How to set up a to scale model of the distances between planets of the Solar System on a football field?
I want to set up a model of the solar system having distances between the planets to scale. This will be on a football field 100 meters in length. So how do I make the distances between the planets to scale on a football field that's 100 meters. I was thinking ratios or proportions but I'm really not sure.
Answer:
Sun is sitting in one goal, Pluto is 100 yards away. (we're still going to count pluto here :)
Now we look up Pluto and find that he's 3670052070 miles away. He's our constant.
Neptune is 2829691160
2829691160 / 3670052070 = .7710
Neptune is 77 meters away from the sun.
Uranus is 1783950480 / 3670052070 = .4861
Uranus is 48.6 meters from the sun.
etc.
You can google up and divide the rest yourself. At the bleachers begin the Oort Cloud, and it runs out past the parking lot. :)
How to get started in astronomy?
When I was younger (5th grade-ish) I was really into astronomy and memorized a bunch on constellations and learned all about our planets, etc. It's been about 8 years and I've forgotten a lot of it.
How would I go about getting back into astronomy? Where is a good place to dive in at?
Answer:
Read, watch astronomy and space related videos, take a class at your local community college.
A really good book to get started -
http://www.amazon.com/NightWatch-4th-Edition-Terence-Dickinson/dp/B00327G1X8
A decent, general purpose astronomy magazine -
http://www.astronomy.com/
Online education -
http://www.khanacademy.org/science/cosmology-and-astronomy
If there is one in your area, check out your local astronomy club or attend a star party -
http://www.go-astronomy.com/astro-club-search.htm
How to determine distances by using apparent and absolute magnitudes?
Good day everyone,
After approximately 7 hours from now I have a final exam on Astronomy class and I`m confusing about how to determine distance by using apparent and absolute magnitudes?
Could you please explain that for me plus the meaning of both M and m by a basic explanation?
Thank you very much I really appreciate what you will offer of help to me.
Answer:
The little m stands for apparent magnitude and is what the celestial object looks like in out night sky. The big M stands for absolute magnitude and is how bright it actually is at 10 parsecs. Now if an object has an apparent magnitude of -1 and an absolute magnitude of +8 you know that it gets dimmer if you are moving it to ten parsecs. Therefore it must be closer than a star that has a m of -1 and a M +4 because star 1 gets dimmer by 9 levels however star 2 only gets dimmer by 5 levels. This also works for stars getting brighter as you move it to 10 parsecs.
How to assemble Orion Spaceprobe 130ST EQ Reflector Telescope?
We received the telescope but no instructions. Hmm.
Please type the instructions down so we can print it out!
Thank you so much.
Answer:
The source has links to the manual and videos showing how to assemble it.
How to calculate the main sequence lifetime of a star ? Give your answer in seconds and in years.?
I have the proper information such as the solar mass and solar luminosity of the star in which I am trying to find its main sequence lifetime, but I need help setting up the equation and converting my answer into both seconds and years. Thanks so much
Answer:
Using the sun's expected lifespan of 10 billion years, it's mass as 1 and it's energy output as 1 it's not too hard to calculate a star's lifespan. To illustrate, Sirius's mass is 2.2 times that of the sun. It's energy output is 25 times that of the sun. Divide 2.2 by 25 and you get .088 .Multiply this figure by ten billion and you get 880 million years. That would be the expected lifespan of Sirius. It must be pointed out that these are only rough figures but it still gives you a general idea of the star's life expectancy.
Interestingly, lower mass stars such as Barnard's star will have far longer lifespans. Possibly in the hundreds of billions of years, maybe even trillions. If you do the math using what I already provided you'll see what I mean.
how to get the satelite view of earth at night?
I tried to see all lights at night time in any goolge earth & google maps but i couldn't do it.So plz help me friends with the solution.
Answer:
I think you're looking to view light pollution as seen from space. The link below has many satellite pictures showing that.
How to Calculate a celestial objects with two observers in different location.?
