How can we tell that the Milky Way is a spiral galaxy?

The clues we have to the shape of the Milky Way are:

1) When you look toward the galactic center with your eye, you see a long, thin strip. This suggests a disk seen edge-on, rather than a ellipsoid or another shape. We can also detect the bulge at the center. Since we see spiral galaxies which are disks with central bulges, this is a bit of a tipoff.

2) When we measure velocities of stars and gas in our galaxy, we see an overall rotational motion greater than random motions. This is another characteristic of a spiral.

3) The gas fraction, color, and dust content of our galaxy are spiral-like.

So, overall, it's a pretty convincing argument. Of course, we have to assume our galaxy is not completely unlike the other galaxies we see--but I suppose once a civilization has accepted that it does not occupy any special place in the universe, arguments about similarity seem sensible.

Can someone explain a Geostationary Orbit in a simpler way?

@Hobbes put the formula:

$V={2pi aover T}$

If you substitute on it $T$ for 1 sidereal day you'll have a one-to-one relation between $V$ and $a$. This seems to imply that you can choose any $a$ and use the corresponding $V$, but this is not true since it does not take into account which force are you using.

The formula above is valid of any object in a circular motion, like a stone on a sling, or a car in a loop. It is also valid for satellites around planets.

But there is another formula stating a relation among $a$ and $T$ when the force is that of Gravity, so you are not free to chose $a$ anymore.

It happens that centripetal acceleration (that of the sling, the Normal force on the loop, or gravity for the satellite) is the Universal Gravity in our case:

$a_c={GMover r^2}$

but it being the centripetal acceleration means

$a_c=omega^2 r$

so we have

$omega^2 r={GMover r^2}$

$omega^2 r^3=GM$

$omega$ is $vover r$ so

$v^2 r=GM$

which gives you a fixed relation between $v$ and $r$ for any planet with mass $M$.

galaxy - Synthesising types of galaxies using various stellar spectra

I have asked this on Physics but Astronomy is probably more appropriate. I have been given the task of synthesising an elliptical galaxy, a starburst galaxy and a spiral galaxy (considering the arms and the bulge separately and adding them) using stellar spectra which I have classified. I need to add different weightings of five types of stars to produce a normalised spectrum of each of these galaxies. I have spectra for giants, main sequence stars and sub-giants with varying stellar types (O, B, A, F, G, K, M). I know that Spiral arms and the starburst are going to be primarily young stars, while the spiral bulge and the elliptical will be of older stars, but I have no idea which types or luminosity classes of stars I would use and their weightings in each. Can anyone offer any information or resources where I would find good information on adding different types of star to 'create' a galaxy?

space time - The multiverse spacetime paradox?

I'm not about to suggest that multiverse(s) do not exist, but merely that it is a fascinating, but highly speculative, topic. The Wikipedia article is a good start. From the wiki:

The structure of the multiverse, the nature of each universe within it and the relationships among the various constituent universes, depend on the specific multiverse hypothesis considered. Multiple universes have been hypothesized in cosmology, physics, astronomy, religion, philosophy, transpersonal psychology, and fiction, particularly in science fiction and fantasy. In these contexts, parallel universes are also called "alternate universes", "quantum universes", "interpenetrating dimensions", "parallel dimensions", "parallel worlds", "alternate realities", "alternate timelines", and "dimensional planes," among others.

From the way you phrase your question, you may want particularly to look at the sections on M-theory.

Does a planet's mass or gravity affect the height of it's mountains?

Yes, gravity definitely affects the maximum heigh of mountains.

Think on a solid bar of steel. It sticks straight because of the electronic forces. But when you make it larger and larger gravity makes it bend: gravity starts being considerable, but still smaller than electronic forces.

If you make the bar larger, there will be a moment in which the weight of the whole bar will be larger than the short-range electronic force: your bar will break purely due to gravity.

