Yes, that is indeed what they are.Unome wrote:Is it something other than this?Magikarpmaster629 wrote:Nice test, AlphaTauri, but what are EGGs? I haven't been able to find what they are.
Also, glad to hear that you liked the test, Magikarpmaster!
Yes, that is indeed what they are.Unome wrote:Is it something other than this?Magikarpmaster629 wrote:Nice test, AlphaTauri, but what are EGGs? I haven't been able to find what they are.
Yeah; my partner and I got a 55%. We missed a lot of the math stuff (never expected Jeans mass to show up ) and the more obscure concept questions.AlphaTauri wrote: Also, glad to hear that you liked the test, Magikarpmaster!
HOLY CRAP HOW DID I MISS THIS thanks so much andrewwski that makes a ton of senseandrewwski wrote:For your first question - let's start by looking at Newton's Law of Universal Gravitation: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?
If is the mass of the star and is the mass of the planet, the force due to gravity can be determined if you know the distance between the two, .
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 . So for any given distance from the star , 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 and , then in theory we can have an orbit for any distance and any mass .
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 is smaller, must be increased (or , or likewise , 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:
where is the standard gravitational parameter .
is the semimajor axis, which is equal to the radius for a circular orbit - so thus as 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 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 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:
or, if we divide these out, the gravitational force will be % 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 km from the sun with a radius of 69,911 km. If we do the same, we find:
or the gravity at the surface of Jupiter closer to the sun is % 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!
Took it, got a 68%. What an awesome (but hard T_T) test. Also yeah magikarpmaster629 it was super cool to have random concepts I had never heard of show up like Poynting Robertson drag. AlphaTauri, thanks for both writing and sharing this test! It was a pleasure to take.Magikarpmaster629 wrote:Yeah; my partner and I got a 55%. We missed a lot of the math stuff (never expected Jeans mass to show up ) and the more obscure concept questions.AlphaTauri wrote: Also, glad to hear that you liked the test, Magikarpmaster!
If you somehow had the radial velocities at apoastron and periastron, then you could find the ratio of the velocity at apoastron to the velocity at periastron to find the ratio (1-e)/(1+e) and solve for e that way. But this requires radial velocity graph to be determined at a very specific orientation of the system with the apoastron occurring at the max positive/negative radial velocity and the periastron occurring at the maximum of the other as other orientations would mess up the relative velocities for the entire orbit. Otherwise I don't know, maybe you could use some calculus to find it.Magikarpmaster629 wrote:Looking at radial velocity plots and this website: here eccentricity obviously affects the radial velocity of a star. However, I haven't been able to find an equation for eccentricity through a radial velocity curve. Is it calculable with just the curve?
I doubt a testmaker would give you a pressure other than 1 atm, so yes.Magikarpmaster629 wrote:Also, if I was asked to find the habitable zone around the star, I would just find the semi-major axes of planets with equilibrium temperatures at 273 and 373, right?
In the event that they do, you can find rough approximation of what temperature it would be using a phase diagram. This is a good one I used a lot in solar system last year:doge wrote:I doubt a testmaker would give you a pressure other than 1 atm, so yes.Magikarpmaster629 wrote:Also, if I was asked to find the habitable zone around the star, I would just find the semi-major axes of planets with equilibrium temperatures at 273 and 373, right?
Whoa really? I seriously never knew this. Looking at your answer, I think we got it correct (-ish), so yay!Magikarpmaster629 wrote:Luminosity is the amount of energy radiated by a star (or any object really) per second, or power in watts.
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