# I Calculating planet's radius using a star's luminosity

1. Feb 10, 2017

### JohnnyGui

Hello,

I’ve been watching a lecture about how in astronomy one would be able to calculate the radius of 2 stars by measuring the velocity of the orbit and then measure the time how long the luminosity of a star dims when one star is behind the other.

After thinking this a bit through, I came up with a scenario of a planet orbiting a star and how to formulate the radius of the planet if we know the apparent luminosity of a star. This is just merely a scenario without taking influencing factors into account and I want to know if the way I’m reasoning this is logical or not.

Suppose we measure the apparent luminosity of a star $L$ and we wait until the orbiting planet is in front of the star right in the middle with respect to us. Since the planet obscures a part of a star, the apparent $L$ of that star would dim a bit by a factor, let us call that dimmed luminosity Ldim. We can say that this factor is caused by a decrease in area of that star. Since luminosity is proportional to the area, we can say that:

Here, Rdim obviously doesn’t mean that the star has reduced in size, but it just states that the area has decreased by a factor because of the planet. We've yet to reformulate Rdim.

We can imagine that the reduction in area of the star is caused by a substraction of the circular area of the planet that is in front of the star, since the obscuring is caused by the planet’s circle area and not its total spherical area from our perspective. Of course the distance of the planet from the star must not be too large otherwise we’d be overestimating the radius of the planet.

So, we can say that the factor by which the star has reduced in its total area is equal to:

In which the small $r$ is the radius of the planet.

Is this reasoning correct? If so, what might be outside influencing factors that make this calculation impossible?

Last edited: Feb 10, 2017
2. Feb 10, 2017

### sophiecentaur

Afaik, one of the current methods of detecting exoplanets is based on just that. Astronomers seem to have high confidence in the method, too. They make all sorts of predictions about the nature of those exoplanets from this data.
Google "detecting exoplanets". There's loads of interesting stuff available from a search along those lines.

3. Feb 10, 2017

### Bandersnatch

In case that's not clear after reading what sophiecentaur suggested, one shouldn't use stellar total luminosity and its total area - but rather, the observed flux and area of the stellar disc. After all, in the transit method one observes and sees the dimming of only the disc, not the whole area of the star.
It's easy to see if we consider a 'planet' with the radius equal to that of the star. It should reduce the flux to 0 (star fully obscured), while according to the equation presented above it does so only by 25%.

Here, I'll drop a couple relevant links as well:
https://www.planethunters.org/
http://astro.unl.edu/classaction/animations/extrasolarplanets/transitsimulator.html

4. Feb 10, 2017

### JohnnyGui

Thanks for the replies

Indeed, one could only measure the flux of the stellar disc after all. However, if we know the relationship between the flux of a stellar disc and the flux of the total spherical area (we know the relationship of spherical area and total flux) and we convert our measured dimmed flux based on the stellar disc into how much the total flux has dimmed, can't one say that my mentioned formula shows how much of an impact an eclips has from the perspective of the total luminosity/flux of the star? So if a planet has the same radius as the star and it completely obscures it, doesn't this mean that it has a 25% impact on the total luminosity of the star?

Also, thanks for the link, I'm going to check them.

5. Feb 10, 2017

### Bandersnatch

No, because it's not blocking 25% of the luminosity.

6. Feb 10, 2017

### JohnnyGui

I may be missing something obvious here but how much of the total luminosity will be obscured then? I'm not talking with respect to the flux from the stellar disc but with respect to the total power of a star when a circular area of pi * r^2 is covered on it.

7. Feb 11, 2017

### sophiecentaur

The relevant ratio is based on the disc areas and not the sphere areas. The observed brightness (relates to Star Magnitude) is not important. [edit - i.e. absolute maximum brightiness]

Last edited: Feb 11, 2017
8. Feb 11, 2017

### JohnnyGui

Yes I understand that now. But I merely wanted to understand the reduction with respect to the total luminosity or star magnitude. Shouldn't that be a 25% reduction if a circular area with the same radius of the star covered the star?

9. Feb 11, 2017

### Bandersnatch

Say, you put the planet at 1 AU from the star. Luminosity is equal to the total flux escaping from an enclosed surface, here - a sphere of radius 1 AU. The proportion of luminosity blocked by the planet will be equal to the area of the planetary disc divided by the area of that 1 AU sphere (and not of the stellar surface). The same planet at 10 AU will block 1/100th as much of the luminosity. The proportion is orbital distance dependent.
In other words, the planet can only block as much of the stellar output as it is receiving, and that goes down with its distance from the star.

10. Feb 11, 2017

### JohnnyGui

Makes sense. But isn't this basically what my formula mentions if you choose $R$ to be any distance from the star to the planet? Again, with respect to the total luminosity of a star.

So, if $R$ happens to be only the radius of the star, then the planet is "touching" the surface of the star. In that case, if the planet's radius is equal to the star's radius ($r$=$R$) then the percentage by which the total luminosity of the star decreases is around 25% (a bit more since the planetary disc area is not "folded" on the star's surface).

11. Feb 11, 2017

### Bandersnatch

1. If the planet is 'touching' the star, then the radius R is not equal to the planet=star radius (r). It's 2r. It'd be R=r only if the planet was colocated with the star (and what that would even mean?).
2. I'm not sure why you want to salvage this luminosity thing. It's not like it's an observable - it's always derived.

