Introduction to Astrophysics - Charles Bailyn - 24 of 24

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 Earth-like." You know, every time they get a new planet, it's always Earth-like, 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 – that's the semi-major axis – 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 is — let's see. Three days is — 3.5 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, 107 kilometers, which is 1010 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 a3 = P2 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 Sun — perfectly respectable mass for a star to be.

And then — let's see. Let's do something else. It's 20 light-years 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.

Let's see. How's this going to work? This has to go over there. That has to come over here. Absolute magnitude, 11.5. Now, you may remember that the Sun has a magnitude — an absolute magnitude of around 5. So, this is much, much fainter than the Sun. That's good, because it's less massive than the Sun, and low mass stars get faint really quickly. How much fainter? Well, we know how to do that, right?

Let's see, that looks — in the easiest format, it looks like this: Mstar - Msun = bstar / bsun. This a minus sign out here in front? Yes. So, that's 10-2/5 (11.5 - 5), which is equal to 10(-2/5)(6.5).