solar system - How long should it take for us to observationally determine if Caltech's Planet Nine is really there?

Those are model calculations, which hint to the existence of a possible body of about 10-times the mass of Earth. Calling this a discovery would clearly be premature. The confidence level is just a little above the "evidence" level of 3 sigma, under the assumption, that the discoveries of the KBO objects leading to the inference aren't observationally biased.
It's a long way to direct observation, since precise orbital data haven't been inferred.
Another option is, that there has been a planet. But it might have left our Solar system. Hence predicting a date for direct observation doesn't appear reasonable at the moment.

planet - How is the mass of solar system objects calculated?

Planets are satellites of the Sun. Most planets have satellites.
Whenever you find a parent body with satellites, you get first good results about their masses and distances by applying Kepler's laws.
A second approach is using parallax measurements for distance estimates.
The distance to some of the planets can be measured very accurately by radar, e.g. for Venus.
The known planets of our solar system have been visited by probes. Using signal travel times, you can calculate the distance of the probe to Earth. One of the methods to determine the distance between probe and the planet is by photogrammetry.
Apply the measured data to the according system of equations. Usually more data than needed are available. Then you get an overdetermined system. A best solution can be found by minimizing appropriate squares.

Subtle deviations from Kepler's law can be used to infer masses of planets which are hard to measure in a more direct way.
Once you know the distance of a planet, use its apparent diameter(s) to calculate the actual diameter(s), and then the volume of a sphere or a spheroid.
With mass and volume you know the mean density of the planet.

Within our solar system, for very distant objects the apparent size of which cannot be determined you may estimate the surface albedo together with distance (by Kepler after knowing the mass of the Sun) and apparent magnitude to obtain a rough estimate of the size of the object.

Better size estimates for objects of small apparent size can be obtained by stellar occultations, meaning measuring the time a star is occulted by the object. Together with the angular velocity of the object you get data about its angular size.

What is the Maximum Speed that can be acheived Because of Acceleration Due to Gravity?

What is the maximum speed that can be achieved because of acceleration due to gravity?

The speed of light. But there's a catch.

In other words which is the strongest and largest gravitational field discovered in the universe till date.

That of a black hole. Sagittarius A* is thought to be the location of a supermassive black hole.

What is the maximum speed an object accelerating in these fields can gain?

The speed of light. See this list. The escape velocity for a black hole is the speed of light, and you can flip this around. If you drop an object from a great height, it's travelling at escape velocity when it reaches the gravitating body. But like I said, there's a catch. Take a look at this by Einstein:

See the second paragraph. The body falls down because the speed of light is spatially variable. If this continued unabated there would come a point where the body is falling faster than the speed of light at that location. But since matter can't faster than this the "coordinate" speed of light, the maximum speed it can fall is at the speed of light at that location. Which is circa half the speed of light at our location.

astrophotography - How do i get focussed images with my T-adaptor and 6mm eyepeice?

Using the eyepiece, and no camera lens, is a very different configuration to the camera+tube+scope (case 3). The configuration you're after is "Eyepiece projection".

I don't know exactly what you have available, but I believe you will need to fit a T-threaded extension tube to the external thread on your T-adapter. Then fit the camera onto that.

Note - before you go out and buy a new extension you can probably do a hand-held test of this by the following:

• Attach the adapter to the scope, with a low-power eyepiece in the adapter (low power makes the test easier to see.)


• Point the scope at anything distant (you can do this in daylight, but NOT at the sun of course!)


• Now hold the camera, by hand, a few inches behind the
  eyepiece+adapter. Perhaps you can use a cardboard tube to keep the stray light out.


• See if you can adjust the scope's focus, and the camera position, to get a sharp image.

If this worked, then you can get a T-threaded extension tube to go on the back of your current adapter, to hold the camera in position.

See for example this link:
http://www.astronomysource.com/2011/10/13/eyepiece-projection/

Orbits in a binary star system

The point you appear to refer to, is called the Lagrangian point $L_1$.
This point is a saddle in the field of gravity, hence not to be considered to be stable in the strict sense.
Two other Lagrangian points, called $L_4$ and $L_5$, can be stable, provided the considered orbiting objects are of small mass in comparison to the two main bodies of the system, and if the masses of the binary components are sufficiently different.