Exactly the same happens to mountains made of solid rock (as opposed to sedimentary ones quoted by Hobbes). There is a point,k depending on the strenght of planetary gravity, where it takes over short-range electronic forces, making the mountain collapse.

This is exactly the force that "rounds up" the planets, as opposed to the non-spherical asteroids.

orbit - Rotational speeds of the material forming planets should increase, not decrease

You need to consider which material the planets are formed of.

Terrestrial planets are formed from less material (way less), and with less momentum (since they were orbiting near from the Sun). Gaseous planets (which is another name for external planets, due to their material), also known as gaseous giants, are formed with a lot more of material with a lot more of angular momentum.

Angular momentum formula is $vec L=vec rtimes mvec v$.

Orbital speed formula is $vec v=vecomegatimesvec r$.

Thus: $vec L=vec rtimes mvecomegatimesvec r$

Since all vectors are perpendicular:

$L=r^2 m omega$

Thus a higher mass $m$ makes a higher momentum $L$, and a higher distance to the Sun $r$ makes a quadratically higher momentum $L$.

colonization - Is it possible to grow plants on Mars?

Some of the rocks on Mars aren't too bad for growth of plants or bacteria. They contain a considerable amount of clay minerals (about 20%), and several other minerals which are considered as essential and sufficient for bacteria to survive, at least.

Probable chlorinated salts, e.g. perchlorates, which are likely and wide-spread in Martian soil, don't favour plant growth, but there are also some soils on Earth containing perchlorates. Some plants, and some bacteria can cope with a certain concentration of perchlorates. But it would certainly be better to clean Martian soil from perchlorates before trying to grow plants on it.

Basaltic rocks are wide-spread on Mars. They need some weathering to be suitable for growth of plants.

Solar energetic particles, and galactic cosmic rays are more abundant on Mars than on Earth. That's not optimal for plants, but probably just acceptable. A severe problem is high ultra violet radiation due to a missing ozone layer on Mars. Plants wouldn't survive this uv level without shielding. Hence at least a protection by glass or something similar would be needed.

Carbon dioxide for the respiration of plants is present on Mars, but the atmospheric pressure is too low for water to stay liquid over a sufficiently long period. Hence a pressurized containment would be needed.

Btw.: There is permafrost in some regions on Mars; you just need to warm it, and keep it confined, to get water.

How does a telescope measure parallax angle?

No, the telescope doesn't measure the parallax. A sextant or any other angle measuring device fit on the telescope does.

And, we don't(can't) directly measure the parallax angle. Instead, we just track the position of the star/object throughout the year. A little bit of spherical astronomy math shows us that the path of a star in the celestial sphere defined by a fixed reference through the year is an ellipse around it's mean postion, called the parallactic ellipse

Here is a link which does the rough math involved.

The semi-major axis of this ellipse is equal to the parllax angle P, while the semi-minor axis is equal to P*sin(b), where b is the stars ecliptic latitude.

EXTRA NOTE: These parallactic ellipses are often rotated and parameters modified as the abberational ellipse is superposed on it. In practice, things get nigh complicated.

black hole - why quantum gravity doesnot exist?

I'm not sure what you mean by saying that quantum gravity "doesn't exist". But because this is the Astronomy SE, I will interpret your question as primarily asking why astronomy hasn't found evidence of quantum gravity. This is a reasonable question; after all, nineteenth-century astronomers have found evidence of funny business in the perihelion precession of Mercury, which in the twentieth century was understood as the relativistic corrections to the Newtonian orbits. Then astronomy one-upped that and found more evidence for general relativity.

