Using Kepler's third law to determine the distance from the Sun of a planet whose period is 5 years.?
928843
see: http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Kepler.27s_third_law_2
110744
Short answer: usam da inverse square law Jar Jar!\n\nLong answer:\n\nNewton simplified things by picturing all of a planets mass gathered in a single point in the center of the planet and the surface as a point on a certain distance from that center of mass. So the gravitationall pull (F) on a body (m) at a distance (r) from the mass (M) can be calculated using this equation:\n\nF=G*((M*m)/r^2)\n\nwhere G is the gravitational constant (6.67 * 10^-11 Nm^2kg^-2)\nEarths (equatorial) radius is 6.378 * 10^6 meters and its mass is\n5.974 * 10^24 kilogrammes\nSo lets test this on a dude weighing 100 kg:\n\n6.668 * 10^-11*((5.974*10^24 * 100)/4.068*10^13)\n==>\n6.668 * 10^-11*(5.974*10^26 /4.068*10^13)\n==>\n6.67 * 10^-11* 1.424*10^13 \n=\n979 N (what I got in my calculator)\n\nSo earth pulls on a guy, with the mass 100 kg, with the force of 979 Newtons. That means 9.79 N/kg, earths gravitational force (g). This should also mean that a mass dropped from a small height should accelerate at the rate of 9.79 m/s^2. It´s a simple experiment to do but where I am the value of g is actually 9.82. This is because I am closer to the earths center. Earth bulges out at the equator because of its rotation. I am farther north. But still a simple experiment can be done to validate this calculation. \nSo if the earths mass was 3 times greater than g shoud be 3 times greater. Right? You betcha. And I have already shown that moving closer to earths center increases g. Moving twice as far from the center should decrease g by half. Right? Wrong! Now what? We where so close. Oh man...\ng is reduced to a quarter if we increase the radius by 2. This is because of the inverse square law, a very important law in astronomy. It means that the apparent brightness of a light source is inversely proportional to the square of its distance. This also aplies to gravitational fields. Double the distance, divide by 4.\nSo lets put this all together. You should be able to do it yourself by now. I got 734 N or 75 % of the dudes weight on earth. And the reason he doesn´t weigh more than on earth is because of the inverse square law. Eventhough the new world is three times as massive we still moved twice as far away from the center. So take the weight of your astronaut, multiply by 3 and divide by 4.
81010
Jupiter is way too close to be using light years.\n\nOne "Astronomical Unit" is the earth-sun distance and is approximately eight light minutes. Jupiter orbits at about five A.U. from the sun.  So even when Earth and Jupiter are at their most distant I would still expect that distance to be less than one light hour.
604131
Extrapolate the formula, using the surface temoerature, and instead of the earth distance, substitute the distance you want to caluclate the temperature at.  \n\n\nMeasuring the Sun's Temperature\nOne method, called Wien's law, uses the wavelength of the peak emission, wavelengthpeak, in the Sun's continuous spectrum. The temperature in Kelvin = 2.9 × 106 nanometers/wavelengthpeak.\n\nAnother method uses the flux of energy reaching the Earth and the inverse square law. Recall from the Stellar Properties chapter that the flux is the amount of energy passing through a unit area (e.g., 1 meter2) every second. From the Inverse Square Law of Light Brightness, you find that the solar flux at the Earth's distance = the Sun's surface flux × (Sun's radius/Earth's distance)2 = 1380 Watts/meter2. Since the Sun's photosphere is approximately a thermal radiator, the flux of energy at its surface = sigma × (the Sun's surface temperature)4, where sigma is the Stefan-Boltzmann constant. Rearranging the equation, the photosphere's temperature = [(solar flux at Earth)/sigma) × (Earth distance/Sun's radius)2]1/4.\n\nThese two methods give a rough temperature for the Sun of about 5800 K. The upper layers of the photosphere are cooler and less dense than the deeper layters, so you see absorption lines in the solar spectrum. Which element absorption lines are present and their strength depends sensitively on the temperature. You can use the absorption line strengths as an accurate temperature probe to measure a temperature of about 5840 K. \n\nRead the link, you'll have to scroll down to come up to the stuff I have cut and paste above.\nst\nHope you find this useful.