According to theorem 4.1 of this paper, $L_4$ and $L_5$ are stable in all directions, if and only if the mass ratio of the two main binary components $frac{m_1}{m_2}geqfrac{25+3sqrt{69}}{2}approx 24.9599$.
According to theorem 3.1 of the same paper all Lagrangian points are stable in z-direction, which is the direction perpendicular to the orbital plane of the binary system.
(Credits for this corrected version go to user DylanSp.)

astrophysics - Would a tablespoon of a neutron star remain intact?

If we take neutron star material and somehow transport it somewhere for examination (say the Earth!), the results would be catastrophic. At say a density of $sim 10^{17}$ kg/m$^{3}$ the neutrons have an internal kinetic energy density of $3 times 10^{32}$ J/m$^{3}$ (calculated using the relevant equations for an ideal gas of degenerate neutrons at this density). So even in a tablespoonful (say 20ml), there is $6times10^{27}$ J of kinetic energy (more than the Sun emits in a second, or a few billion atom bombs) and this will be released instantaneously.

The energy is in the form of around $10^{39}$ neutrons travelling at around 0.1-0.2$c$. So roughly speaking it is like half the neutrons (about a billion tonnes) travelling at 0.1$c$ ploughing into the Earth. If I have done my Maths right, that is roughly equivalent to a 50km radius near-earth asteroid hitting the Earth at 30 km/s.

The neutrons in a dense neutron star gas are relatively stable (beta decay is blocked by electron degeneracy). The expansion described above would allow beta decay into protons and electrons, but as this happens on timescales of 10 minutes, it is hardly relevant to the initial destruction. However, you would end up after a few tens of minutes with an expanding cloud of ionised hydrogen a few light minutes across.

The minimum possible size to gravitationally bind neutron star material is thought to be around $0.15 M_{odot}$ (see here). The equilibrium electron density (there are always some electrons and protons present in neutron star material) for lower masses is too low to block neutron beta-decay.

How to calculate the temperature of a star

I need a way to calculate the effective temperature (surface temperature) of a star for a stellar model. I need something in the form Te=....

I have:

• Radius in m


• mass in kg


• the composition of particles (eg H 90%, He 8% etc)


• the combined stored thermal energy of the body in J

Constants (any really but I'm using these for now):

• G=gravity constant=6.67408E-011


• k=kbolzmann=1.3806485279E-023


• s=sbolzmann=5,67036713E-008


• PI=pi ~3.14...

Example of the sun:

• mp=average mass of a particle=1,7E-027


• M=total mass of the body=2E30


• r=radius of the body=700000000

I'm using this equation to estimate the core temperature :

(G*mp*M)/(r*(3/2)*k)

which nets 15653011 for the sun which is close enough given that that is the only star core temperature known (afaik).

I'm using this to estimate the luminosity L:

4*PI*(r^2)*s*(Te^4)

which results in an error of ~1-5% with 90% of my sample stars which is close enough. For the sun this results in 3,95120075975041E+026 W which is only 2,7% off.

The problem is I need Te for the 2nd formula which I don't have in my scenario.

Due to the formula for L being dependent on the surface temperature to the power of 4 this value has to be relatively precise.

Assumptions of my model:

• uniform distribution of particles: so every slice of the body has the same composition as the entire body.


• perfect sphere: every body is a perfect sphere, no handling for elliptic bodies needed.