So why not now? Because gravity is weak. Very weak. Weakness per se isn't a big problem theoretically; it makes it possible to calculate quantum gravity corrections when they aren't too large, which can be interpreted as a running Newton's constant:
$$G(r) = Gleft[1 - frac{167}{30pi}frac{Ghbar}{r^2c^2} + ldotsright]text{.}$$
For example, for the Sun's gravitational field near the Sun's surface (mass $M_odotsim 2times 10^{30},mathrm{kg}$, radius $R_odotsim 7times 10^8,mathrm{m}$), the relativistic corrections are of order $GM_odot/rc^2sim 10^{-6}$, a few parts per million, while the quantum gravity corrections are on the order of $Ghbar/r^2c^3sim 10^{-88}$, which is so much smaller as to be completely hopeless.

Making the Sun a black hole would not improve things by much, since in that case $r = 2GM_odot/c^2$ means while the relativistic corrections become appreciable, the quantum gravity corrections are on the order of $10^{-76}$. Note that larger black holes are actually worse, which is sensible because large-scale quantum gravity must agree with general relativity.

An interesting overview by Cliff P. Burgess can be found Living Rev. Relativity 7 (2004) 5, here.

Heat from other Stars - Astronomy

Does the Earth receive any heat at all from the millions of other Stars in our Galaxy ?

Effectively, no. Stars are too few and far between.

Qualifying that "effectively, no": From http://stjarnhimlen.se/comp/radfaq.html#10, the stellar magnitude from total starlight is -5. Compare that to the -26.7 magnitude of the Sun as viewed from the Earth. That difference of 21.7 means that starlight is responsible for one part in 10²² of the heating of the Earth. Another way to express one part in 10²² is "effectively none".

Another way to look at it: The Earth would eventually cool to 2.7 kelvins if the Sun and stars magically turned off. If it was only the Sun magically turned off, the Earth would cool to 3 kelvins. Compare that to the nice balmy 287 kelvins we experience thanks to the Sun.

Is it light that is bringing the heat and perhaps it cools down on the long journey in Space getting to the Earth and that's why no significant amount of heat is getting here ?

The stars in our galaxy are extremely close to us in a cosmological sense. Even the Andromeda galaxy is extremely close. The light we see from stars in our galaxy is more or less the same as emitted.

Over very, very long distances (much, much longer than the distance to Andromeda), the cosmological expansion of space means that light is redshifted. How much light is redshifted offers a clue as to the distance to some remote object.

We do receive a minuscule amount of energy from the cosmic microwave background. That radiation was not emitted by stars. It marked the transition from the very early hot and opaque universe to a cooler and transparent universe. The universe transitioned from opaque to transparent when the temperature dropped below 3000 K or so. Now that the light has an effective temperature of only 2.725 K.

light - In the chromoscope, what is the X-ray source between Ophiuchus, Libra and Scorpius?

I have absolutely no idea what the object is. Well, my guess was Scorpius X-1, but I haven't been able to prove or disprove that, so I'll leave that out for now.

The X-ray source is extremely bright; it dominates that region of the sky. In front of it, however, is a thick lined-shape object, in black. Unfortunately, I don't have a true X-ray baseline because of all the background radiation; however, my guess is that black represents a region practically devoid of X-rays - or a region where X-rays are blotted out. Therefore, I hypothesize that the black band is a disk surrounding the object - not necessarily an accretion disk - that is blotting out X-rays.

The circular formations in the hydrogen-$alpha$ spectrum appear to be unrelated. I have
three reasons to believe this.

• They're around stars. Switch quickly to the visible-light image, and you'll find that there appear to be visible-light sources, presumably stars, near the center of these formations. I suspect these are nebulae, reflecting some light from the stars.


• They're not unique. These objects are in many parts of the sky, mostly around stars.


• They're not properly aligned with the X-ray source. If these were polar jets, bipolar outflows, or something else related to the X-ray source's axis, they should be emitted at roughly right angles to the disk. This is clearly not the case; they are at perhaps 30 degree angles with respect to the disk.

fundamental astronomy - How many astronomers are there in the world today?

"The worldwide community of professional astronomers is only about 10,000; most are located in the us (with about 1,000 in the UK and 250 in Australia)." From So You Want to Be an Astronomer by Duncan Forbes.