451278
here is a calculator for determining planet and orbit sizes to scale.  i hope this helps.\n\nhttp://www.exploratorium.edu/ronh/solar_system/
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The relative size of Earth, Jupiter, Sun = 1 : 10 : 100
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Mean distance from\nEarth 149.6×106 km\n(92.95×106 mi)\n(8.31 minutes at the speed of light) \n\n--\nThe Sun is the spectral type G2V yellow star at the center of Earth's solar system. The Earth and other matter (including other planets, asteroids, meteoroids, comets and dust) orbit the Sun. Different latitudes of the Sun rotate at different rates; a point on the equator takes 25 days to complete a rotation, while a point at a pole takes 36 days. The resultant torsion creates the Sun's very strong magnetic field, which in turn drives the 11-year cycle of solar activity. The Sun accounts for more than 99% of the solar system's mass. Energy from the Sun has supported almost all life on Earth through photosynthesis since the appearance of the first organisms. Humans use sunlight for eyesight, warmth, growing crops, and powering solar cells.\n\nThe Sun is a ball of plasma with a diameter of 1.392 million km (864,950 mi) and a mass of about 2.0×1030 kg, which is somewhat greater than that of an average star in the Milky Way galaxy. About 74% of its mass is hydrogen, with 25% helium, and the rest made up of trace quantities of heavier elements. The Sun is about 4.6 billion (109) years old and is about halfway through its main sequence evolution, during which nuclear fusion reactions in its core fuse hydrogen into helium. About 5 million tons of matter are converted into energy within the Sun's core every second, producing neutrinos and solar radiation. In about 5 billion years, the Sun will evolve into a red giant and then a white dwarf, creating a planetary nebula in the process.
787745
The distance varies (a little bit) depending on the year.  I'll assume you are referring to 2006.  The perihelion for 2006 occurred on January 4th.  The distance at this point in our orbit was aproximately 91,376,859.108 miles (with July 5th being the aphelion of 94,537,401.539 miles).  Based upon the perihelion and aphelion we can calculate the dates you have outlined:\n\n03-21-06: 92,730,130.038 miles\n06-21-06: 94,319,132.034 miles\n09-21-06: 94,310,401.253 miles\n12-21-06: 92,703,937.698 miles\n\nAs this illustrates, the orbit of the Earth around the Sun is elliptical.
687280
Not counting the sun, the closest star to earth is Proxima Centauri, which is about 4.1 light years away from the earth. Light travels at 300 000 km/s so you can now use this to calculate the distant. 4.1 light years x 365 days per one light year x 12 months per 1 year x 30 days per one month, x 24 hours per one day x 60 minutes per one hour x 60 seconds per one minute = 4,654,713,600 seconds. Multiply this figure by 300 000 km/s and you get an answer of 1,396,414,080,000,000 Kilometers.
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dr_islam_safwat is exactly correct, as far as he goes. The problem is that Kepler's 3rd law does not tell you how far Neptune is from the Sun, it only tells you how much farther from the Sun it is than the Earth. The real trick to to find out how far the Earth is from the Sun. Actually, knowing the absolute between any two planets would be enough information, because then Kepler's 3rd law could then be used to find all the other absolute distances from the ratios.
914608
By Keplers 3rd Law T² α R³\n\nOr T1²R1³ = T2²R2³\nFor earth R = 1 and T = 1 and for Jupiter R = 5.2\nSo T2²/5.2³ = 1²/1³\nSo T2 = 5.2^(3/2)\n≈11.86 years
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The first step was to use Kepler's laws to get relative distances between the planets and the sun. So, it was known that mercury orbited at a distance of 40% of the earth's before the actual distances were known.\n\nAfter that, the actual distance to *one* object orbiting the sun was enough to give all the distances. Historically, that was done by paralax with asteroids that came close to the earth. Today it is done by bouncing radio beams off of other planets such as Venus and measuring the time it takes for the reflected beam to get back to us. Various probes also give a distance calibration.
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The mean radius is 6371.0 km.  The radius at any latitude is equal to that times the cosine of the latitude.  The Earth turns once on it's axis in 23 hr 56 min 4.0989 sec, or 23.93447 hrs.\n\nIt takes that long to travel 2(Pi)(6371)(cos(lat)).  Divide that distance by the time above and you're in the ball park (neglecting the difference in polar and equatorial radii).\n\nNote: The figure given for the Earth's rotation is correct according to Astrophysical Data: Planets and Stars.  The 24-hour day is measured from noon to noon - in relation to the Sun.  The Earth kind of rolls around its orbit, so after it has turned thru one rotation it has also moved along its orbit.  It has to turn for another 53.9 sec past the full rotation for the Sun to be back to due south again.
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Period is proportional to R^3/2.\nThis is an answer I posted to bombay2pac2000 under Astronomy & Space 8 hours previously, and it applies here as well.\nKepler's 3rd law says: Orbital period T = 2*pi*R^1.5/SQRT(GM)\nwhere R is the radius (in the case of a circular orbit), G is the universal gravitational constant and M is the mass of the central body.  In orbit, centrifugal acceleration CA balances gravitational acceleration GA.  CA=w^2R (where w is omega, the orbital angular rate which = 2pi/T). So CA = (2pi/T)^2*R. From Newton we know GA = GM/R^2. Put 'em together, \nCA = GA\nR*(2pi)^2/T^2 = GM/R^2 \nT^2 = R^3*(2pi)^2/GM\nT = 2*pi*(R^3/2)/SQRT(GM)
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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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It takes longer since it is furthest away from the sun.  since a year is how long a planet takes to orbit the sun, the farthest away you are, the longer it takes.
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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.