My sample values (first line is the sun with a core temp of 15000000):

     emitted energy Surface temp    radius       mass 
     (in Lsun)      (in K)          (in m)       (in Msun)
     1              5800            700000000    1
     8700000        53000           25200000000  265
     6300000        50100           23100000000  110
     2900000        42000           23660000000  132
     2000000        44000           16800000000  80
     1260000        13500           140000000000 45
     57500          3600            618100000000 12.4
     78             5700            6440000000   2.56
     78.5           4940            8540000000   2.69
     15100          7350            51100000000  9.7
     1.519          5790            858900000    1.1
     0.5            5260            605500000    0.907
     370000         3690            994000000000 19.2
     123000         33000           7560000000   56
     2200000        52500           12600000000  130
     200000         10000           151900000000 22
     446000         19000           43330000000  42.3
     25.4           9940            1197700000   2.02

Errors in luminosity to actual value (the maximum error is about 100% which I can live with since it might just be inaccurate measurements for the sample stars)

2.74%
6.71%
-1.13%
11.29%
-2.00%
-4.27%
106.76%
3.99%
2.51%
-6.50%
1.12%
4.00%
-8.27%
2.10%
1.57%
113.75%
1.64%
2.15%

gravity - Does more ocean on an Earth analog produce a different rotation period?

Would an exoplanet that has more ocean than Earth rotate at a
different speed as a result of this?

Basically no. Rotation speed, or angular velocity doesn't measurably change with your proposed example.

To explain this in more detail, there's two important concepts. First, is Angular Momentum, (Short explanation or longer) and the 2nd concept is Moment of Inertia.

The basic formula is that Angular velocity (time for 1 full rotation) = Angular Momentum divided by Moment of Inertia. Formulas and explanations are in the link(s) above, and if I was to explain it, it would get wordy, but your question seemed to be more general, less about doing the mathematical calculations, so I'll skip the formulas.

In nutshell, talking about a planet, the state of matter doesn't affect the angular momentum. Angular momentum is conserved and that, divided by moment of inertia determines rotation speed. Now if you get something spinning fast enough the angular rotation can overwhelm the gravity and when this happens, the planet can begin to fly apart, which happens more easily with water than a rocky surface which has some cohesion, but ignoring crazy super fast rotations, a water world, a deep ocean world, a shallow ocean world and a rocky world all obey the the angular velocity law and composition doesn't matter. The angular momentum is conserved. The moment of inertia of a planet can change but for the most part, doesn't change much.

The Earth's moment of Inertia, for example, changes as glaciers grow or shrink, or when there's an earthquake. Even, every time we a tall building is built the Earth's moment of inertia increases a teeny tiny bit, similar to a skater extending their arms to slow down coming out of a spiral.

When there's an earthquake which, for the most part, settles the Earth, there's a small increases in the Earth's rotational speed. The total angular momentum and total mass remaining the same but shifting of material changes the moment of inertia. (Granted space dust and tidal effects change the Earth's moment of Inertia, but quite slowly).

Would the amount of water impact the weight, gravitational pull,
and/or tidal forces and cause a difference in the exoplanet's rotation
period?

This is harder question to answer precisely because adding water changes the mass of the planet and changing the mass changes the moment of inertia, but sticking to the principal of your question, there's no measurable effect.

Lets take a somewhat simpler example without changing the planet's mass. Ice ages. When the Earth is in an ice age there's less liquid oceans and more ice at the poles but the total mass is unchanged. More mass at the poles and less mass in the oceans decreases the Earth's moment of inertia because the bulk of the Earth's moment of inertia is around the equator, so, as a result, the Earth rotates slightly faster during an ice age and slightly slower after an ice age. Over time, the Earth's crust has a tendency adjust for this effect but that takes tens of thousands of years. Parts of the Earth's crust is still rebounding from the last ice age.

Gravitational pull isn't relevant. Neutron Stars with enormous gravitational pull can rotate very fast and the planet with the fastest rotation in our solar-system is Jupiter and the one with the slowest rotation is Mercury. Angular velocity has no direct correlation to mass or gravity though there is an indirect correlation. As a star, for example condenses it's rotation speeds up, because the angular momentum is conserved but as it settles the moment of inertia decreases. That's why young stars, White Dwarfs and Neutron stars can spin very fast.

Tidal forces can create drag on rotation but the effect is slow, taking millions or hundreds of millions of years. With enough time, tidal forces cause a planet or moon to stop rotating and become tidally locked but there's no short term affect. (I'll say a bit more on this later).

