Can an astronaut standing on the moon observe earth's rotation with the naked eye?

Yes, as long as one was patient.  If one was on the Moon and stared up at the Earth and Washington DC was in the center of their view, in about thirty minutes, Lexington, Kentucky will have taken its place and in less than an hour Chicago, Illinois will have taken its place.  Add another thirty minutes and he or she would be looking at Topeka, Kansas.

Is there another planet like Earth?

NASA's Kepler Space Telescope, astronomers have discovered the first Earth-size planet orbiting a star in the "habitable zone" -- the range of distance from a star where liquid water might pool on the surface of an orbiting planet. The discovery of Kepler-186f confirms that planets the size of Earth exist in the habitable zone of stars other than our sun.
While planets have previously been found in the habitable zone, they are all at least 40 percent larger in size than Earth and understanding their makeup is challenging. Kepler-186f is more reminiscent of Earth.

The diagram compares the planets of our inner solar system to Kepler-186, a five-planet star system about 500 light-years from Earth in the constellation Cygnus. The five planets of Kepler-186 orbit an M dwarf, a star that is is half the size and mass of the sun.

"The discovery of Kepler-186f is a significant step toward finding worlds like our planet Earth," said Paul Hertz, NASA's Astrophysics Division director at the agency's headquarters in Washington. "Future NASA missions, like the Transiting Exoplanet Survey Satellite and the James Webb Space Telescope, will discover the nearest rocky exoplanets and determine their composition and atmospheric conditions, continuing humankind's quest to find truly Earth-like worlds."
Although the size of Kepler-186f is known, its mass and composition are not. Previous research, however, suggests that a planet the size of Kepler-186f is likely to be rocky.

Can neutrinos travel faster than the speed of light?

On July 12, 2012 the OPERA collaboration published the end results of their measurements between 2009–2011. The difference between the measured and expected arrival time of neutrinos (compared to the speed of light) was approximately 6.5 ± 15 ns. This is consistent with no difference at all, thus the speed of neutrinos is consistent with the speed of light within the margin of error. Also the re-analysis of the 2011 bunched beam rerun gave a similar result

For how long do I have to aim a laser at the moon to see a dot?

There are two possible questions here.  The first is simply how long would it take for a laser to travel to the moon and back.

The moon is around 384,000 km away.  The speed of light is 299,792,458 m/s.  The journey is round trip, so:

However, if the question really is about seeing the dot - you won't.  Ever.  

Beams of light diverge.  Take a flashlight (torch) for example.  Why is it that if you are 20 meters from a wall the light will illuminate it, but if you are 40 meters from that wall, the light won't illuminate it?  

What's happening here?  Is something stopping the light from traveling that distance?  No, the light is traveling unimpeded, 40 meters is nothing for a photon.  But the beam of light gets larger and larger with distance.  So, the extremely dense cross section of photons at the lens of the flashlight is very bright, but as the beam gets wider, those photons are distributed over greater area.  The light density gets less and the beam gets dimmer.

Very expensive lasers are designed to minimize this beam spreading, called divergence - but they can't stop it.  And we do reflect lasers off the moon.  Well, more accurately, we reflect lasers off mirrors that the Apollo astronauts left on the moon.

Although the moon looks bright to us, that's just because the sun is radiating it with so much light.  The moon is gray like charcoal.  It only reflects about 7% of the visible light that hits it.  So, even the best lasers combined with the best telescopes aren't going to be effective at reflecting visible light off of the surface.  But those mirrors are highly reflective.

Even so, very few of the photons from the lasers aimed at those mirrors actually make it back to the telescope.  There is a project called APOLLO (Apache Point Observatory Lunar Laser-ranging Operation) that fires laser pulses at those mirrors and measures the returned signal to calculate extremely precisely the distance to the moon.  They use a powerful laser and yet only 1.7 in 1E17 of the photons from their laser are sensed upon return.

That's 1.7 in 100,000,000,000,000,000 photons.  With their system, that means the returning signal consists of 5-10 photons.  A giant 3.5 meter telescope can only detect 5-10 photons.  Your eye isn't going to have such luck.

Here's a picture of APOLLO shining its laser on the moon.

What makes raindrops bigger or smaller?