For example, person 1 is in Paris with latitude of 48.48 N and longitude of 2.20 E and Person 2 is located at Osaka, Japan with 34.32 N and 135.30 E. Person 1 observed the object with the Azimuth of 200 degrees and an altitude of 70 degrees. What might be the azimuth and altitude of the object in Person 2's location?
Answer:
My way may be slow, but I'd do it by first trying to find out
the geographical point where the object would be directly over head.
Let's call that point S.
The length of arc PS is 90-70 = 20 degrees,
and the angle NPS is 200 degrees,
where "N" is the north pole.
Arc NP is 90 - 48.48 = 41.52 degrees.
You can use the law of cosines to find arc NS:
cos (NS) = cos(NP) cos(PS) + sin(NP) sin(NS) cos(angle NPS)
All of the quantities on the RHS are given above; I get
cos (NS) = 0.49052
which would make NS = 60.625 degrees,
so the latitude of S would be 29.375 North.
Next, the longitude of S can be found by using the same law again:
cos (20) = cos(60.625) cos(41.52) + sin(60.625) sin(41.52) cos(angle PNS)
where all arcs are in degrees, and angle PNS is the longitude difference
between Paris and S.
Note that S will be somewhere EAST of Paris, not west.
Once you have the latitude and longitude of S,
construct a new spherical triangle ONS.
You know NO = 90-34.32 = 55.68 degrees, NS = 60.625 degrees,
and angle SNO is the longitude difference between Osaka and S.
Then
cos (OS) = cos(55.68) cos(60.625) + sin(55.68) sin(60.625) cos(angle SNO),
where the last term was found in the previous part of the problem,
and cos (OS) will give you the altitude of the object as seen from Osaka,
because OS = 90 minus the altitude of the object.
Lastly, the azimuth in Person 2's location will be obtained by solving
cos (NS) = cos (NO) cos (OS) + sin (NO) sin (OS) cos (angle NOS), or
cos (60.625) = cos (55.68) cos (OS) + sin (55.68) sin (OS) cos (angle NOS).
Here the arc OS was found in the preceding step,
and the angle NOS will give you the azimuth from Person 2's location;
you just have to pay attention to whether the object is east or west of
Osaka (I'm GUESSING it's west), to know if the azimuth is bigger than 180 degrees,
or if it should be considered negative.
How to calculate distances within our Universe?
I need to create an A4 fact sheet on the main ways to calculate distances within our universe. Its for my science coursework.
Answer:
Here's a whole bunch of ways -
http://www.astro.ucla.edu/~wright/distance.htm
How to estimate distance to star cluster?
So I am given a chart of stars, their B-V colour and their Apparent magnitude.
I am asked to plot the colour - magnitude diagram and estimate the distance and age of this cluster.
How do I do this as you need the absolute magnitude for the HR diagram?
I made a diagram with the colour and apparent magnitude, but could you please elaborate on 'aligning the vertical scale to give the distance modulus'.
Thanks,
Answer:
As it's a star cluster the choices are fairly limited.
If the stars in question are mainly yellow or red, then it's a globular cluster and it's quite old. If they are mainly white or blue stars, then it will be an open cluster and quite young.
Without the distance, it's not possible to deduce the absolute magnitude.
However, the distance can be deduced from the brightness and comparing it to a similar star of known brightness and distance.It's called brightness measurements
The age will be deduced from it's spectral type. Red old, blue young.
How to show constellations in a different perspective?
I am doing a project in which I have to show two constellations from a different perspective, and I'm not sure how to do this. I thought about making a diorama inside a box with both the top and bottom cut out so that I could just show the constellation in my perspective, then turn the box to see it from behind...But I'm not sure if that's a good idea or not. Perhaps a clear box would be better, so one could see it from all sides?
Any ideas would be great!
Answer:
Actually, that sounds like a pretty good idea!
You might show this as the different perspectives that Northern and Southern Hemisphere residents have for a couple of the constellations.
One grouping in particular that you might show would be Orion, Canis Major, Canis Minor and Lepus.
If you can draw the pictures of the greek representations of the constellations (or use a good copy of them from a star atlas or program such as Stellarium), and then have the major stars shown, with different sized holes (larger for brighter), you could have quite an interesting diorama!