Another source stated that the number of professional astronomers is about the size of a small town.

Is there an orbit that could cause regular debris showers like in the movie Gravity?

If I understand you correctly (and I haven't seen the movie ^_^), you want to know if it is possible for two different objects in orbit to come close to each other periodically, with roughly the same period as one of the object's orbital period.

Yes, that is possible, but not probable.

In ordinary two-body Keplerian celestial mechanics, an orbit's semi-major axis is the only orbital element that determines the orbital period. Therefore, one orbit may be circular and the other elliptic and inclined in all sorts of ways, but as long as the semi major axis of both is the same, they will have the same period.

However, this whole scenario is completely unstable for any real-world celestial body. The (relatively small) asphericity of the Earth, as well as the presence of the Moon for example, causes all orbits around it to drift in some way or another. If you put two objects in orbits like you describe, they will not likely meet up very often before never seeing each other again for centuries.

Would space-faring vehicles in interstellar space be in pitch darkness, show reflections from the nearest star or be bathed in light completely?

The brightness of the spaceship follows (almost) the inverse square law, meaning twice the distance from the star, the brightness will be a quater.
In the middle of nowhere, but within a galaxy, it would look like in a moonless, and cloudless night, far away from any artificial light source.
It wouldn't be pitchblack, but much too dark to read a newspaper. You would see the stars in the background, and the fuzzy band of the Milky Way. The starship would occult the stars; that way you would see the black silhouette of the starship.

At 630.1-fold the distance of the Earth from the Sun (about 3.65 light-days), it would look like at full moon (derived from apparent magnitudes of full moon and Sun)

But maybe the starship is illuminated artificially.

What do shape of universe mean?

I think it's useful to realize the meaning of "shape" is not the same as in "what is the shape of this object I'm holding." Because we are a part of the universe, "shape" is about how things work (eg, how light moves through the universe), rather than "is the universe a sphere, cube, hypersphere or some other 'shape'."

This question is probably too broad to be answered here. I suggest reading the WikiPedia article on the 'Shape of the Universe'.

gravity - Tide on the Moon

The tide would be locked in place, roughly, because the Moon always shows the same side to the Earth (tidally locked). But the height of the water at the location facing the Earth and the opposite side should be greater than the mean height, and at the limb, as we see it, the height will be lower than the mean but by half the magnitude of the high tide.

Imagine installing a pipe with cross-section 1 cm$^2$ from the top of the ocean at its high point down to the center of the moon and then back up to the top of the ocean at a point where the net tidal field from the Earth is 0. Fill it with water until one side is equal with the water level on one side and its level on the other side will automatically be equal to the water level there. The weights of the water on both sides need to be equal if it is static. For now, we will ignore that g changes when one gets deep into the moon. Therefore, roughly speaking $rho$gh should be the same in both sections of the pipe, and:
$$
rho(g_M - g_t)(h_t + r_M) = rho g_M r_M
$$

where $g_t$ is tidal acceleration
$$
g_t(Moon) = 2frac{GM_Er_M}{d^3}.
$$
Subscript M means Moon and E means Earth and d is the separation between the Earth and Moon. Solving for h$_t$ gives:
$$
h_t = frac{g_tr_M}{g_M - g_t} sim frac{g_t}{g_M}r_M
$$
We could substitute values in here, but if we do this in comparison with the height of the tide on the Earth, 54 cm, we cancel out some of the error in the approximation and get an even simpler formula, namely,
$$
frac{h_t(Moon)}{h_t(Earth)} = frac{M_E}{M_M} frac{g_E}{g_M} Big(frac{r_M}{r_E}Big)^2
$$
The mass of the Earth is 81.3 time greater, the acceleration at the surface of the Moon is 1/6.25 of Earth's g force, and the diameter of the Moon is 0.2725 of the Earth. So, that is 81.3*6.25*.2725$^2$ = 37.7 time higher than the Earth's tide or 37.7*54 cm = 20.4 meters. Note that the height of the ocean does not matter as long as it is considerably more than 20 meters.

