Astronomy C

Post by **Skink** » Sun Jan 24, 2016 8:43 am

Good stuff, syo, thanks...tell you what, re: the National test images, I'll have to check in with my group to see where we had trouble when working through them and come back later. The test had a lot of 'Pick which image from all of these that I'm walking about...'-type of questions, so sometimes we were on target while, others, it was a guessing game. Like, speaking of light curves, there was 34. I think we reasoned through them working backwards but cannot remember offhand.

sciolymom · Post by **sciolymom** » Mon Jan 25, 2016 10:25 am

Regarding high mass versus low mass stars and their affect on planets:

First question - one of the objects connected a short orbital period with the low mass of the host star. I'm not understanding how that is connect.

Second question - I always thought that higher mass stars were more likely to have tidally locked planets due to their significant gravity. But I read this elsewhere:
On the opposite extreme, stars with less than half of Sol's mass are more likely to tidally lock planets that are orbiting close enough to have liquid water on their surface too quickly, before life can develop (Peale, 1977).

I have a feeling these two questions might be related... I would think that if low mass stars tend to have planets closer to the host, the closeness is what causes the tidal locking. But what is it that makes them end up closer to the low mass star than to a high mass star?

jkim117 · Post by **jkim117** » Mon Feb 08, 2016 12:05 am

Has anybody else noticed that the DSO page for Astronomy 2016 appears to have a lot of mistakes?
Just a few examples:

M42 or the Orion Nebula is in the constellation Orion and not Lepus
51 Pegasi b is in the constellation Pegasus and not Hydra
51 Pegasi b is 50.9, not 176 ly away
51 Pegasi b's right ascension and declination is 22h 57m 28.0s/+20 degrees, 46', 08'' not 11h 01m 52s/-34 degrees, 42', 17''
55 Cancri is 40.3 ly, not 172 ly away
55 Cancri equatorial coordinates inaccurate
AB Aurigae equatorial coordinates inaccurate
AB Aurigae is in constellation Auriga, not Aquila
and so on.

I could be be wrong (though I'm pretty sure that the Orion nebula is in Orion) but could someone check this out?

Thanks!

Post by **syo_astro** » Mon Feb 08, 2016 3:00 pm

Sorry for not replying to questions faster. I meant to, but I figured someone else would, and I was busy...

Considering that, sciolymom I will give it another week or two before putting together a response, unless you're hard-pressed for a response with some competition coming up or something.

To jkim, yeah, astro wiki...could use some edits. But that's where users like you come in

. Or hey there's always the onwards blog har har...too bad nobody wanted to continue helping me edit it cough :/ (and nobody PM'd me or Alpha as I've heard to take it over totally?). Typically just go by the guidelines that editing information is okay as long as you're not just omitting it all without replacement. If you are confused with editing the wiki, I believe there are pages on it, or you could PM JohnRichardsim. Also, yes, M42 is most definitely in Orion!

Post by **andrewwski** » Mon Feb 08, 2016 5:28 pm

Adi1008 wrote:
If you use the semimajor axis of an elliptical orbit, isn't it the same as using the radius of a circular orbit? I don't know why, but I've heard that it has something to do with the energy of an orbit staying constant. Not sure though, but thanks a ton!

Oops - neglected to check this thread, sorry! (Astronomy usually isn't my thing, but orbital dynamics is.)

The answer to your question is no - it isn't the same. If an elliptical orbit is circular (think of a circle as an ellipse with zero eccentricity), then the semimajor axis is equal to the radius of the circle. Moreover, however, the distance between the two masses is always constant and equal to the radius. For an elliptic orbit with nonzero eccentricity, however, the radius is not defined, but that distance (between the two masses) is changing continuously with time.

When you formulate the equation for energy in an orbit, it's the distance between the two bodies that you're concerned with, which happens to be constant and equal to the radius in the circular case - but not in the eccentric case. Let's look at the formulation:

Define $r$ to be the distance between the two bodies. Note that this is also equivalent to the radius of a circular orbit, but this is not the formal definition.

If we assume that mass $M$ is stationary (inertially fixed), then the total kinetic energy of the system is:

$T=\frac{1}{2}mv^2$

or just due to the motion of $m$ . The potential energy is:

$U=-\frac{\mu m}{r}$

Note that here, I've used $\mu=G(M+m)$ , which is known as the "gravitational parameter." This is common in astrodynamics use, as for any given body, $\mu$ can be determined much more precisely than G. If $M >> m$ , as is the case the orbit of a planet around a star, etc, then we can say $\mu \approx GM$ . I'll proceed with this definition, but analytically you could proceed forward with either. (Note: $\mu$ is more commonly used in astrodynamics than G, because for a given body, it can be determined to much greater precision than the universal constant G. I've never seen $\mu$ used in high school physics courses or the like - they usually stick to the formulation with G, but it's what orbital dynamics people use.)