In science we learn that one question often leads to another, or several others. Before we can discuss raindrop sizes, we must understand what a raindrop is. How is a raindrop made? How big can a raindrop be?
In order to have rain you must have a cloud--a cloud is made up of water in the air (water vapor.) Along with this water are tiny particles called condensation nuclei--for instance, the little pieces of salt leftover after sea water evaporates, or a particle of dust or smoke. Condensation occurs when the water vapor wraps itself around the tiny particles. Each particle (surrounded by water) becomes a tiny droplet between 0.0001 and 0.005 centimeter in diameter. (The particles range in size, therefore, the droplets range in size.) However, these droplets are too light to fall out of the sky. How will they get big enough to fall?
Picture a huge room full of tiny droplets milling around. If one droplet bumps into another droplet, the bigger droplet will "eat" the smaller droplet. This new bigger droplet will bump into other smaller droplets and become even bigger--this is called coalescence. Soon the droplet is so heavy that the cloud (or the room) can no longer hold it up and it starts falling. As it falls it eats up even more droplets. We can call the growing droplet a raindrop as soon as it reaches the size of 0.5mm in diameter or bigger. If it gets any larger than 4 millimeters, however, it will usually split into two separate drops.
The raindrop will continue falling until it reaches the ground. As it falls, sometimes a gust of wind (updraft) will force the drop back up into the cloud where it continues eating other droplets and getting bigger. When the drops finally reach the ground, the biggest drops will be the ones that bumped into and coalesced with the most droplets. The smaller drops are the ones that didn't run into as many droplets. Raindrops are different sizes for two primary reasons.

1. initial differences in particle (condensation nuclei) size
2. different rates of coalescence.

The last thing Albert Einstein was working on before he died.

At his hospital bed just after he passed away; he had left his pen and papers on top of the little side table.  The papers were on his passion, his Unified Field Theory. The evening before he died, Einstein was hard at work writing and scribbling his thoughts and workings on this theory. A little later he stated to his nurse, "...I think I will rest for a while" and he placed the items on the table.

There is a nice story this nurse enjoyed telling; one I too enjoy recounting.  She was, in fact, the last person to have a conversation with Einstein. The nurse had wheeled Einstein over to the hospital window to admire the view of the little round garden from the bedside window. 
“Professor Einstein, do you think God made the garden?”
He replied, “Yes, God is both the gardener and the garden” to which the nurse replied ,”Oh I’d not thought of it that way” to which Einstein replied, “Yes, and I’ve spent my whole life just trying to catch a glimpse of him at his work”.

The Earth is not exactly spherical. Why?

Earth is rotating about its axis since 5 billion years, when earth was still molten and some giant object hit it tangentially. Due to the centrifugal force produced by rotation, the earth got bulged at the equators. 

It has retained its shape till now, when the crust formation has taken place so we see it flattened from the poles(or bulged at the equator, whichever way you say that). And it is likely to remain so, because earth is still rotating about its axis.

I hope it answered the question.

What does the ISS do that can't be done from Earth?

The ISS provides an international laboratory in which scientists can have experiments that require the removal of gravity, as an influence, conducted.  It can be difficult for scientists to study isolated variables in a system when one or more variables has an overwhelming influence.  On the ground, scientists can build experiments that remove the influence of variables like temperature, light, moisture, pressure, and sound.  But, they really can’t do much, other than small scale centrifuge use, to remove the influence of gravity.

The ISS provides an international laboratory in which we can study the impacts of the space environment on payloads.  Gravity isn’t the only difference between operating on Earth and operating in space.  We need to learn how variables such as solar and galactic radiation affect items.  We need to learn about how the space environment will affect the items we send to space for future missions, such as spacecraft equipment.  The crew are also payloads.  If we are going to one day venture farther out into the solar system, we need to better understand what happens to people that are in space for a long time.  What happens to their bones and muscles?  Are they psychologically altered?  Does their ability to perform a task diminish with time?  Through observing the crew over the last 16 years, we’ve learned a lot about long duration spaceflight.  We’ve made exercise and dietary changes that have reduced the bone and muscle loss.  We’ve observed changes that weren’t predicted, such as changes in eyeglass prescription as the shape of the eye changes during long duration spaceflight.

The ISS provides an international platform to mount experiments that can study Earth and the space environment without needing to construct dedicated spacecraft.  Cameras and other sensors can be delivered to space by a cargo vehicle and then mounted to the ISS by the crew.  Those items can receive their power from the ISS and provide their data to the ISS.  They can be kept safely in orbit by the ISS, and if needed, maintained and upgraded by the crew.  They can also be operated by the crew.  The crew are often called upon to perform Earth observation activities to support people on the ground during major events such as hurricanes, earthquakes, floods, and fires.  Just today, the crew have been asked to collect data about the tropical cyclone that is threatening Japan.

The ISS provides an international platform to mount Earth observation and radio frequency medium experiments in low Earth orbit without needing to construct dedicated spacecraft.  For example, there is an experiment called SCaN that is composed of software defined radios.  The experiment is utilized to conduct various ground station to SCaN radio and SCaN radio to satellite experiments.

The ISS serves as a technology testbed.  Equipment for future space missions needs to be validated in space before primary use.  Such equipment can be sent to the ISS for testing and then safely returned to the ground.  While in space, it can sometimes be utilized by the crew, in a way similar to its eventual use for its mission.

Is the mass of the Earth constant?

