How can I calculate a scale model of the solar system only knowing that the sun is 6 cm in diameter?
112968
Here is a site where you can find all the needed data.\nhttp://seds.lpl.arizona.edu/nineplanets/nineplanets/nineplanets.html
680640
Ask the designer of the Solar System he probably know.
112212
First solve for the mass of the planet M\n\nM = 5200 kg/m^3 x 4/3 x 3.14 x (1600 x 10^3 m )^3\n  = 5200 x 4/3 x 3.14 x 4.10 x 10^18 kg\n  = 8.93 x 10^22 kg\n\nNext, use the formula for centripetal force, and set it equal to the gravitation force between the object and the planet.\n\nG (Mm)/r^2 = m v^2/r , where r is the total radius of the object, that is 1600 km  + 630 km = 2230 km.\n\nDividing out the common factors of m and 1/r, we have\n\nG M /r = v^2\n\n(6.67 x 10^-11)N m^2/kg^2 (8.93 x 10^22 kg) / (2.230 x 10^6 m)\n\n2.67 x 10^6 N m /kg = v^2\n\n1.63 x 10^3 m/s = v\n\nCheck my math to make sure, but the physics is correct.
774198
i'm not sure mabey u can do a model of the solar system and show haow it works
1044084
That's easy enough to calculate with some basic trigonometry.  If you draw the angle (a) you are looking for, you see that you can bisect it to form two right triangles whose smaller angle is (a/2) and where one leg equals 5 light years and the other leg is half the diameter of the star.  The diameter of the sun is about 1.39x10^6 km, and a light year is about 9.46×10^12 km, so using these values, we find that the angle a is:\n\ntan(a/2) = 1.39x10^6km / 5*(9.46x10^12km)\n\nSo...\n\na = 2*tan-1 [ 1.39x10^6km / 5*(9.46x10^12km) ]\n\n= 5.877x10^(-8) radians\n\n...or about 0.000003367 degrees.
732650
People usually don't do the Sun to scale when they do their model. They do it smaller. And they'll use distances between planets that are not to scale for the real distances from the sun.\n\nThat's the straight answer. Use artistic license, since using scale is not practical.
799579
To get the Centripetal force(the force acting inwards to the center of the radius in this case the sun)\n\nwe use the formular angular force=mass*Acceleration\n=mass*(angular velocity^2/radius)\nsince you have already provided the nessesary requirements lets get started>\n\n=5.9742*10^24*(29.8/93,000,000 Miles)\nsince the sun is93,000,000 Miles away from the earth we use the distance from the earth to the sun as the angular radius.\n from mile to kilometers we convert to get\n93 000 000 mile = 149 668 992 000 meters \n\nnow centripetal force=5.9742*10^24*(29.8/149 668 992 000 )\n\n=1.991060379427156160709627816562e-10kg\nthat is the force that acts on the earth inwards
13641
There is no actual picture of the entire solar system. If one could be made, it is very unlikely that you would be able to see all the planets. I am sure you would only see The gas giants Jupiter and Saturn - as dots; maybe you will see Neptune and Uranus, that is about it. From earth if you look at Jupiter in the night sky it is as big as a star - though one of the brightest objects in night sky. From a position where you could see all the planets (the solar system is bigger than that, but let us restrict ourselves to only Pluto's orbit) you would see it as much smaller. Jupiter is 1,600 times bigger than earth (volume). Sun is a million times bigger. From most cities you can see Mars, Saturn, Jupiter and Venus. From a position where you could take a picture of all the planets you wont see all of them.\n\nEDIT: btw, I just remember I was talking about visibility of sun and planets. Actually if a photograph were taken jupiter and sun will be visible. The Sun is a star it will be visible not due to its size but due to the amount of light it gives. If we plotted it on a graph based only on its size and not its luminosity, it will be as visible as anything on the graph and no more. So if we drew an illustration 1024 pixels wide (most 17 inch computer screens are 1024x768 pixels) and put sun on the left most pixel and Neptune on the right most pixel - none of them will be visible. The distance between sun and earth is 1 AU (1 astronomical unit). Sun's Diameter is 1/150 AU, Neptune is 30 AU away from sun. If we scale 30AU to a 1024 pixel screen every pixel is equal to 0.029 AU, anything smaller than 0.029 AU can not be represented on this screen, the sun is 0.006AU wide - 1/5th of a pixel. If you open MS Paint and put a dot with a pencil that is one pixel imagine 1/5th of that - that is sun's size scalled to a 1024x768 screen.\n\nAnyway, Cronides' second link is the answer you are looking for.
