what is the scientific notation of the distance of each planet and the sun?
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Usually Astronomical Units (AU's). \n\n1 AU is the mean distance of Earth from the sun.\n\n1AU = 92,955,800 miles\n\nMercury = .39 AU\nVenus = .72 AU\nEarth = 1 AU (of course)\nMars = 1.52 AU\nJupiter = 5.2 AU\nSaturn = 9.54 AU\nUranus = 19.18 AU\nNeptune = 30.06 AU
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Here is the largest scale model of the Solar System, complete with driving directions between the sun and planets.\n\nhttp://www.bradley.edu/las/phy/astronomy/Scale_Model.html
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Kepler's Third Law states that the ratio of the squares of the periods of any two  planets is equal to the ratio of the cubes of their average distances from the sun.  Here, the "period" simply means the length of time it takes for a planet to orbit the sun once.\n\nIf R1 and R1 are the orbital radii (i.e., the distance from the sun) of planets 1 and 2, and T1 and T2 are the orbital periods, then the Kepler's Thrd Law has the mathematical form:\n\n\n(T1/T2)^2 = (R1/R2)^3\n\nHere, we know that for the Earth, T1 = 1 year, and R1 = 1 AU (1 AU is one Astronomical Unit, which is defined as the average distance from the Earth to the Sun).  We want to find R2 if T2 = 5 years.  So:\n\n(1 yr /5 yr)^2 = (1 AU/R2)^3\n1/25 = (1 AU)^3/(R2)^3\nR2 = (25 AU^3)^(1/3)\nR2 = 2.924 AU
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That is a really intersting question.\n\nI'm pretty sure atoms are more spaced... let's find out!\n\nSolar system in km\n\nPlanets in our solar system (diameter)- About 4880 Merc (I hate pluto and refuse to call it a planet) to 142,984 Jupt\nSun- 1.392*10^6km\nSun to planets distance (semimajor axis) - from 5.79*10^7 km to 4.5017*10^9 km\n\nDistance to planet size ratio -\nBig planet farthest - 31,483.94 (it's about 30,000 planet lengths)\nSmall planet closer - 11,864.75 (it's about 12,000 planet lengths)\nDisance (half way point) to sun size ratio - 1,637.79 (it's about 1,500 star lengths)\n\n\nHydrogen atoms in pm (pico meter)\n\nAn electron: 10^(-6)pm\nH nucleus - .01 pm\nSize of H atom (distance apprx) - 250 pm\n\nDistance to electron size ratio:\n2.5*10^(8) (it's a hell of a lot of electron lengths)\n2.5*10^(4) (it's about 25,000 neucleus lengths)\n\nSo there is much more space in an electron... the distance compared to the size of things is on a much higher order.  Plus we must remember that I used the smallest atom (others would have other shells of electrons a little further out but slightly bigger nuclei as well) and I used our solar system... other solar systems we have detected have big planets much closer to their stars.  Perhaps the electron's distances  more comperable to the distances between galaxies???  Wild guess.  My... calculator is too tired to figure it out. haha
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The link below has great overview info and statistics (including density) for all the planets, asteroids, comets, etc. \n\nCheck it out!
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Mercury is the closest planet to the Sun. the asteroid belt is located between the 5th planet (Mars) and the planet Jupiter
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Astronomers determined the size and shape of our galaxy and measured distances to other stars using the “parallax method” (and eventually other methods using the star’s light spectrum).  Unfortunately, these methods do little to help us find out the exact distance to our very own star, the Sun.  Surprisingly, to get the distance to the sun, astronomers first had to determine the exact distance to Venus!\n\nWith the invention of radar, scientists were able to measure the distance to Venus very precisely.  By timing how long it took the radar beam to travel at the speed of light to Venus and back, the total distance to the object can be determined from the equation:\n\nDistance = (Speed of Light) X (Total Time)/2\n\nThe reason the total time is divided by two is to get just the distance from the Earth to the object.\nEarth-Sun Distance = (Earth-Venus Distance) / (Cos q)\n\nUsing trigonometry, astronomers now know that the “Astronomical Unit” = 92,955,820.5 miles (149,597,892 kilometers).  This incredible degree of accuracy is possible because the speed of light is known very precisely and extremely accurate clocks are used to time the radar reflection.  Radar can’t be used to determine the distance to the Sun directly because the Sun doesn’t have a solid surface to reflect the radar efficiently.
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You may ask similar questions. So I answer for all.\nDistance from the sun. All in tera mete. tera means 10^12.\n \nMercury----0.05791 Tm \nVenus------0.1082 Tm \nMoon & Earth--0.1496 T m\nMars------0.2279 T m\nJupiter---0.7783 Tm\nSaturn----1.427 Tm\nUranus----2.869 Tm\nNeptune---4.498 Tm\nPluto-----5.900 Tm
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it's pretty hard to remember this off the top of one's head, because there are a lot of terms, and a lot of the numbers are orders-of-magnitude, and who can keep track of all that?\n\nanyway the equation is N = R* fp * ne * fl * fi *  fc * L\n\nwhere\n\nN = The number of communicative civilizations\nR = The rate of formation of suitable stars (stars such as our Sun)\nfp = The fraction of those stars with planets. (Current evidence indicates that planetary systems may be common for stars like the Sun.)\nne = The number of Earth-like worlds per planetary system\nfl = The fraction of those Earth-like planets where life actually develops\nfi = The fraction of life sites where intelligence develops\nfc = The fraction of communicative planets (those on which electromagnetic communications technology develops)\nL = The "lifetime" of communicating civilizations
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Do mean "difference" or "distance"?\n\nIf it's difference, well, Venus is the 2nd planet from the sun, Mars is the 4th planet from the sun, and Jupiter is the 5th planet from the sun.\n\nVenus is a rocky planet just slightly smaller than earth with a hot, dense, atmosphere.\nMars is a rocky planet just over half the diameter of earth with a cold, atmosphere with a pressure of about 1/10 that of earth.\nJupiter is a gas giant that could contain many earths.\n\nI'm guessing that your question has more to do with the distances between the orbits of the planets rather than the actual distances between the planets because the distance between the planets can vary greatly depending on where each is it's orbit.\n\nThe site linke below has the distances in astronomical units and kilometers (km) from the sun for the orbits of each planet.  Maybe from that you can figure out the answer to your question.
