how the perimeter and density of a distant planet are measured?
606341
It depends on whether you are talking about a planet in our solar system or an extra solar planet. I'll describe both.\n\nFor a planet in our solar system, the apparent size can be directly measured optically. One way is to take a picture of the planet through the telescope and measure the size of the picture with a ruler. Calculations based on the focal length of the telescope will tell you how many degrees (or fractions of a degree, since it will be less than 1) wide the planet appears to be in the sky for a given measured size on the photograph. Then the distance to the planet must be measured, which is not so easy. Basically it comes from triangulation to nearby asteroids combined with kepler's laws. Knowing the distance and apparent size allows you to calculate the real size and volume. To get the density, you need to know the mass. If the planet has Moons, the orbital period of the moons gives you the planet's mass from orbital mechanics calculations. If the planet has no moon, like Mercury and Venus, then theoretically their mass can be gotten from the effect their gravity has on the orbits of other planets, asteroids and comets, but this is not so accurate since those other objects don't usually pass close enough to the planet we want to measure to show much of an effect on their orbits. Fortunately we have sent space craft to Mercury and Venus and measured their mass directly by the effect their gravity had on the space craft's path.\n\nFor a planet orbiting another star, it is not so easy, because such planets cannot even be seen. Their presence is deduced from tiny Doppler shifts of the star they orbit. A planet does not just orbit a star, the star and planet orbit each other around their common center of gravity. Since the star is much heavier than the planet, it doesn't move much, but it does move, and modern instruments can measure the tiny Doppler shift in the star's light to get the motion of the star. Stellar evolution models tell us how heavy the star is and orbital mechanics tell us how heavy the planet is. (We can calculate the mass of double stars using orbital mechanics and the Doppler shift of each star's light, but we have no direct knowledge of the mass of single stars.) If the planet also happens to orbit in a plane parallel to our line of site, it will cross directly in front of the star once each orbit, like Venus sometimes crosses the face of our sun as seen from Earth. When it does, it blocks a small percentage of the star's light and the star looks slightly dimmer. If the star dims by 1%, then the disk of the planet has 1% the area of the disk of the star. Again, from stellar evolution models, we can deduce the diameter of the star and so get the diameter of the planet. (We can measure the diameter of some large, nearby stars, directly with the largest telescopes and adaptive optics, but for most stars we have no direct knowledge of their diameter.)\n\nThere is much more detail, but enough is enough, don't you think?
1059357
Thats why they asked theoretically.\n\nsame radius as the solar system\n1267650600228229401496703205376 kms\n\nOf which you allready knew, otherwise u would not be asking
289592
You can't-you dont have enough information.  You need to know the mass of the planet as well, to know the gravity.  Orbital velocity is determined based on calculating the gravity based on the mass, assuming all the gravity originates in the exact center of the planet.  Then, if you know how far the surface is from the center (the radius of the planet) you can calculate what the gravity is at this point...gravity decreases as the square of the distace from the center of gravity.  Now, you know the gravity at the surface, and can calculate escape velocity.
730461
near zero. \n\nExcept that there is some mass in space, - like space dust. And there's the solar wind pushing outwards.
931645
first answer was close to right.\n\nthere is one qualification that seems need.\n(do this all in meters for easyness of calculation)\n\nR = radius of earth = 6400 km = 6.4 * 10^6 meters\nd = distance to horizon in meters\nh = height in meters\n\nSimilar triangles:\n\nR / d  = d / 2h\n\nd^2 = 2Rh\n\nd = sqrt (2Rh)\n\n....  works pretty well for small "h"\n\neg 2 meters height =>\n\nd = sqrt (2 * 6.4 * 10^6 * 2 ) = sqrt (25.6 * 10^6) \n=5.06 km\n\n[[ online calculator gives 5.5 km ]\n\nmine's a LOT easier to use ...\n\nI use "2h" instead of "h" for similar triangles\nbecause of discrepancy of right angle at the\nradius from eearth-center to observer and\nr-a from earth-center to distant object.\n[intersection happens "h" below earth surface]\n\nthat sounds confusing but I can't draw a picture\n\nmy eqn is easiest possible tho ..
294195
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.
