What visible differences do Geminids have to other meteors

Well done on your observations, I've been completely clouded out.

You are quite right, the Geminids are noted for a variety of colours, and a good number of fireballs, though most observers note a predominiance of yellow meteors This is partly due to their composition and density (coming from a peculiar "rock comet") and partly to their velocity (35km/s) which is fairly fast.

There is a good chance of seeing some good trails with this shower. Brighter meteors and most fireballs will leave a trail, and the Geminids have more of these than most other showers.

senescence - How do caspase proteins kill a cell?

Caspase do not directly kill the cell, but rather activate a process known as apoptosis, or programmed cell death. The programmed part is there to distinguish it from other types of cell death, such as necrosis, which are more aspecific death processes.

Coming back to caspases, they are a series of proteasis, that can activate in cascade in response to a pro-apoptotic signal.

To simplify matters a lot, the pro-apoptotic signal first activates initiator caspases, such as caspase 8, 9, 10 (plus a series of other proteins).

These, in turn, cleave other effector caspases, which are the ones who effectively do the dirty job: caspase 3, 6, and 7.

Now there are many things happening at the same time (see this PDF from AbCam to get an idea of the complexity of the system).

At the nuclear level you have, for instance:

• activation of DNA cleaving enzymes, such as CAD (caspase-activable DNAse), which is normally inactive, due to binding to its inhibitor I-CAD. The latter is a target of effector caspases and its degradation causes CAD to go in the nucleus and starting fragmenting DNA.




• cleavage of PARP (Poly(ADP-ribose) polymerase), a nuclear enzyme involved in finding and repairing single-strand breaks on DNA.




• cleavage of lamins (by effector caspases + other proteases), proteins that are important for the formation of nuclear lamina, and stability of the cell nucleus




• cleavage of U1, a nuclear protein necessary for processing of mRNA.

Very complex events also happen at the mitochondria level: here you have a series of protein, which belong to the Bcl-2 family that control apoptosis mostly by regulating the levels of Ca2+ in the cell. There are 25 members of this family, that can be divided into two "sides": pro-apoptotic proteins such as Bax, Bak, and BAD and anti-apoptotic proteins such as Bcl-2, Bcl-xL, and Bcl-w.

The functioning of the Blc-2 proteins is very complex and I am not up-to-date with the last literature but to simplify, essentially what happens is that when the balance between anti-apoptotic and pro-apoptotic proteins is shifted towards the pro-, there is:

• liberation of cytochrome c from the mytochondria into the cytoplasm, which can activate effector caspases.


• induction of MPTPs (Mitochondrial Permeability Transition Pores), proteins which increases mitochondrial permeability which causes all sorts of trouble, eventually leading to swelling of mitochondria and loss of their function

Different events also bring to an increase in cytoplasmic calcium concentration, which can, for instance, induce the activity of calcium-activated endonuclease or other pro-apoptotic calcium-dependent proteins.

This is far from being an exaustive list of what happens, but it gives you the idea of the complexity of the system.

observation - When do Mercury/Venus reach greatest elevation at sunset/twilight for a given location?

On what day does Mercury reach its greatest elevation (in degrees from
the horizon) at sunset a given location?

The obvious answer is the day of Mercury's greatest elongation from
the Sun, but, since the ecliptic is slanted with respect to the
horizon, I'm not convinced this is correct.

In other words, on the day after greatest elongation, Mercury's total
angular distance from the Sun will be smaller, but it's vertical
distance in elevation (for a given location) at sunset might be
higher.

Same question for Venus, and for when the sun is 6 degrees below the
horizon (ie, civil twilight), and for sunrise/dawn.

I'm guessing the date might vary based on position (mostly latitude) since the ecliptic's slant varies at different locations.

I googled and found nothing. My (preliminary and possibly wrong)
expierments with stellarium show that Mercury's elevation at sunset IS
higher 1-3 days after its greatest elongation, but by less than 1/2
degree.

So, it's possible that the date of greatest elongation is a close
enough approximation.

What would happen if those gravitational waves were much stronger?

Based on a post by an Astrophysicist my understanding is as follows:

1. Gravitational waves would still be faint to detect irrespective of the size of the Black Holes. BTW aren't black holes meant to be more concentrated rather than bloated?


2. The distance from ground zero is what matters rather than the size of the colliding black holes and even if they were a billion times closer they would still not measure up to 1 mm.
   
   
   3. There wouldn't be significant impact from the waves themselves rather the Black Holes are the ones that would do the damage.
   
   
   4. For the shift to be 1 mm we would require a force enormously stronger than that of a gravitational wave.
   
   

Here's the link to the article: http://www.forbes.com/sites/briankoberlein/2016/02/13/could-gravitational-waves-ever-be-strong-enough-to-feel/#1605e7fa4aac

Disclaimer: All my points are a result of my learning from the above link rather than any native knowledge or original research that I profess to have done.

volcanism - Volcano activity on moon

The grey areas are known as "maria" (singular:"mare") and they were formed from flood basalts. These are a type of volcano and they have formed on earth too, examples include the lake eruption on Iceland, and the Deccan traps, that played a role in the mass extinction 65 million years ago.

On the moon, the flood basalt is old. About 3-4 billion years old, with some regions probably younger, but nothing less than 1 billion. Volcanism is driven by the heat of radioactive decay. On the moon there are no longer enough radioactive elements to melt rock, and so the moon is no longer a volcanic active body.

There are no active volcanoes on the moon, and there is no prospect of any in the future.

Why do post main sequence stars enter the red giants branch?

I am an early graduate student in astronomy and have hard time understanding why do post-MS stars move up the RGB.

Here is what I understand about post main sequence evolution of stars. As their hydrogen core is exhausted, the core shrinks under its own gravity. This lets a region of previously too cold hydrogen enter hotter regions, thus starting a hydrogen shell burning process. While this process can keep the luminosity constant, it expands the envelope, hence lowering the temperature and moving the star to the right in the HR diagram.

Then suddenly, the star starts moving up on the RGB: slight or no temperature change, but sudden increase in luminosity. What makes the star move up the HR diagram all of the sudden? This is before the helium flash when the core starts burning helium.

Thank you!

How to calculate the expected surface temperature of a planet

The formula

$$
4 pi R ^ 2 ơ T ^ 4 = frac{pi R ^ 2 L_{sun}(1 - a)}{4 pi d ^ 2}
$$

is correct, if you want to calculate the radiative equilibrium temperature. You only need to use the right units. We can further simplify the formula to

$$
T ^ 4 = frac{ L_{sun}(1 - a)}{16 pi d ^ 2 ơ};.
$$

You should input the luminosity in watts, the distance to the star in meters and the Stefan-Boltzmann constant as
$$
σ = 5.670373 × 10^{−8} ;mathrm{W}; mathrm{m}^{−2}; mathrm{K}^{−4}.
$$

The albedo is dimensionless. The resulting temperature will be in Kelvins. Let me make an example for Earth:

$d = 149,000,000,000 ;mathrm{m}$

$L = 3.846×10^{26} ;mathrm{W}$

Albedo of Earth is 0.29. (The Bond albedo should be used.) You will get

$$
T ^ 4 = frac{ 3.846×10^{26}(1 - 0.29)}{16 pi times (149,000,000,000) ^ 2 times (5.670373 × 10^{−8})}=4,315,325,985 ;mathrm{K}^4;.
$$

After powering this number to 1/4, we obtain temperature 256 K, which is -17° C. This looks reasonable. The real average temperature on Earth is closer to 15° C, but the greenhouse effect is responsible for the difference.