Now, we can combine the two together to get the total energy:

$E_{mech}=T+U=\frac{1}{2}mv^2-\frac{\mu m}{r}$

Great, right? Well yes, but we still have $v$ , or the velocity in the equation. We know that for an elliptic orbit, this is related to $r$ by other orbital parameters.

Let's now look at the Vis-Viva equation. Note that this is actually derived from the fact that energy and angular momentum must be constant throughout an orbit - so the energy at apogee is equal to the energy at perigee, and the angular momentum at apogee is equal to the angular momentum at perigee. I'll skip the derivation here, but Wikipedia's derivation is clear if you're interested. Anyway, the Vis-Viva equation:

$v^2=\mu(\frac{2}{r}-\frac{1}{a})$

where $a$ is the semimajor axis.

Now, let's substitute the Vis-Viva equation into the total energy equation:

$E_{mech}=-\frac{\mu m}{r}+\frac{1}{2}\mu m(\frac{2}{r}-\frac{1}{a})$

This simplifies to:

$E_{mech}=-\frac{\mu m}{2a}$

So total energy in an orbit is not a function of the current position $r$ , but of the semimajor axis $a$ ! This makes sense, as we know the total energy cannot change in an orbit.

Now, if your orbit is circular, then this becomes $E_{mech}=-\frac{\mu m}{2r}$ as $r = a$ is constant (only for the circular orbit). Then, we know the potential energy at any point (since $r$ is constant!) is $U=-\frac{\mu m}{r}$ - or total energy is half the potential energy. So we've proven this for the circular case! But what happens in the elliptical case?

The answer is, since $r \neq a$ , we can't say anything about the kinetic or potential energies for an arbitrary point, except that they must equal the total mechanical energy, unless we know $r$ . If we're concerned about actually propagating the orbit, there are a few ways to determine $r$ , using a choice of anomaly (true, eccentric, or mean) to determine the position vector, and then taking the magnitude of the position vector. For periapsis and apoapsis, however, we have the relations:

$r_p=a(1-e)$
$r_a=a(1+e)$

We can use these to determine the kinetic and potential energy components for periapsis and apoapsis. At periapsis, potential energy is

$U_p=-\frac{\mu m}{r_p}=-\frac{\mu m}{a(1-e)} = \frac{2E_{mech}}{1-e}$

and at apoapsis, potential energy is

$U_a=-\frac{\mu m}{r_a}=-\frac{\mu m}{a(1-e)} = \frac{2E_{mech}}{1+e}$

So yes, it has to do with the energy (and momentum) of an orbit being constant, which will give you the Vis-Viva equation. Substitute this into the definition of kinetic and potential energies, and the result follows.

Hope this makes sense!

Post by **andrewwski** » Tue Feb 09, 2016 11:02 am

sciolymom wrote:Regarding high mass versus low mass stars and their affect on planets:

First question - one of the objects connected a short orbital period with the low mass of the host star. I'm not understanding how that is connect.

Second question - I always thought that higher mass stars were more likely to have tidally locked planets due to their significant gravity. But I read this elsewhere:
On the opposite extreme, stars with less than half of Sol's mass are more likely to tidally lock planets that are orbiting close enough to have liquid water on their surface too quickly, before life can develop (Peale, 1977).

I have a feeling these two questions might be related... I would think that if low mass stars tend to have planets closer to the host, the closeness is what causes the tidal locking. But what is it that makes them end up closer to the low mass star than to a high mass star?

For your first question - let's start by looking at Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{r^2}$

If $M$ is the mass of the star and $m$ is the mass of the planet, the force due to gravity $F_g$ can be determined if you know the distance between the two, $r$ .

Now, this doesn't necessarily say anything about an orbit. It applies whether or not the objects are in orbit. If we were to apply Newton's Second Law to this, we could derive the equation of motion for the two-body problem, and then determine analytically the conditions for which there is an elliptical orbit - but that would be beyond the scope of this event.

However, if you think about the notion of a circular orbit, you will realize that the gravitational force must balance with the centripetal force, which is $\frac{mv^2}{r}$ . So for any given distance from the star $r$ , the gravitational force can be determined, and thus there is one velocity that corresponds to a circular orbit at this distance. This concept could be extended for an elliptical orbit, but let's think about the circular case as it's more straightforward to visualize.

Understanding this, if we were to take the pure two-body problem - i.e. we assume that the only two bodies undergoing gravitational attraction are those corresponding to $M$ and $m$ , then in theory we can have an orbit for any distance $r$ and any mass $M$ .

But, we never encounter the perfect two-body problem in the universe. There are always other "perturbing" forces - for example, gravity due to other objects, or force due to radiation pressure from the star, etc. These forces are in addition to the gravitational force to the central star.

If the gravitational force between the planet and star are high compared to the perturbing forces, then the perturbing forces will have a minimal effect. However, if the gravitational force is not high compared to the perturbing forces, then they will have a large effect. At some point, they will make it such that the object is no longer orbiting the central star.

So, for a low-mass star, this means that you expect to see objects orbiting closer to it than a high-mass star. Since $M$ is smaller, $\frac{1}{r^2}$ must be increased (or $r^2$ , or likewise $r$ , must be decreased) - so these planets will be closer to the star!

