# Do I have the right formulas?

1. Oct 15, 2011

### s3a

Basically, this thread is just about (basic) Astrophysical theory because the Math is quite simple. I am attaching a set of word problems and all I am asking for is to get feedback on whether or not I am correct or not about the formulas I am planning to use for each question. If I am wrong about one of my choices of formulas, it would be appreciated if you could explain why I am wrong and provide me the correct formula as well as the related background theory. I'm most uncertain about the last question (#3).

I think I know what Very Long Baseline Interferometry is in theory but, to apply the math, is it the same formula as #2? Is a baseline analogous to a diameter of a telescope from #2? Are baselines the distances between the antennas? Are antennas telescopes in this context? (I am a bit confused about that part). Also, I am not asking for you guys to solve my problem(s) but you can if you want to :tongue:.

#### Attached Files:

• ###### Questions.jpg
File size:
23.8 KB
Views:
119
File size:
39.2 KB
Views:
103
2. Oct 15, 2011

### cepheid

Staff Emeritus
You equation for 1a seems fine. It's just the Wien-Displacement Law.

For 1b, your equation is totally wrong. E = hc/λ is the equation for the energy of a single photon of wavelength λ. You want to know the energy emitted (typically per unit time and per unit surface area) by a perfect thermal blackbody emitter at temperature T. For this, you need to look into the Stefan-Boltzmann Law.

More on why you are wrong: A blackbody is going to emit many photons over a very large range of wavelengths (it has a continuous, broad spectrum). The λ from 1a is just the wavelength at which peak emission occurs.

Your equation for the flux in 1c is fine, provided that you get the luminosity correct from part b. (Recall that luminosity is just the astronomy term for the total power in EM radiation emitted by a source, so it's the thing you are looking for in part b).

For 2, your formula for the size of the diffraction disk is approximately right, assuming that you want the result in arcseconds.

Yes, the baseline is the distance between antennae. Yes, the terms "antenna" and "radio telescope" are not any different in this context. (In fact, if you look closely at a satellite receiver dish, you'll notice that it is essentially a prime focus telescope, which is one that has a primary mirror, but no secondary, since the detector/receiver is right at the focal point in front of the primary). And yes, if I recall correctly, the "effective" diameter of your telescope can be considered to be the length of the longest baseline in your interferometer (that's why you do interferometry -- to achieve much better angular resolution than you could with a single dish telescope at any practical size). If that's true, then you're right, you would just use the same equation for the diffraction-limited resolution as in part 2, but with D being the "effective" diameter. However, you will want to double-check this last claim of mine, because it's been a while for me since I read about interferometry.

I know you were half joking about someone doing the problems, but I just wanted to emphasize that, no, we won't do your homework for you, we don't do that here at PF (it is, in fact, against the rules). But we're more than happy to help you learn. Also, for future reference, this post should have been in the Homework Help section of the forums, not in the Astronomy subforum.

Last edited: Oct 15, 2011
3. Oct 18, 2011

### s3a

For 1b, assuming you are talking about S = σT^4, how is it implied by the question that a rate of energy is wanted. I thought a value with a plain old energy unit was required (and nothing more, the units of S are W/m^2 which is flux/luminosity).

For 2b, how do I know if the HST can resolve a certain object? Is there a certain numerical value for spatial/angular resolution I compare to for each calculation? Assuming 2b does use 2a's formula, is the wavelength 1km whereas the distance is 385 000 km for the first example, i? Also, for 2b, Stefan Boltzmann's law (assuming I mentioned the correct equation earlier) doesn't relate the radius of a star so it means that the rate of energy is independent of the size of the star but only depends on its temperature (kind of like having masses cancel out in certain energy equations or the smaller mass cancel out when trying to find the gravitational acceleration of something), right?

What about the rest of the set of problems (including #3)? Is there anything else I am attempting incorrectly (use of incorrect formula, etc)? For #3, the formula is the same but it's just the theory that is different in that I need to understand that the telescopes are “working together,” right? Also when #3 says “minimum angle,” does it basically imply that the equation from 2a is outputting the minimum resolvable angle?

Sorry if I repeated what you said a bit, I am just trying to make sure I grasped everything correctly. Thank you for the previous response and I wasn't intentionally trying to break the rules. Also, do all homework help questions go in the homework help section regardless of the topic?

4. Oct 18, 2011

### cepheid

Staff Emeritus
It makes no sense to talk about the "energy" emitted, since stars continously emit energy, so if you were to ask me, "how much energy does the sun emit?", I would have to ask you in return, "over what time interval?". The longer you wait, the more energy you will receive.

Clearly the quantity we want in order to compare the two stars is the rate of energy emission (i.e. energy per unit time). In other words, we want the power, which astronomers call luminosity.

NO, be careful here! Flux and luminosity are not the same thing. Luminosity is radiated power, and hence it is measured in watts (W). Flux is power emitted per unit of surface area (or power received per unit surface area, depending on the context), and hence it is measured in W/m2

Now, if you know the surface flux S from the blackbody temperature, then you can figure out the total luminosity by integrating the flux over the surface area over which it is emitted:

L = 4πr2S = 4πr2σT4

where r is the radius of the star. You don't know r, but you are explicitly told that both stars are the same size. Therefore, they both have the same surface area, and the ratio of the their luminosities is just equal to the ratio of their surface fluxes. So, just compute their surface fluxes using the Stefan-Boltzmann law, and be done with it! The one with the higher flux will be the more luminous one.

5. Oct 18, 2011

### cepheid

Staff Emeritus
You computed the angular resolution of Hubble in 2a. In 2b, you're given both the physical sizes of the objects, and their distances. Therefore, you can figure out their angular sizes. If their angular sizes are smaller than the angular resolution of the telescope, then you cannot resolve them. The angular resolution tells you the smallest angular separation[ that you can discern.

No, of course the wavelength is not 1 km! That is the size of the crater. The wavelength λ is a property of the light (electromagnetic radiation) that the telescope is receiving from the source. In 2a you are specifically told to assume light of 500 nm wavelength. Hubble is a telescope that operates at visible light wavelengths (400 nm - 700 nm). Wavelengths of 1 km would be really really really really long wavelength radio waves, and I don't think even radio telescopes observe at such long wavelengths. You really need to think more critically when doing these problems.

Are you actually referring to 1b here? If so, no, the rate of energy is NOT independent of size. As I explained in my previous post, the Stefan-Boltzmann law gives you the flux, which is the power emitted per square metre of surface area. Therefore, a larger star at the same temperature will have a larger surface area. So, although its flux will be the same, its total power emitted (or luminosity) will be larger.

6. Oct 18, 2011

### cepheid

Staff Emeritus
Yes. As I said above, the angular resolution is the angular size of the smallest thing you can resolve.

Yes, but the Homework Help section of the forums consists of several subforums:

Introductory Physics
Precalculus Mathematics
Calculus and Beyond
Other Sciences
Engineering and Computer Science

I would have posted this as a thread in the Intro Physics homework help forum, because it is not advanced physics, and there is no subforum dedicated to astronomy homework help. "Other Sciences" refers to chemistry and sciences that are not physical sciences (like biology) so that category doesn't fit either. Besides, Introductory Physics is by far the most frequented homework help forum, so the chances your thread will be viewed and responded to are much higher there.

7. Oct 22, 2011

### s3a

I just wanted to say thank you again .