So, a planet with large oceans wouldn't rotate any slower than a planet with no oceans because liquid or solid can have equal angular velocity, but over time, tides will slow a planet with oceans more quickly than a planet without them.

Because the Earth has oceans, the Moon's gravitation on the Earth's tidal bulge does slow down the Earth's rotation, but this has been happening for 4 billion years and the Earth still rotates every 24 hours - one of the faster planets. If the Earth had more water the Moon's tidal tug would slow the earth down a bit faster, but it would still be very gradual.

Jupiter, which is basically a ball of gas, is the fastest rotating planet and Mercury, basically a rock, the slowest, so those are 2 examples of composition not being a factor, though Mercury's slow rotation is in large part due to the strong tidal forces it receives from the Sun.

Now, I Understand the logical approach to your question, as there's something apparent about water resisting rotation - touched on in this question, but the fact that water doesn't spin with a glass when you spin a glass is an example of conservation of angular momentum, not an argument against it. On a planet, the oceans are rotating with the planet and the angular momentum is already there.

Hope that wasn't too long, but that's the gist of it. I can try to clean up or clarify if needed.

gravity - Is the moon moving further away from Earth and closer to the Sun? Why?

Yes, the moon is moving away from Earth at around 1.48" per year. According to the BBC:

The Moon is kept in orbit by the gravitational force that the Earth exerts on it, but the Moon also exerts a gravitational force on our planet and this causes the movement of the Earth's oceans to form a tidal bulge.

Due to the rotation of the Earth, this tidal bulge actually sits slightly ahead of the Moon. Some of the energy of the spinning Earth gets transferred to the tidal bulge via friction.

This drives the bulge forward, keeping it ahead of the Moon. The tidal bulge feeds a small amount of energy into the Moon, pushing it into a higher orbit like the faster, outside lanes of a test track.

So, tidal forces are ultimately what causes this to happen.

Also, there is a Wikipedia article on tidal forces:

Tidal acceleration is an effect of the tidal forces between an orbiting natural satellite (e.g. the Moon), and the primary planet that it orbits (e.g. the Earth). The acceleration causes a gradual recession of a satellite in a prograde orbit away from the primary, and a corresponding slowdown of the primary's rotation. The process eventually leads to tidal locking of first the smaller, and later the larger body. The Earth–Moon system is the best studied case.

Michigan's Pole Star - Astronomy

It would depend on what time the Earth started to spin around Michigan. There is no way to answer this without saying what time, time of year, and which century (or at least millennium) it happens.

Edit: Hi Dayna. Thanks for the the time of year the Keweenaw Peninsula becomes the North Pole. One could be omniscient except for that and not know this. But we also still haven't ruled out a single star since your OP. With the time though we could tell you exactly what star. The decade is not important. What century or part of history it happens, yes but not the decade. (yes, astronomy is weird).

Why is it always planets orbiting stars?

Planets often circle stars because they have a very strong gravitational pull. Infact, our moon orbits the earth and we can look at a solar system in the same way. The earth has enough matter inside it to keep the moon in orbit, and the sun has enough matter inside it to create a gravitational pull strong enough to keep a entire solar system in its orbit.

You can't have several stars circling a planet, the pull of a single planet wouldn't be strong enough (the planet's center would be so dense nuclear reactions would start occurring and it would be a star anyway) . You can have binary star systems though, which are where 2 stars orbit around their common barycenter.

So basically, things orbit other things with a stronger gravitational pull.

universe - Why when we look through a telescope in space, do the billions of stars not block our view from seeing further?

Yes they do, or rather not the stars but the dust and gas nebulae that are between them. However this is only a serious problem when we are looking through our own galaxy towards more distant objects.

Look at this apod from Jan 16 2016: It shows a relativly nearby galaxy, and one that would be a fairly easy object for amateur telescopes, if it were not aligned to the edge of the milky way in Cameleopardis. It can be seen, but only through a mess of nearby stars and gas. Galaxies that are behind the galactic centre in Sagittarius would be essentially invisible.

