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Frontiers and Controversies in Astrophysics: Lecture 24 Transcript
April 26, 2007
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Professor Charles Bailyn:
Just in time for our last class, we get this in yesterday's
New York Times
, and all
over the rest of the media. "New Planet Could Be Earthlike." You know, every time they get a new planet,
it's always Earthlike, but this one might really be true. It was found by the standard Doppler shift method,
which you guys all remember, and turns out to be the lowest mass planet that's been discovered in that
particular way.
And so, as a kind of final farewell calculation, I thought we'd check the
New York Times
' numbers on this.
You can go back and read the article for yourself. The information given there is that the orbital period is
about 13 days. The distance between the planet and its star â€“ that's the semimajor axis â€“ is given as 7
million miles. These are, of course, not the world's best set of units. The distance to the system is 20 light
years. That's really close. Somebody's quoted in the article as saying, you know, we could go there. Not so
much. And I looked up the apparent magnitude of this star, which turns out to be about 10.5.
And so, what can we do with information of this kind? Well, let's first put this into some kind of sane set of
units, here. Thirteen days islet's see. Three days is3.5 days would be 1% of a year. So, this is, like, 3 x 10
2
of a year. Seven million miles. That's something like 10 million kilometers, 10
7
kilometers, which is 10
10
meters. So, that's something like 7 x 10
2
Astronomical Units. And you know right away what to do with that,
or, at least, you will if you go back in your notes for a few months.
This is in good units to use
a
3
=
P
2
M
.
M
, then, shows up in solar masses. So let's figure out the mass of this
star. Let's see, (7 x 10
2
)
3
/ (3 x 10
2
)
2
. That'll be the mass in solar masses.
7 x 7 = 50, times 7 is 350, times 10
6
over 10 x 10
4
[(350 x 10
8
) / (10 x 10
4
)]. That's 35 x 10
2
.
.35, which is 1/3. Okay? So, this star is 1/3 of the mass of Sunperfectly respectable mass for a star to be.
And thenlet's see. Let's do something else. It's 20 lightyears away. Twenty light years, that's something like
6 parsecs. So, it's really nearby, as these things go. Not that you would want to take a spaceship and go there
or anything, but it is one of the closer stars. And then, we can do this thing. We can figure out the absolute
magnitude of this star. I've written down the apparent magnitude. So, that's that equation. Let's see.
5 log (6/10). Let's call that 2 x 3 x 10
1
, yeah? That's 6/10  five.
And then, you know, this thing about logs, if you multiply them inside the bracket, you can add them outside
the bracket. So, this is log of 2, plus log of 3, plus log of 10
1
. Log of 10
1
is 1, and the other two, I happen
to know. The log of 2 is .3. The log of 3 is .5. Minus one, that's 5 x .2 = 1. And so, 10.5 minus the absolute
magnitude would be 1.
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This note was uploaded on 02/06/2012 for the course ASTR 160 taught by Professor Charlesbailyn during the Spring '06 term at Yale.
 Spring '06
 CharlesBailyn

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