I run into not knowing m. I guess I could use the apparent magnitude from Earth (≈-12.6), but when I plug that into the equation using the 10pc I get back -12.6 for the absolute magnitude and that does not work for me. I can only presume I am missing a piece of the puzzle and I don't know where it is. Any help you can offer me would be appreciated.
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snorkack
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#2
Sep12-13, 07:31 AM
P: 353
Simplify it: skip the 10 pc.

The piece of puzzle which you have not mentioned is the distance of Moon from Earth. Since the distance of Mercury from Moon is conveniently in AU, which is of convenient magnitude, find out distance of full Moon from Earth in AU.

Then ratio of distances Moon-Earth and Moon-Mercury. Which you can then get logaritm from.
rtfirefly
rtfirefly is offline
#3
Sep12-13, 08:51 AM
P: 2
Right, thank you, but I was only given this one ready assignment that sort of talks about this stuff and the homework is based on it and did not read anything about this. The closest that it talks about ratios is be using the equation: m -n = 2.5log(Ln/Ln), where m and n are two apparent magnitudes, and Ln and Lm are their brightness. So I really don't know what equation I should be using to solve this problem. I realize that I may have to think outside the text (though in my experience teachers don't like you coming up with stuff from other places), but I did anyways and I still am confused as too what I am suppose to do. I think I'm thinking to hard now and everything at this point is becoming muddled.

snorkack
snorkack is offline
#4
Sep12-13, 11:31 AM
P: 353

Apparent magnitude of the Moon from Mercury?


Quote Quote by rtfirefly View Post
The closest that it talks about ratios is be using the equation: m -n = 2.5log(Ln/Ln), where m and n are two apparent magnitudes, and Ln and Lm are their brightness.
You have (Ln/Ln) in your equation - no Lm, and twice Ln, so that expression is always 1.

Just consider that the apparent luminosity decreases with inverse square of distance - for the same source like full Moon and viewed from the same direction,
Lm*dm2=Ln*dn2,
where dm is the distance where object has brightness Lm, and dn is distance where the object has brightness Ln.

Divide that expression with
Ln*dm2
and we get
(Lm/Ln)=dn2/dm2
Now take the logarithm
log (Lm/Ln)=log((dn/dm)2)=2log(dn/dm)
so
2,5 log(Lm/Ln)=5log(dn/dm)

Can you continue from that point?


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