As you get closer to the star (the orbital radius shrinks), the orbital period decreases. Note that orbital period is given by:

$T=2 \pi \sqrt{\frac{a^3}{\mu}}$

where $\mu$ is the standard gravitational parameter $\mu \approx GM$ .

$a^3$ is the semimajor axis, which is equal to the radius for a circular orbit - so thus as $r$ decreases, so does the period. This is intuitive - the velocity must be faster as the orbit becomes smaller.

Your second question follows from this. Let's consider why tidal locking happens - it is due to a gravity gradient within the planet. If the mass of the planet is not uniformly distributed, gravity will have a stronger pull on the part with a higher mass concentration. Since planets tend to not have perfectly symmetrical density, or be a perfect sphere, gravity will act more on one part of the planet than another.

But the question is - how much more? Let's look again at Newton's Law of Gravitation - we see that as M increases, F increases linearly, but as r increases, F increases to the power of two! So the distance from the star has more of an effect than the mass of the star!

Thus, the distance of the point from the center of mass of the central star (assuming M >> m) has the greatest effect on the gravitational force. As you get closer to the star, the effect of moving one unit closer to the sun in $r$ has a greater and greater effect on the gravitational force.

Let's look at the solar system, for example. If we look at Mercury, for example, we see that it is $57.91\times10^6$ km from the sun, and has a radius of 2440 km. Then, let's look at the difference in gravitational force on the surface of Mercury closest to the sun:

$\frac{1}{(57.91\times10^6-2440)^2}-\frac{1}{(57.91\times10^6+2440)^2}=-1.7267\times10^{-8} km^{-2}$

or, if we divide these out, the gravitational force will be $1.73\times10^{-6} \%$ greater at the closer surface than the far surface.

Now, let's take Jupiter, for example, which is much, much larger than Mercury but also much further from the sun. It is about $778.5 x 10^6$ km from the sun with a radius of 69,911 km. If we do the same, we find:

$\frac{1}{(778.5\times10^6-69,911)^2}-\frac{1}{(57.91\times10^6+69,911)^2} = -1.2844\times10^{-9}km^{-2}$

or the gravity at the surface of Jupiter closer to the sun is $1.28\times10^{-7} \%$ greater than at the far surface of Jupiter. Compare this to Mercury, even though Jupiter has a radius over 28x larger than Mercury!

So, you can see that the closer a planet is to a star, the greater the gravity gradient, and thus the more likely it is to be tidally locked. As we discussed before, a lighter star is more likely to have planets orbiting closer to it.

Hope this makes sense!

Post by **syo_astro** » Thu Feb 11, 2016 5:05 pm

Great news. A few tests (NY Invite and MIT Invite) were posted on the national website: https://www.soinc.org/astronomy_c

Hopefully they serve as good practice/discussion!

sciolymom · Post by **sciolymom** » Fri Feb 12, 2016 8:48 am

syo_astro wrote:Great news. A few tests (NY Invite and MIT Invite) were posted on the national website: https://www.soinc.org/astronomy_c

Hopefully they serve as good practice/discussion!

Woohoo! Thanks for sharing!

FuzzyLogic · Post by **FuzzyLogic** » Sat Feb 13, 2016 6:46 pm

Just went to regionals. After spending hours on my binder they used last year's test! It was about variables and didn't have a single question about the DSO's! We were pretty frustrated.

I'm not sure some of the math questions were even possible. One question gave us the radius of an exoplanet and the distance from its star but not the luminosity of the star. Correct me if I'm wrong, but I don't think that's possible. At this point we just wrote down "well, this problem requires the luminosity of the star to be answered. Unfortunately it was not provided." The next question asked if the planet was habitable. We wrote "perhaps."

But hey,we got second! I think everyone else was just as clueless. The team that won actually didn't study. They just used last year's binder. Probably the only time not studying pays off.

Post by **syo_astro** » Sat Feb 13, 2016 6:56 pm

FuzzyLogic wrote:Just went to regionals. After spending hours on my binder they used last year's test! It was about variables and didn't have a single question about the DSO's! We were pretty frustrated.

I'm not sure some of the math questions were even possible. One question gave us the radius of an exoplanet and the distance from its star but not the luminosity of the star. Correct me if I'm wrong, but I don't think that's possible. At this point we just wrote down "well, this problem requires the luminosity of the star to be answered. Unfortunately it was not provided." The next question asked if the planet was habitable. We wrote "perhaps."

But hey,we got second! I think everyone else was just as clueless. The team that won actually didn't study. They just used last year's binder. Probably the only time not studying pays off.

A shame people use last year's test :/. Also by variables do you mean variable stars? Technically variable stars are a major component of the event still just because of being able to use light curves to distinguish different pre-main sequence stars or to identify exoplanets via transit method.

As for the math problem, what were they asking for? Something like app. mag or temperature?

Ignoring all that, congrats on the 2nd!

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