On the other hand, the water would begin boiling since it is at 0 pressure until a water atmosphere is created, but the moon is too small to hold an atmosphere for long and the whole ocean would be gone in a while.

cmb - Units of angular power spectrum

I am not sure whether this question might be better suited for crossvalidated or stackoverflow, but I will give it a try here:

I have a map of the full sky in in the healpix format, units are $rm MJy, sr^{-1}$ (definition of $rm Jy$ here). To compute the angular power spectrum i.e. the $C_l$, I use the function anafast from the healpy library.

Now, I think that the resulting $C_l$ have units of $(rm MJy,sr^{-1})^2, sr$, but I could not find anything on the units in the literature so far.

I'd very much appreciate any help regarding the units or links to familiarise myself further with this topic.

gravity - Could there be dark matter black holes?

The problem with trying to form a black hole with dark matter is that dark matter can only weakly interact (if at all) with normal matter and itself, other than by gravity.

This poses a problem. To get dark matter concentrated enough to form a black hole requires it to increase its (negative) gravitational binding energy without at the same time increasing its internal kinetic energy by the same amount. This requires some sort of dissipative interaction between dark matter and normal matter (or itself).

The following scenario should make this clear. Suppose we have a lump of dark matter that gravitationally attracts another lump of dark matter. As the two approach each other, they accelerate and gain kinetic energy. The kinetic energy gained will be exactly enough to then separate them to a similar degree to which they started, unless some dissipative process takes place.

An example is to suppose that dark matter is weakly interacting massive particles (WIMPs). WIMPs are gravitationally drawn towards the centres of stars. If the weak interactions happen sufficiently frequently then it might be possible for them to accumulate in stars, rather than shoot through and out the other side.

It has been hypothesised that black holes could be made like this near the centre of a Galaxy, seeded by dense neutron stars. The density of neutron star matter, combined with the enhanced density of dark matter near galaxy centres could result in dark matter accumulation in the neutron stars, leading to the formation of black holes.

Once a black hole is formed then any dark matter that enters the event horizon cannot emerge regardless of what kinetic energy it gains in the process. However, there is still a problem. Material in orbit around a black hole has less angular momentum the closer it orbits. To pass inside the event horizon requires the dark matter to lose angular momentum. Normal matter does this via an accretion disc that can transport angular momentum outwards by viscous torques, allowing matter to accrete. Dark matter has almost zero viscosity so this can't happen.

So building a supermassive black hole from a smaller seed would be difficult, but forming small black holes out of neutron stars might be easier. It has been proposed that a relative lack of pulsars observed towards our own Galactic centre could be due to this process.

Distance between Earth and where The Big Bang's happened?

The distance to where the big bang theoretically happened would be "0" (and this should be true for any point within the universe). The entire universe was theoretically produced in the big bang, with every point in space being "inside" the big bang. The whole universe is very special, it is amazing what is out there (including black holes, earth, stars, planets, nebulae, etc...).

Whether or not there is even a "center" to the big bang / the universe, however is another related interesting question. Current models seem to indicate that there is no center. It would be interesting if there is a true center of the universe, and would be interesting to know if there is anything special about the center if it exists.

The question of whether or not there is a center of the universe is addressed in the following links:

What is in the center of the universe?

http://www.spaceanswers.com/deep-space/is-there-a-centre-of-the-universe/

orbit - Do all orbiting bodies eventually collide?

Yes.

Two bodies in orbit around each other will inevitably collide. The reason for this is that the system will give off energy in the form of gravitational waves. This effect is commonly cited in binary neutron star systems, where the two stars are isolated and close together. One of the most famous of these systems is the Hulse-Taylor binary.

