How are rotating masses around each other calculated in space.?
448583
what do you mean?\nhow to calculate their speed or wat?
233618
1 mass\n2 magnitude of separation
508636
same mass, same weight
990074
Picture a weight on a string that you spin overhead. The shorter the string, the faster you have to spin the weight to keep it horizontal. It's the same with an orbiting body. A geostationary satellite is placed at the proper orbital radius that its rotational velocity is the same as that of the Earth, so it remains placed over the same spot on the globe as it speeds forward in its orbit.
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
1102952
Gravitational time dilation caused by a spherically symmetric, nonrotating massive body can be computed exactly by using the Schwarzschild solution to the Einstein field equations. This is also a good approximation to a slowly rotating, roughly spherical body like the Earth. The time dilation given is:\n\nt0=t1√(1-2GM/(rc²)) where:\n\nt0 is the time between two events for an observer within the gravitational field\nt1 is the time between those same events for an observer well outside the gravitational field\nG is Newton's  gravitational constant\nM is the mass of the object creating the gravitational field\nr is the distance of the observer in the gravitational field from the center of said field\nc is the speed of light
643815
angular momentum is given by m*v*r\nso know, we have the units are\nkg*m/sec*meter=kg meter^2 per sec\nits dimensional formula will be ML^2T^-1
504321
Mass (according to Einstein) coz mass changes into mass.
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.
840333
You need to consider conservation of angular momentum\n\nI0*w0 = If*wf\n\nI0 is the initial moment of inertia, If is the final moment of inertia, w0 is the initial angular speed, and wf is the final angular speed.\n\nI = Iss + Io\n\nwhere Iss is the (constant) moment of inertia of the student and the stool, and Io is the moment of inertia of the objects. This general relation applies to both the initial and final moments of inertia. Io can be computed from\n\nIo = 2*m*r^2\n\nwhere m is the mass of the object and r is the distance from the axis of rotation. The factor of 2 accounts for the two different objects. The first equation then becomes\n\n(Iss + 2*m*r0^2)*w0 = (Iss + 2*m*rf^2)*wf\n\nwhere r0 is the initial radius (1.0 m) and rf is the final radius (0.40 m). Solving for the final angular speed you have\n\nwf = (Iss + 2*m*r0^2)*w0/(Iss + 2*m*rf^2)\n\nUsing this expression it is easy to compute the final angular speed.
1076161
An object orbiting another in space can be considered to be in freefall around the center object.  You have probably heard someone explaining circular motion like that of an astronaut or space station or the moon as being in freefall towards the earth.  It's a balancing act between the gravity between the two objects (causing the free fall) and the inertia of the planets, moons, satelites, etc that keep them moving in a direction perpinduclar to the inward falling direction.  \n\nEventualy the orbiting object will have to fall toward the center object or away from it, debending on whether the inertia or the gravity is stronger.  \n\nOur moon is moving away from us.\n\nAll the stars in the galaxy are moving around the center of the galaxy and all the galaxies are moving around the center of the universe but we will never fall in because not only is the universe still expanding, the rate at which all the galaxies are moving away from the center of the universe is getting faster.
735689
Basically the equation is \n\n[Sum of (mass*Distance X)]/mass gives the X coordinate,\n[Sum of (mass*Distance Y)]/mass gives the Y coordinate\n[Sum of (mass*Distance Z)]/mass gives the Z coordinate\n\nFor arbitrary shapes that cant be broken down into  point masses,you can use integral equations, but most of them simplify into simple equations, this site has a good list\n\nhttp://mathworld.wolfram.com/GeometricCentroid.html
366583
This is because of their centre of gravity.\n\nAll of their constituent mass is  pulled with an even force into the centre of the object....thus the surface of the celestial body is equidistant from the centre...ie. almost spherical.\n\nMany are in fact oval due to the fact that they spin around. This moves some of the mass to the equator of the object....rightly due to centrifugal force.
406596
no idea...but I think you use the formula (Gravitational constant x mass 1 x mass 2) / distance squared
103036
We often use the terms "mass" and "weight" interchangeably in our daily speech, but to an astronomer or a physicist they are completely different things. The mass of a body is a measure of how much matter it contains. An object with mass has a quality called inertia. If you shake an object like a stone in your hand, you would notice that it takes a push to get it moving, and another push to stop it again. If the stone is at rest, it wants to remain at rest. Once you've got it moving, it wants to stay moving. This quality or "sluggishness" of matter is its inertia. Mass is a measure of how much inertia an object displays. \n\nWeight is an entirely different thing. Every object in the universe with mass attracts every other object with mass. The amount of attraction depends on the size of the masses and how far apart they are. For everyday-sized objects, this gravitational pull is vanishingly small, but the pull between a very large object, like the Earth, and another object, like you, can be easily measured. How? All you have to do is stand on a scale! Scales measure the force of attraction between you and the Earth. This force of attraction between you and the Earth (or any other planet) is called your weight. \n\nhttp://www.exploratorium.edu/ronh/weight/
25445
The earth is APPROXIMATELY round due to gravitational forces.
701530
Plug the numbers into the Newtons Equation :\n\nF =  (GMm) / r^2\n\nF = Gravitational Force / weight  \nG = Universal gravitational constant   6.673*10^-11 (note the - sign on the exponent)\nM = mass of earth (kg)                 5.98*10^24 kg\nm = mass of satellite (kg)             150kg\nR = radius of earth    (m)             6,378,137 m\nh = height above earth (m)             12,800,000m\nr   = h+R\n\nThe answer will come out as the weight being @ 162.67N\n\nNot weightless, although someone standing on the satellite would feel weightless as its in continuous free fall.\n\nEver been in an elevator and felt lighter as it falls or been weighless on the big dipper as it plummets?\n\nThe gravitational force (weight) of 162.67N is what keeps it in orbit and prevents it from flying off at a tangent. Basically it pulls the satellite towards the earth but the satellite is travelling so fast it misses and continues round in its orbit.\n\nAdded:\n\nWeight is measured in Newtons and is the attractive force between the two masses. (M & m)\n\nIn this case the earth and the satellite. A 1 Kg mass on the surface of the earth will be pulled towards the earth's centre with a force of 9.8N, therefore its weight at ground level is 9.8N. If you plug take m as 1kg, h as 0 and plug them into Newton's formula should get the answer 9.8N, which is where we get the constant g from. \n\nAs you get further away from the centre of the mass the gravity field decreases in inverse proportion to the distance squared. Hence the divide by r^2 on Newtons fornula. Therefore the weight of a 1Kg mass at 12,800km is 1.08N.\n\nBecause the satellite is in a stable orbit the opposing forces of the satellite trying to escape orbit and the gravitational pull towards the earth are equal so yes the weight is 162.7N and the centrefugal force is 162.7N. This is the correct answer to the problem. \n\nAs an aside; for simplicicities case in this scenario the mass of the satellite can be regarded as constant due to its relatively low velocity compared to the speed of light. \n\nHowever as you come closer to the speed of light mass increases according to Einstein's theory of relativity, eventually reaching infinity at light speed.