However, if we are not looking through our galaxy, but out of it, the nearby stars don't block our view this is because stars are so small compared to interstellar distances: If the sun were the size of a football, then Alpha Centuari would be on the other side of the world.

amateur observing - Can the Milky Way be seen with the naked eye? Does this apply to any galaxy? If yes, then how and when?

Yes, we certainly can see the Milky Way from Earth. I saw it through an airplane window at night. But you don't need to do that, either. Find some place with not much pollution, especially light and smoke, and gaze up at night. You should see a beautiful band of stars. Proof? In the 1990's, there was a major electricity blackout in Los Angeles. People looked up and saw it, but many of them thought that aliens were invading or something . . . so they called up 911.

All you need to know is that you can, you should, and hopefully, you will. As for other galaxies, you won't see them in quite the same way, because we're not situated in such a viewpoint. But you can see them, like the Andromeda Galaxy, if you find a dark place without pollution. Go to the countryside!

It's also possible at any time. Because the milky way is shaped like a bulging disc, and we're in a great spot, we can see it at any time; it spans across the whole Celestial Sphere!

multiverse - Existence of multiple universe?

We don't know. The Multiverse Theory is really just a hypothethis; sure, it makes sense, but it cannot be scientifically proven. We really have no reason to doubt that this is our only universe. Wormholes are just as hypothetical, since they would require infinite energy or negative energy to be created, neither of which we can get. Also, keep in mind that String Theory is, for the most part, unprovable. It cannot be confirmed through scientific experiments and is mostly guesswork.

Also, it will take trillions of years before the Universe gets too cold for life, and we will be long gone by then. However way you put it, it's unlikely that humans will ever reach other habitable planets before we go extinct, should our planet be destroyed. The closest potentially-habitable planet is nearly 480 lightyears from us – as in, it would take us 480 years to get there, if we travelled at the fastest speed in the Universe. We surely won't live to see the end of the Universe.

the sun - Procession of the Equinox

The sun does not orbit another star. A star, even a very small brown dwarf, would have been seen, if not in visible light then in the infra-red by the WISE survey.

However there is no mystery about precession. The Earth is non-spherical, it is wider around the equator, and because the Earth is tilted with respect to the ecliptic, there is an asymmetry in the force of the sun, and the moon, on the Earth. This force pulls the axis of the Earth's rotation in a circle. The basic maths of this is not so hard, and is in the Wikipedia page on Axial precession. This explanation of axial precession was known to Newton. There is nothing controversial about it.

The sun does orbit the galaxy it takes about 230000 years to complete one orbit. This has nothing to do with axial precession.

The page you link to is worthless. It offers an "alternate explanation" of a phenomenon that is very well understood, and pretends that there is some mystery, where there is none. It the uses this to justify a lot of astrological nonsense.

To repeat.

• There are not "many theories" explaining Axial precession.


• Axial precession is a well understood phenomenon, due to the Earth not being a sphere and the axis of the Earth not being perpendicular to its orbit


• The sun does not orbit another star. We know this because we could have seen it.


• No astrology based on the presence of another star is valid.

orbit - Where is Luna 1?

It currently is orbiting the sun, on an orbit that takes 450 days. It is an orbit that takes it close to Earth's orbit, and so is not stable in the long term. It may eventually collide with the Earth, but perhaps only after many thousands of years. It is not functioning (it doesn't have any way of generating power) and is pretty small, much to small to be picked up in Earth telescopes. We don't know exactly where it is now.

X-ray flare questions - Astronomy

I search the reference papers about flare, but nobody said something about below clearly. Could anyone give some help?

If a single X-ray decay without eclipse or other irrelevant factors is not able to fitted well by an exponential decay, can we say it is not a flare?
Can exponential decay be a criteria? Could somebody specify a flare with a weird shape or non-exponential decay? How large is the decay index (e**(-tau)) tau acceptable?

What kind of ratio(rising time/decay time) is acceptable?

If we cut a timing series of a flare into several parts, can we tell whether they are from the same flare by their X-ray spectra ?

spaceflight - How would FTL travel appear through window of ship?

