29th March 2013

Photo

Ikertorony of the day

Erdős Pál vs. Jeff Goldblum

Erdős Pálnak meghatározó szerepe volt a matematika hasznosíthatóságában: bár ezt még ő is csak később ismerte fel, valójában már jóval azelőtt szoftvereket készített, hogy a számítógépet feltalálták volna.

Tagged: matekikertoronyof the day

15th February 2013

Photo reblogged from Sandor's Tumblr with 4 notes

borcsok:

Nap képe: üdvözlet a Masatnak Marokkóból

“Nagyon meták vagyunk: a ma egy éve az űrben keringő Masat-1 műholdnak a Marokkóban ténykedő Puli Space rover kameráján át küld jókívánságokat az ÖWF szimulált marsi asztronautája. A Masat csapat ma tart összejövetelt a jeles alkalom megünneplésére. (Amin persze én nem tudok ott lenni, mert a Piszkéstetőn vigyázok épp a távcsövekre.) Egyébként a Masaton kívül csak a spanyol Xatcobeo érte meg épségben az egy évet, illetve a lengyel PW-Sat is működik még, de az több hardverhibával is küzd, és a kommunikáció igen nehézkes vele.

Ami meg a Pulit és a MARS2013-at illeti, a rover a marokkói sivatagban randalírozik nagy lelkesen, az eddigi fejleményekről pedig ezt illetve ezt a posztot érdemes elolvasni.”

http://cydonia.blog.hu/2013/02/13/nap_kepe_udvozlet_a_masatnak_marokkobol

magyar űrkép of the day

Tagged: of the dayűr

8th December 2012

Photo reblogged from Exploring space, for all mankind. with 52 notes

sagansense:

Does the Moon Orbit the Sun or the Earth?

image: NASA

Everyone knows the moon orbits the Earth and that the Earth orbits the Sun. But what about the path of the moon around the Sun? What does that look like? I will go ahead and state that this is a difficult thing to show. Why? Scale, that’s why. Let me give some values for the sizes of these things, then I will make some sort of diagram.

Radius of the Sun: 6.95 x 108 m. Radius of the Earth: 6.38 x 106 m. Orbital radius of the Earth around the Sun: 1.5 x 1011 m. Radius of the moon: 1.7 x 106 m. Orbital radius of the Moon around the Earth: 3.48 x 108 m.

Here is one attempt at drawing these three objects.

The objects all have the correct scale – but the Sun is in the wrong place (hopefully, this is obvious). The Earth and moon are the correct distance apart relative to their size. What about the Sun? In this diagram, the Earth and moon are about 11 cm apart (at least on my monitor). If you could see the whole Sun, it would be about 40 cm across. Where would the Sun be? You if you had a cut out of Sun, that piece of paper would have to be 43 meters off to the side. Yup. I said 43 meters. The sun is pretty far away.

And this is the problem. How do you show the orbit of the Earth and moon around the Sun? You really can’t, at least not to scale. Most textbooks end up making a plot where nothing is to scale. Here is something you might see.

It sort of works, right? It shows that the Earth orbits the Sun and the moon orbits the Earth. But what would it look like to scale? I am not sure what the best way to show this will be. Let me first assume perfectly circular orbits for both the moon and the Earth. I’m not going to show the Sun – here is just a part of their path.

This just shows half a month. If I wanted to show a longer time period, the motion of the Earth and moon around the Sun would make it super-difficult to see the motion of the moon relative to the Earth.

Maybe it will help if I plot the distance from the Sun for both the Earth and the moon. Here is that plot over about 1 month.

You might notice that the distance from the Earth to the Sun changes. I put in an initial velocity to give the Earth a circular orbit. However, I also included the gravitational force on the Earth from the moon. This causes a bit of a wobble (but it isn’t really important for this discussion).

The real question that I want to look at is: does the moon orbit the Earth more or the Sun more? Which is more important? Let me try another plot. Here is the radial component of the acceleration of the moon over one month. Remember, from my last moon post we can break the forces (and thus the acceleration) into two types. There is a radial component that changes the direction of the momentum and a parallel component that changes the magnitude of the momentum. So, this is just the magnitude of the radial component.

But what does this even mean? Well, this says that no matter where the moon is in relationship to the Earth it has a radial acceleration in the direction of the Sun. It does not accelerate away from the Sun. If it did, it would have a negative radial acceleration component. Why does this look so similar to the radial position plot? Think of this. When the moon is on the far side of the Earth (so farther away from the Sun than the Earth is), it has both the Earth and the Sun pulling it towards the Sun. With this greater force comes a greater acceleration AND it is further away from the Sun. So, the plots look similar but they are not the same.

Let me show this with a diagram (not to scale).

Here I have shown the gravitational force on the moon from the Sun is greater in magnitude than the gravitational force from the Earth. Is this actually true? To calculate this, I need to use the following model for the gravitational force:

This shows the magnitude of the gravitational force on the moon due to the interaction with the Sun. G is the gravitational constant (6.67 x 10-11 N*m2/kg2) and r is the distance between the Sun and the moon. Even though the moon moves around the Earth, this Sun-moon distance doesn’t significantly change. If I use this gravity model, I can calculate the force per unit mass for an object in the location of the moon due to both the Sun and Earth.

Gravitational force per mass due to the Sun = 0.00589 N/kg Gravitational force per mass due to the Earth = 0.00270 N/kg The Sun wins. But wait. How close would an object have to be for the gravitational force from the Earth to be greater? Here is a plot.

At an orbital distance of about 2.6 x 108 m, the force from the Earth and the Sun would be equal. This means that when an object was on the Sun side of it’s obit, it would for a moment have a zero radial acceleration. For objects orbiting closer than this, the gravitational force from the Earth would be greater. This means that it would be accelerating towards the Earth and not the Sun at some parts of the orbit.

Two Answers

There are really two questions here. Let me answer both of them.

Does the moon orbit the Sun? I would say yes. The interaction between the Sun and the moon has a greater magnitude than that of the moon-Earth interaction. The moon moves around the Sun at the same time it moves around the Earth. Perhaps the best answer is to say the the moon interacts with both the Earth and the Sun at the same time. This is what we call “physics”. I don’t think you could say that the moon just orbits the Earth.

How do you represent the path of the moon as it goes around the Sun? You don’t? I don’t know. How do you represent the scale of the solar system? Again, this is a tough problem. You really can’t do it in a textbook, can you? If I had to make a recommendation, I would tell introductory astronomy textbooks to NOT draw that squiggly moon path diagram. I don’t think that helps anyone understand anything important.

szemfelnyitó csillagászati cikk of the day - és még csak 00:04-van.

Tagged: űrof the day

Source: sagansense

30th November 2012

Photo reblogged from Űrrepülés with 2 notes

urrepules:

Szaturnusz, valamelyik pólus 

(via Cassini Solstice Mission: Swirling Storms on Saturn)

ez a kép …

Tagged: űrof the day

17th November 2012

Photo reblogged from Exploring space, for all mankind. with 177 notes

egyértelműen quote of the day

Tagged: of the day

Source: coloradocosmonaut