On a local perspective the mass of earth is both increasing and decreasing:
Increasing : because of the meteorites that collapse on earth each seconds and add its matter ( hydrogyn, water, metals ) to the mass of the earth
Decreasing : because of the gazes that escape the atmosphere, actually the gazes that have very light mass like hydrogyn and helium are barely affected by the gravity of the earth and because they are light , they are pushed up to the moment they are no longer affected by the gravity , Also the earth radiates energy to space , and using Einstein formula you can conclude that there are some mass wasted in this radiation

On the global perspective , unfortunately earth is missing mass more than it is gaining so yeah the mass of the earth is decreasing
The good news thought, that the sun will become a red giant and vaporise the earth before that the decreasing of its mass become noticeable

What is the moment of inertia tensor? How is it derived?

Let’s talk about where the moment of inertia tensor came from. One of the confusing aspects, I think, is that it seems like a completely separate quantity. What really happens is that when you compute the kinetic energy of a rigid body rotating through space, you’ll find that a specific choice of coordinates allows you to split up the total energy into two components:translationaland rotational. Note, I’ve borrowed a lot of the mathematical rigor from Hand & Finch.

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Figure 1: A diagram of a rigid body. By definition, a rigid body is such that (r⃗ i−r⃗ j)2= constant  (the distance between any two points is a constant).

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The velocity of the ith  point in the rigid body as expressed in the space system is 
            v⃗ i|space=R⃗ ˙+ω⃗ ×(r⃗ i|body) 
Remember, the vector r⃗ i=0 when the point is located at the origin of the body system K. I'm going to drop the body subscript since I'll only be using the r⃗ i from the body frame K.

Total Kinetic Energy

To an outside observer (in the space frame K′), the total kinetic energy, by definition

Remember, the total mass M is equal to the sum of the point masses of the rigid body, so

                     M=∑imi              ∑imir⃗ i=Mr⃗ cm
The last expression should be incredibly familiar. It follows straight from how we compute the location of the center of mass (weighted average).

The interesting thing about our rigid body is that we’ve generalized it completely. We can make some simplifying assumptions about r⃗ cm. In particular, we can choose a symmetry such that the center of mass of the rigid body lies in the origin of K space, such that r⃗ cm=0. Then, the above expression for kinetic energy has two pieces.
            T=12MR⃗ ˙2translational+12∑imi(ω⃗ ×r⃗ i)2rotational
The translational kinetic energy is the same as if all the mass were located at the center of mass of the rigid body. Thus from now on, we assume the origin of K is located at the center of mass.

The Moment of Inertia Tensor

Tensors are just a fancy way of really saying "matrices and vectors" (at least for the level of this answer). When we talk about a tensor, they can be m by ndimensional. They contain properties and help us talk about maps between spaces. For now, when we talk about a moment of inertia tensor, just think about a matrix.

Imagine a pen thrown up in the air. Any of its points will follow a rather complicated motion, but if we talk about the motion in terms of the center of mass of the pen... we see that the center of mass follows a parabolic trajectory. It’s just like the motion of a point particle, this is boring. So let’s focus on the rotational part.
            (ω⃗ ×r⃗ i)⋅(ω⃗ ×r⃗ i)=ω⃗ ⋅(r⃗ i×(ω⃗ ×r⃗ i))
from the scalar triple product (it is invariant under a circular shift of its operands). Then
            ω⃗ ⋅(r⃗ i×(ω⃗ ×r⃗ i))=ω⃗ ⋅(r2iω⃗ −(ω⃗ ⋅r⃗ i)r⃗ i)
from the vector triple product. This is known as the triple product expansion. Now, let's expand out this product in terms of the components. Since we'll be talking about 3D space, it'll be useful to just write with indices α,β=1,2,3for x,y,z components respectively. That is
            x⃗ =(x1,x2,x3)=∑3α=1xαe⃗ α=xαe⃗  α
And likewise, the dot product between two vectors
            a⃗ ⋅b⃗ =∑αaαbα=aαbα
Simplify that expression we obtained earlier by rewriting it in a component form:
            ω⃗ ⋅(⋯)=∑αr2i,α∑βω2β−∑αri,αωα∑βri,βωβ
The first term is from the dot product between ω⃗ ⋅ω⃗  and the second term is from the dot product between ω⃗ ⋅r⃗ i. Use the Feynman method, think really hard and write down the answer:
    Trotation=12∑imi(ω⃗ ×r⃗ i)2
                = 12∑imi(∑αr2i,α∑βω2β−∑αri,αωα∑βri,βωβ)
                   =12∑α∑β∑imi(r2i,αω2β−ri,αri,βωαωβ)
                  =12∑α∑βωαωβIαβ
where we defined the moment of inertia tensor I with components Iαβ given by
             Iαβ=∑imi(r2iδαβ−ri,αri,β)
and
            δαβ={10if α=βotherwise

In matrix form

Notice some interesting features: Iαβ=Iβα which is a symmetric tensor (we often state this by saying I[αβ]=0. Because of this, we can write the total kinetic energy in a more compact form!

            T=12MR⃗ ˙2+12ωTIω