886238
Remember the third law:  P^2=ka^3, where k simply is telling you to use consistent units (e.g. Earth years and a.u.)  So the square of 5[years]=cube of distance in a.u., or cube root of 25 is the distance you want, expressed in astronomical units.  I'll let you finish this, but as a hint, it falls in the outer part of the asteroid belt (Ceres has a period of 2.72 years).
41352
Well, the diameter of the planet is 12,756.3 km.  That would make the radius approximately 6,378 km.  So the radius shoud be an approximation of the distance between the inner core and the surface.\n\nAh ha!  I just look it up and I was right! "Inner core (5000 km - 6370 km below the Earth's surface)."\n:o)
137156
Well it depends on how old they are. I would get a big ball of plasticine. I would make it big and round, and then show them how it started out spinning. Then as it spun, it spread out like a plate. Take a lump and roll it for the sun, and smash the rest flat so it is a big flat plate. Cut out a hole in the middle and place the sun in it. Then get a knife an cut it into annuli, and then roll those annuli into balls, and place them at different distances from the "sun" ball. There, you have made yourself a big spinning solar system with a sun and planets. \n\nAnd you have had fun rolling the plasticine into balls and smashing it and cutting it etc.
467892
The *relative* distances between the planets can be determined from Kepler's laws and Newtonian physics. This will give the ratios betwen all the distances in the solar system. Once this is done, only one *absolute* distance needs to be found to get all of the rest. Until fairly recently, absolute distances were determined through parallax. It is possible to watch a transit of Venus from two different locations on earth and get a parallax measurement of the distance to Venus in that way. This gives the actual distance between the earth and Venus, and so the distance from the earth to the sun.\n\nOther methods involved parallax measurements of asteroids that pass close enought to the earth to get good measurements. Now, it is much easier to use radio bounced off of other planets (or probes) and the speed of light to get the required absolute distances.
800147
Average radius (in Astronomical units) of the orbit of:\n\nMercury = 0.38710 AU\nVenus = 0.72333 AU\nEarth = 1.0000 AU\nMars = 1.52369 AU\nJupiter = 5.20283 AU\nSaturn = 9.53876 AU\nUranus = 19.19139 AU\nNeptune = 30.06107 AU\n(Pluto) = 39.52940 AU\n\n1 Astronomical Unit (AU) = 1.49597870 E11 meters
1075138
The Earth is about 13 thousand kilometers (8000 miles) wide, whereas the Sun is roughly 1.4 million kilometers (900,000 miles) across. This means it would take more than 100 Earths to span the width of the Sun! If the Sun were a hollow ball, you could fit about 1,300,000 Earths inside of it!
109975
There are two very good websites.\n\nhttp://solarsystem.nasa.gov/index.cfm\nhttp://www.nineplanets.org/
9885
To answer this question, I'm going to assume the Earth is a perfect sphere.\n\nThe Earth's average radius is about 6.37*10^6 m. Then you can use trignometry to figure out the radius at 38 degrees north.\n\ncos(38 degrees)=r/6.38*10^6 m\nr is about 5.03*10^6 m\nTherefore the diameter is about 1.01*10^7 meters.
731974
You should be doing your own homework.\n\nYou need to know the diameter of Jupiter in meters.\nConvert that number of meters to AU.\nUse the definition of "sin" (or "tan").\nOne second of arc is 1/3600 of a degree.