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General Characteristics of the Sun\n\nThe sun is actually a star of about medium size; it appears larger than the other stars because of its relative nearness to the earth. The earth's distance from the sun varies from 91,377,000 mi (147,053,000 km) at perihelion to 94,537,000 mi (152,138,000 km) at aphelion (see apsis). The mean distance is c.92,960,000 mi (149,591,000 km); this is taken as the astronomical unit (AU) of distance used for measuring distances within the solar system. The sun is approximately 865,400 mi (1,392,000 km) in diameter, and its volume is about 1,300,000 times that of the earth. Its mass is almost 700 times the total mass of all the bodies in the solar system and 332,000 times that of the earth. The sun's surface gravity is almost 28 times that of the earth; i.e., a body on the surface of the sun would weigh about 28 times its weight on earth. The density of the material composing the sun is about one fourth that of the earth; compared with water, the sun's average density is 1.41. At its center, the sun has a density of over 100 times that of water, a temperature of 10 to 20 million degrees Celsius, and a pressure of over 1 billion atmospheres.\n\nObservations of sunspots and studies of the solar spectrum indicate that the sun rotates on its axis from east to west; because of its gaseous nature its rate of rotation varies somewhat with latitude, the speed being greatest (a period of almost 25 days) in the equatorial region and least at the poles (a period of about 35 days). The axis of the sun is inclined at an angle of about 7° to the plane of the ecliptic.\n\nThe bright surface of the sun is called the photosphere. Its temperature is about 6,000°C. The photosphere appears darker near the edge (limb) of the sun's disk because of greater absorption of light by the sun's atmosphere in this area; this phenomenon is called limb darkening. During an eclipse of the sun the chromosphere and the corona (the outer layers of the sun's atmosphere) are observed. Also of interest is the high-speed, tenuous extension of the corona known as the solar wind.\n\nProduction of Solar Energy\n\nThe vast and continual production of solar energy cannot be attributed merely to combustion, to the gradual cooling of a hot body, to the fall of meteorites into the sun, or to gradual shrinkage with transformation of potential energy into heat (a theory proposed by Helmholtz). The theory of relativity with its implication of the equivalence of mass and energy led to the assumption that energy stored in the atoms constituting the sun's gases is constantly being released by conversion of some of the masses of the atom's nuclei during nuclear transmutations (see nuclear energy). H. A. Bethe proposed a cycle of nuclear reactions known as the carbon cycle, or CNO bi-cycle, to account for the nuclear changes. In this cycle carbon acts much as a catalyst, while hydrogen is transformed by a series of reactions into helium and large amounts of high-energy gamma radiation are released. It is now thought that the so-called proton-proton process is a more important energy source; this process begins with the collision of two protons and ends with the production of helium, while gamma radiation is released throughout.\n\nSee nucleosynthesis; stellar evolution.\n\nThe Study of the Sun\n\nBy means of the spectroscope much has been learned about the composition of the sun. There are numerous dark lines of varying widths in the solar spectrum. These were first intensively studied by Joseph Fraunhofer and are commonly known by his name. From a study of the lines the chemical composition of the sun is determined on the basis of the discovery by Kirchhoff that the dark lines correspond in position to the bright lines characteristic of the spectra produced by elements in the laboratory. The darkness of the lines in the sun's spectrum is attributed to the presence of a slightly cooler layer of gases above the phot
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the average distance between the Earth and the Sun (approximately 150 million km or 93 million miles).
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First ... assume an average diameter (or radius) for the planet.\n\nThe average radius normally given is 3963.1 miles (6378 km).\n\nCircumference of a circle = 2 * pi * r = pi * d\n\nCircumference = (2) * (3.14159) * (3963.1) = 24900.9 miles\n... which is approximately 40074 km.
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Their distance from the Sun
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The average distance from Mars to the Sun is 1.52369 astronomical units.  One astronomical unit (AU, the average distance from the Sun to the Earth) is 1.49597870E11 meters.\n\nConverting those numbers, then, means that Mars is:\n\n1.52369 * 1.4959787E11 meters = 2.27941E11 meters, or, converting into English units:\n\n141,635,833 miles (on average -- remember that the orbits of the planets around the sun are not circular, but elliptical)
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Planets: Venus and Mars\n\nMoons: Titan, Europe
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Here are some quick facts:\n\nhttp://www.sir-ray.com/ Enter the site, click astronomy then click on Space Dimensions & Planetary Facts (this site is excellent)\nhttp://www.dustbunny.com/afk/planets/\nhttp://www.kidzone.ws/planets/index.htm\nhttp://nssdc.gsfc.nasa.gov/planetary/planetfact.html\n\nThese sites should be everything that you need~the first site should do it but the other three have little bits of information that might be useful~