How did Isaac Newton calculates the mass of each of the planets?
953783
did he? \nhe talked about attraction force (F) between bodies (m1, m2, staying d distance apart) though\nF = G (m1. m2) / d^2
859149
http://en.wikipedia.org/wiki/Isaac_Newton
1095815
Mass is the amount of matter present in a body and is an intrinsic property of the body. Mass of an object remains the same always at any place.\n\nWeight on the other hand is the force which a given mass feels due to the gravity at its place. Weight is measured in units of Force like Newton (which is the SI unit of Force).\nIf your mass is 60 kgs then your weight is approximately 60 x 10 = 600 Newtons. This is because\n\nForce = mass x acceleration (From Newton's second Law)\n\nThus, weight = mass x acceleration due to gravity
50990
newtons are the unit of force.  the second law of motion says that F=ma, where F is the force, m is the mass and a is the acceleration of the mass.   so, you need to know the acceleration!\n\nif you are talking about an object on earth, then the acceleration is "g", which is 9.8 m/s^2.  so multiply the mass times that acceleration to get the force with which gravity pulls on that object, in newtons.
72896
Newtons is the unit for measure of force
22095
Let's try to anser the question without looking it up and without using a calculator.\n\nWe know Newton's gravitation law: F = G*m*M/r^2\nAnd we know the gravity on earth: F = m*g\n==> m*g = G*m*M/r^2\n==> M = r^2*g/G\nNow, the acceleration on the earth surface is approximately 10 m/s^2\nThe earth's radius is approximately 6*10^6 m\nAnd I remember G to be approximately 7*10^-11 Nm^2/kg^2\n\nM = 6^2*10/7 * 10^12/10^-11 kg\n\nThe earth weights about 5*10^24 kg\n\nIf you look it up in Wikipedia, it says 5,9736*10^24 kg. \n\nSo we're close enough if you take into account that we used rough estimates for the constants.
677265
The Earth and our body exert forces on each other according to Newton's law of gravity.  Essentially mass attracts other mass.
916976
I think the truest indication of Newton's genius was his discovery (not exactly intuitive) that the same force that causes an object to fall to the ground was none other than the same force that keeps the moon and planets in their orbits.
662080
Not quite correct.  You would learn essentially the EARTH's mass by watching the moon in its orbit (actually the mass of the Earth + moon combined).  You can learn the MOONs mass most easily aqnd precisely by flying small low mass objects around each of them.  \n\nOur most precise mass estimates of solar system bodies are deduced by the impact they have on things passing near them.  Sometimes nature provides (think about the mass of Jupiter via its satellites) sometimes we do a flybay or closed orbit with a spacecraft. \n\nBy observing the motion, via doppler shifted radio signals from the spacecraft for instance, you get the velocity as a function of time.  From there it is SIMPLE to get the acceleration, and for a small mass a=GM/d^2 where M is the mass of the thing yanking on it. \n\nStars of similar mass in binary systems are trickier, though if you look at enough binary systems involving a star of one stellar type you can deduce the mass of that particular stellar type.\n\nYou can't tell the mass of a solo star directly.  You have to infer it based on the masses of other stars like it and stellar models. You can tell something about a star's  "surface" (photosphere) gravity however by the broadening of spectral lines. If you can then deduce its radius via distance and brightness measurements you can calculate the mass. This method of measuring surface gravity is a bit imprecise.  Don't expect more than one significant figure if that, which is still pretty good for astronomy.\n\nGalactic mass is easy.  Since you get the motions of individual stars  or gas clouds orbiting them, and each star has insignificant mass in comparison to the whole Kahuna, your estimates are much more precise, usually limted by how accurately you  can estimate the distance between the center of mass and the object in question.
526084
This goes back to the famous story where Newton was sitting under a tree and he saw an apple fall.  Whether or not that is true, it says that he made the connection that the same force that caused the apple to fall was the force that caused the moon to orbit the Earth.  From there, he developed the Universal Law of Gravitation.  \n\nThis was all published in his masterpiece: Principia Mathematica.  It was published in 1687, although he had the work finished a lot earlier.  He was afraid to publish it.
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.
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.
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?
260486
im pretty sure itz derived from newtons laws. his F=ma.
609693
weight = mass * gravity\ntherefor\nmass = weight/gravity\n\nsorry naughty but you're wrong, if you go to the moon or jupiter you will have the same mass you have in earth but you'll have a different weight
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.
355325
Newtons, or sometimes kilograms