Star Treck's warp drive assumes being able to establish a warp bubble around the space vessel to move slower than light in the local space, but faster than light in the surrounding space, to work around the limitations of special relativity.
A warp bubble would show some similarity to a worm hole. To get an idea of spacetime distortion caused by a wormhole you may like to watch some of the videos the University of Tübingen (Germany) is hosting.

The same site provides a simulation of a walk through the city with just below the speed of light.

In both cases you get visual geometric distortions.
But you get changes of the color by the relativistic Doppler shift, too, as simulated here.

When travelling faster than light you would get additional strange effects, depending on the method you overcome the speed of light barrier.
One well-known effect is Cherenkov radiation caused by charged particles travelling faster than the speed of the light in the medium, which may still be slower the vacuum speed of light. Vacuum speed of light in a flat spacetime can only be achieved by massless particles like photons.

Particles moving faster than the vaccum speed of light (tachyons) have been hypothesized, but have not yet been found.

Overcoming the speed of light would resemble somewhat falling into a black hole. When faster than light, light cannot reach you from the back. So this region would look black.

Travelling faster than light in a medium is physically feasible, at least for subatomic particles. Travelling with the vacuum speed of light or faster leads to serious problems, with the exact vacuum speed of light being the biggest challenge, since you either need to get rid of your rest mass, if travelling in flat space, or you need to warp spacetime. The latter needs huge amounts of energy causing devastating damage to the environment.

coordinate - Finding a location using the direction of shadow, UT and date approximately

Does anyone know how to find a location by using :

1. the direction of shadow relative to the normal (for example azimuth
   angle)


2. the Universal time (UT) and date

I have a satellite image and I know north, time and date. One of the cues that I can use is, for example, in the northern hemisphere the direction of shadow is north.

Would anyone know if it were possible, and how, to calculate the location based on these parameters?

the sun - What do we mean by Sunrise and Sunset?

Sunrise is the moment when the point you are located on, on Earth, gets to receive its part of the sunlight (due to the Earth's rotation you spoke about). And sunset is the moment when the point you are located on, on Earth, has made its turn and faces away from the Sun, then preventing you from receiving its light.

When the Earth will have made another half of its rotation, you will get to see another sunrise.

The Earth makes a full rotation in a 24h cycle. This is why you get a sunrise and a sunset everyday :)

space - Can an asteroid have a molten core?

Yeah, entirely possible.

There are a couple of examples of asteroids that we've identified which may have, or recently had, molten cores. 21 Lutetia is a notable example. Vesta is another.

The main issue is that because of their size, the ratio of surface area to volume is higher than the average planet. This means it cools much faster. So while some may have molten cores, they won't stay that way for as long as a planet might. Additionally, most aren't very near large gravity wells like a moon might be, so they don't have the same tidal forces adding energy to their interiors.

All this makes it less likely to find an asteroid with a molten core, but it's not impossible.

What happens when an Ultra Massive Black Hole cannibalize another?

From this article:

“Bigger black hole masses are in principle possible – for example, a
hole near the maximum mass could merge with another black hole, and
the result would be bigger still. But no light would be produced in
this merger, and the bigger merged black hole could not have a disc of
gas that would make light.”

The theoretical size limit has to do with the ability to form an accretion disk, not the hole itself. While it's probably an extremely rare occurrence due to space expansion, if a 50 billion solar mass black hole was to merge with another 50 billion solar mass black hole, you'd have a 100 billion solar mass black hole. There might be some interesting spiraling in towards each other and some gravitational waves made in the process. It would be cool to study, but in such a scenario, there's no force that would prevent the two black holes from merging into a bigger black hole.

asteroids - How long does it take Dawn to orbit Ceres?

As of March 6, 2015, Dawn has entered orbit around Ceres. But it's really so far only "captured by Ceres' gravitational pull". Then, it was still 61,000 km from Ceres.

It's slowly spiraling down into an orbit that JPL's Dawn Journal calls "RC3", which will be 13,500 km above Ceres. This orbit will last 15 days per revolution.

It will take about 15 days to complete a single orbital revolution at this altitude.

The first news link above indicates that Dawn will reach RC3 on April 23, 2015.