12. Feb 11, 2017

### JohnnyGui

1. Ah of course, forgot to take that into account. So an estimated 25% decrease in total luminosity is only the case if a huge flat disc with the same radius as the star is touching the star. Right?

2. I'm not focusing on what's observable, just trying to understand the relationship between the two entities.

13. Feb 11, 2017

### sophiecentaur

The actual spacing between the star and planet will hardly make a difference at the observation distance. Surely all you need is the visible area of the star and the projected area of the planet on the star.
As you say, Luminosity is not relevant.

14. Feb 11, 2017

### Bandersnatch

For the observed flux, yes. But the OP was asking for the proportional contribution to the total luminosity.

Yeah. With that caveat you mentioned, where the disc is not 'folded' into a partially-spherical shape.

15. Feb 11, 2017

### sophiecentaur

Do the sums. What measurable difference would there be with the planet 'touching' the star or separated by a tiny fraction of the distance from us? If you draw a picture on a piece of paper, the angles and distances involved will give a false impression of the reality of the situation.

16. Feb 11, 2017

### Bandersnatch

@sophiecentaur you're still missing the OP's question. The question was: given a star of luminosity L and planet of radius r, what fraction of L does the planet block/receive?

17. Feb 11, 2017

### sophiecentaur

OK but, as you say, it is irrelevant. We do not observe the Luminosity of the star. We can 'infer; the luminosity by making the assumption that it is an isotropic radiator. It is just the ratio of areas that affects what we see. That area ratio is (I am pretty sure) the only relevant thing and the calculation is very easy and, as long as we can know the radius of the star, we have the radius of the planet (assuming that the whole of the planet's disc is transitting).

18. Feb 11, 2017

### sophiecentaur

You worried me for a minute but I re-read the OP.
He uses the term "apparent Luminosity" which is a fair enough term but it's not relevant. All we are concerned with is the flux arriving at the Earth. If a solar planet moves across the star, the luminosity hasn't changed. The flux in other directions is of no consequence. The formula he wants to use is not relevant to Observations. It appears to give the proportion of the total radiated power that is intercepted by the planet. How is that of any use? I am confused that he doesn't consider the radius of the star itself.
We're in the same situation as discussing total and partial eclipses of the Sun. If the Sun were a point source, there would only be total eclipses and its significant radius gives us a penumbra and partial eclipses. Can you put me right on this? I can't see how I'm going in the wrong direction.

19. Feb 11, 2017

### Bandersnatch

Oh, I don't think it is. :) The closest to anything useful this is, as far as I can tell, is a somewhat convoluted way to calculate solar irradiance received by a planet.
The OP seems to agree, btw, as it was later indicated that it's merely a fun exercise to do. It certainly was for me - took me way too long to make sure I'm not going to say something stupid.

20. Feb 11, 2017

### sophiecentaur

Don't go too far down that road. It could inhibit some good conversation.
I sometimes worry about that but then - what the hell - it usually stimulates some extra sense to be injected by someone else.
I still think that the original (two star) scenario is not likely to need anything but simple geometry to give the received radiation at a planet with any expectation of life. The 'Penumbra' situation is going to be what applies.

21. Feb 11, 2017

### JohnnyGui

I do think such exercises really trains logical/good reasoning, even if it's something we can't use in practice. It helps to be able to reason things that are of use. I tend to do such exercises quite a lot and as long as I'm aware of their limits in practice, it should do only good.

22. Feb 12, 2017

### JohnnyGui

@Bandersnatch : I might be in some blackout phase at the moment, so apologies if I'm asking a dumb question that we might have discussed before. I was playing with the simulator you linked to and I noticed that they're calculating the intensity/flux ratio by the following:

$$\frac{πR^2 - πr^2}{πR^2} = \frac{I_{dim}}{I}$$

Just like we agreed on. Then something hit me; I noticed it doesn't take the distance difference from you to the planet and from you to the star into account. So if the planet is so close to you that it looks equal in size as the star it would still give the same flux ratio. Shouldn't they actually use the angular size ratio instead of the radii of the star and the planet?

23. Feb 12, 2017

### Bandersnatch

It doesn't, because the distance to the star is so much greater than the dimensions of the observed system, that the light rays you receive are effectively parallel.

Say, you look at a star:

As the distance $d$ increases, the angle $\alpha$ goes down. At astronomical distances, this means that even for large stars the angle is almost 0, and the light rays coming from the edges of the star are practically parallel in its closest neighbourhood:

So the amount of light blocked by a planet transiting at e.g. either 1 or 10 AU is so close to equal, that any difference gets swamped by instrument inaccuracies anyway.

24. Feb 12, 2017

### JohnnyGui

Thanks, this clears it up. Here's the thing though, and again this might be because I'm blacked out like hell right now. If different distances of the planet doesn't really matter, then shouldn't this ratio...

$$\frac{4πD^2 - πr^2}{4πD^2}$$

..., where D is the distance between the star and the planet, not only give the ratio between $L / L_{dim}$ but also the flux ratio $I / I_{dim}$? Such that

$$\frac{πR^2-πr^r}{πR^2} = \frac{4πD^2 - πr^2}{4πD^2}$$

(I might need some sleep..)

25. Feb 12, 2017

### Bandersnatch

No, it shouldn't. But you're right - you should get some sleep and think it over. ;)