The time it will take for the objects to collide can be calculated:
$$t=frac{5}{256}frac{c^5}{G^3}frac{r^4}{(m_1m_2)(m_1+m_2)}$$
where $r$ is the initial radius, $m_1$ and $m_2$ are the masses of the bodies, and $c$ and $G$ are the familiar constants, the speed of light in a vacuum and Newton's universal gravitational constant.

However, tidal acceleration could offset some of the effects.

What is the brightest star (relative magnitude) in M31?

The Variable stars that Hubble studied in M31 when he showed that it was a galaxy are among the most luminous. They include Var-A1, a luminous blue variable, and magnitude about 16.5. Var A-1 is one of the most luminous stars known.

Nasa has a catalog of stars in the m31 field, The brightest star in this catalog is an 11.4 magnitude star. But I think that database needs careful interpretation.

Did big bang happen only one time or it might happen again somewhere million time away from our observable universe?

We don't know it, because the question isn't feasible thus far to experiments or observation.

But it's possible to develop theories far beyond our observable universe, even far beyond the multiverse theories described here in Wikipedia. There exist theories about a tranfinite hierachy of meta-levels of theories in mathematical physics, each level comprising all possible theories on the level below (beyond any notion of infinity) and all their interconnections (including hierarchies of interconnections between interconnections). To explain this in reasonable detail would need hundreds of mind-boggling pages, at least. So I won't even try to explain more details. At the end it's not even clear, whether there is a difference between mathematics (including a transfinite number of meta-levels) and physics (including a transfinite number meta-levels).

The anthropic principle tries to answer the follow-up question "Why do we live in exactly this universe with exactly these laws of nature?". But this kind of answer isn't satisfying from a scientific point of view, when looking for cause and effect.

If you're looking for a final answer comprising everything: There is no, because the notion "everything" leads to paradoxa, when used in an absolute sense. (See Gödel's incompleteness theorems, if you need more details.)

the moon - The RA and Dec of lunar poles

The "small vibration" is Libration, and it does not imply that lunar axis changes orientation. It is caused by Moon's movement not being a perfectly circular orbit parallel to Earth's Equator with aligned axis.

Here's a description of the causes of libration from the Royal Observatory (check the linked page for graphics of each source):

Synchronous rotation and libration

The 27.3 days it takes the Moon to complete one orbit around the Earth is about the same as the time taken for it to complete one rotation. This synchronous rotation means that it always shows the same face to the Earth.

However, libration means that it is possible to see around 59% of the surface. It has three components:

i) Diurnal libration

As the Earth rotates the view of the Moon changes. An observer on the surface of the Earth sees slightly around the eastern limb of the Moon at moonrise. At moonset the same observer sees slightly more around the western limb of the Moon.

This arises from the changed position of the observer because of the rotation of the Earth. A parallax effect means that the Moon looks slightly different at moonrise and moonset.

ii) Longitudinal libration

The Moon moves in an elliptical orbit around the Earth. As a result it does not move at a constant speed (this is Kepler's second law)

However the Moon rotates at a constant speed. When it is moving faster in its orbit, this allows observers on Earth to see around the trailing edge of the Moon. When it is moving slowly it is possible to see around the Moon's leading edge.

iii) Libration in latitude

The Moon's axis is tilted by 6.7° to the plane of its orbit around the Earth. This means that sometimes the Moon's North Pole is tilted towards the Earth and sometimes tilted away.

As a result it is sometimes possible to see beyond the lunar North Pole and the South Pole is hidden. Conversely it is sometimes possible to see beyond the South Pole and the North Pole is hidden.

The combination of the three kinds of libration allows observers on Earth to see 59% of the lunar surface. However 41% remained hidden from view until the space age.

So in order to get Moon's poles position you just need to know where the Moon's center is, which is quite standard, and the position of the poles referenced to that center, which is always the same (not in RA/Dec but in true space).