This picture from the Dawn Journal gives us a picture of Dawn's path: it's not completing any revolutions around Ceres yet; it's still approaching the RC3 orbit as of this writing (March 10, 2015).

So as of now, March 10, 2015, there is no orbital period yet. But when RC3 is achieved, it will be a slow orbit -- 15 days.

There are other orbital altitudes planned past RC3.

According to Dawn's schedule, the "survey orbit" will only last 22 days before it spirals in to the "high altitude mapping orbit" in August 2015, although it's unclear exactly when the "survey orbit" will take place. The "low altitude mapping orbit" will be in November 2015. In between orbits, when spiraling down to the next scheduled orbit, Dawn's orbital period will slowly decrease. At the last orbit, Dawn will be left in this orbit after its batteries and hydrazine fuel are exhausted.

---

Update (Oct 23, 2015)

Excerpt from latest news update from the Dawn mission (Sept 30, 2015):

Dawn is currently orbiting Ceres at an altitude of 915 miles (1,470
kilometers), and the spacecraft will image the entire surface of the
dwarf planet up to six times in this phase of the mission. Each
imaging cycle takes 11 days.

Starting in October and continuing into December, Dawn will descend to
its lowest and final orbit, an altitude of 230 miles (375 kilometers).
The spacecraft will continue imaging Ceres and taking other data at
higher resolutions than ever before at this last orbit. It will remain
operational at least through mid-2016.

radio astronomy - Which frequency should be used to communicate with a cube satellite?

I am trying to build a cube satellite using a raspberry Pi. I am trying to figure out which radio frequency could be used to communicate with the cube sat to transfer data like images, videos, different observations etc.

How to gain access to that frequency band? Is it a global organisation or you need you do that in your local divisions?

gravity - A camera and time dilation?

For simplicity, let's say that the black hole is isolated and non-rotating (and uncharged), so that the situation is described by the comparatively simple Schwarzschild spacetime. Let's also suppose that the camera free-falls radially into the black hole.

What is the camera looking at? Suppose it is looking at some stationary object that does something with a known frequency. Your question is basically how at what frequency it will be observed on the video feed emitted by the camera.

Without loss of generality, we can suppose that the camera is looking at us, and that we're shining a laser beam at it: the 'doing something at a known frequency' would be the oscillations in the electromagnetic wave of the laser beam. We can do this because time dilation affects every physical process, so we might as well pick one that is more convenient to think about.

At this point, it is straightforward why the camera feed will not show any time dilation: being equivalent to a reflected laser beam, the gravitational blueshift when going inward will be cancelled by the gravitational redshift going outward.

the sun - Is it possible to move a planet out of its orbit? At least a lighter planet?

yes it is possible by very few different ways.

https://www.uwgb.edu/dutchs/pseudosc/flipaxis.htm

Nothing acting solely from on or within the Earth could change its orbit or seriously alter its rotation.
One way to move an object is to throw mass in the opposite direction, the way jets or rockets do.
If we think really big and imagine blasting a chunk out of the Earth as big as North America and 100 miles thick so that its final speed, after escape, with respect to the Earth is 25,000 miles an hour, we will have expelled only 1/500 of the total mass of the Earth. The Earth would move in the opposite direction 1/500 as fast or 50 miles an hour. The speed of the Earth in its orbit is about 67,000 miles an hour. We will not change the orbit of the Earth very much--if we apply the impulse to speed up the earth in its orbit we would put the Earth into a new orbit with its most distant point about 70,000 miles further from the Sun than now--and the Earth's distance from the Sun varies now by three million miles over the course of a year! Exactly the same arguments apply to changing the orbit of the Earth through the impact of a large asteroid. The largest asteroid, Ceres, about 600 miles in diameter, is only about as massive as our hypothetical chunk of Earth above. Changing the orbit of a planet is a tall order. An impact big enough to have even a tiny effect on the Earth's orbit or rotation would almost certainly destroy all life on Earth as well.

http://usatoday30.usatoday.com/news/science/astro/2001-02-15-orbit.htm

hope this two links provides enough satisfactory explanation to you. if not do say so i will make further search to clear you the answer.