265928
1.27 X 1022 KG (0.0021 Earths)
1160107
Since we can't actually see a planet orbiting a distant star, the way we can look for planets is by their influence on the star they orbit. This influence is due to the force of gravity. Even though a planet may be small in comparison to its star, it will nonetheless exert a gravitational force, gently "tugging" at its star. As the planet(s) orbit the star, they will tug at it from different sides. The net effect then, if we carefully watch this star for many years, is that we will see this star "wobble" back and forth! \n\nThere are two basic methods of detecting a planet's gravitational influence on its star: \n\n\n  a. Astrometric Detection \n  b. Radial Velocity Detection \n\n\nThe first method - Astrometry - was hinted at in the previous paragraph. We observe a star and measure its position relative to distant background stars (hence the word "metric" in the name). As an orbiting body tugs at its companion star, we observe a change in position of the star. Measurements of a periodic change in position, back and forth, can indicate that something is possibly orbiting the star. Careful analysis of these measurements can tell us about the orbiting object. \n\nThe second method - Radial Velocity - is a completely different kind of measurement, but relies upon the same principle. That is, gravity. As an orbiting body tugs on its companion star, the light from the star will experience a Doppler shift. If the planet pulls the star slightly away from us on Earth, the starlight will be shifted towards longer wavelengths and appear more red; pulled slightly towards Earth, the starlight is shifted towards shorter wavelengths to appear more blue. To accurately measure this Doppler shift, we chose a known spectral line and observe its shift from red to blue and back. \n\n\n\n\ncheck in here.....I think you'll have your questions answered.\nTo my knowledge,  have been discovered 197 planets out of our solar system until now (update 16 june 2006)\n\nhttp://exoplanet.eu/catalog.php\n\nhttp://isi9.mtwilson.edu/~david/planets.html
939439
The frequency is irrelevant, just divide the distance by the speed of light, which is 3 x 10^8 meters per second.\n\nIf the distance between the two planets is approximately 8.00 x 10^7 km, that is 8.00 x 10^10 meters, so:\n\n8.00 x 10^10\n----------------\n3.00 x 10 ^8\n\n= 8/3 x 10^2\n\n= 2.667 x 10^2\n\n= 266.7 seconds
86906
Check this site out,\nhttp://www.onlineconversion.com/weight_on_other_planets.htm\nand/or this one,\nhttp://www.daviddarling.info/encyclopedia/W/weight.html
765953
Saturn. It's density, if I remember right is something on the order of .97 gm/cc. I remember seeing an illustration in an old book of Saturn floating in a sea.
417578
"Astronomers can use parallax to find distances to objects much farther even than planets. To calculate the distance to a star, astronomers observe it from different places along Earth's orbit around the Sun. If they measure the object's position several months apart, their "two eyes" will have a separation of well over 100 million miles! \n\nNow astronomers have technologies to measure distances to other planets more directly. When we have a spacecraft at another planet, we know the time it takes a radio signal to travel between Earth and the spacecraft. We can also send a powerful radar signal toward a planet and time how long it takes for the echo to return. Astronomers know how fast these signals travel (the speed of light), so measuring how long they take makes it easy to calculate the distance very accurately."
149911
Density = Mass / Volume\nIf you know any 2 of the above quantities, you can find the other.
544990
Well as far as a formula i can't help you there. i just happen to own a program that allows me to draw perfect shapes of any size and diminsion them accordingly (auto CADD). and at the top of your 300 ft house, or where ever it is that your measuring from the horizon at sea level is a tad over 21 miles away (21.20389 to be analy exact). first you must know the the earths circumfrance (i know i can't spell) which is 20,890,637 ft, next you have to know the distance you are away from the earth (which you do, 300 ft). now you have to know the formula for finding a line that intersects your hieght and crosses tangentally across the earths surface. know you have your triangle and apply the Pythagorean theorem. you have an answer.
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.
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.
1161756
It's been a while since I took Chem, but if I remember correctly you can measure the mass of a non uniform rigid body using displacement. For example, if you have a beaker filled with water fill to the 50ml level and you insert a non uniform object into the beaker the the water level will rise. This change in the water level will equal the volume of the non uniform object. Then you can use the equation p=m/v and solve for the mass assuming that you know the p the density.