There is a secondary effect of "wobbling" called Nutation, which is caused by the Moon on the Earth, not a movement on Moon itself.

redshift - Wher can I find a list of blueshifted regions in the Milky Way?

I don't think there is a catalogue containing information about blue/red shift per region of the sky. You probably need a catalogue containing blue/red shifts per object and then calculate the average blue/red shift per region.
Most catalogues with blue/red shift are however targeted towards galaxies. They may also contain stars but you will have to filter the stars from the background galaxies. If I remember correctly, the Sloan Digital Sky Survey (SDSS) contains a flag specifying whether an object is a galaxy or a star. So you might use the SDSS data to get a large set of stars with blue/red shift. This flag has been set automatically using pattern recognition. I do not know how good the classification was.

You can access the data at the SDSS SkyServer. There are a lot of objects in SDSS so you will need time to acquire the data and knowledge about SQL to query the database.

Even then, you still don't have information about the spiral arm in which the object is located although you might use a method such as was used by Belokurov et al. (2006) to detect the field of streams to separate the different spiral arms.

galaxy - Which structure is represented in this video (if any)?

I think it is supposed to be a representation of the Centaurus A/M83 Galaxy group.

https://en.m.wikipedia.org/wiki/Centaurus_A/M83_Group

These are a bunch of galaxies between 12 million and 15 million light years away.

The big central dust lane in Cen A is very recognisable.

I guess the sequence then continues through the Virgo cluster and supercluster which are roughly in the same direction.

comets - Why all the photos from 67p are black and white?

Regular digital cameras color filter arrays that put the pixel sensors so that it only detects a particular color. Therefore, there is a trade-off, because without the filter array, you would have a significantly higher-resolution black-and-white image.

However, most scientific digital cameras work differently. Rather, the usual technique is to put filters over the entire image, and switch out the filters if a different color is required. Subsequent differently-filtered images can then be combined into one image. But there's still a trade-off: putting in filters decreases the dynamic range of the camera significantly. Additionally, this approach does not require one to use the usual red, green, blue scheme; e.g., and IR filter would allow one to emphasize different features.

It is very likely that they simply didn't bother to put color filters in the cameras. Why would they? The colors are probably rather drab on the comet anyway, and it unnecessarily increases the complexity of the cameras (which is already a quite complicated panoramic array).

Red cresent moon - Astronomy

Yesterday night i witnessed something very strange when i looked outside the window. I saw the moon (crescent) but it was dull red and right on the horizon ,which is strange considering that it is usually on the upper right of the night sky and white in colour. On further inspection with my binoculars i noticed it was lowering down until it was hidden by the mountain range (5km away) next to my building, this all occurred within a few minutes (about 5).

Tonight i saw the moon (crescent) had again returned to its normal position.

Please explain the cause for this, i'm completely baffled!

(Sorry for the poor wording, i'm not familiar with all the astronomical terms!)

How are the speed, distance and time of India's MOM and USA's Maven Mars missions calculated?

First of all, trajectories of interplanetary missions are not designed to minimize the TRAVEL TIME but to minimize the COST, which is directly related to the fuel required to transfer the probe between the Earth and another planet. One can easily find a solution on how fast and in what direction should the probe travel, and how much time and energy will be spent after departure from the near Earth orbit by solving the gravitational force equations. But that's far not enough to launch a real probe. Scientists have built accurate kinetic models and databases of the motions of solar system objects, such as DE405, by NASA Jet Propulsion Laboratory. Some of them even take account of the perturbations of all major planets in the solar system. Engineers should also build models for the probe itself. For example, how much acceleration will the spacecraft get when turn on the booster, and how does the solar radiation pressure change the attitude angle during the trip. What's more, every probe carries at least one gyro system. Optical guiding telescopes are also widely used to calculate the attitude angle. The speed of the probe can be calculated by the Doppler shift of the radio frequency.

For more details, see the book Interplanetary Mission Design Handbook