Can a 3 body problem be solved?
477263
There are several cases where it can be solved.  There are a few solutions to cases of the circularly restricted 3-body problem, and situations where one of the bodies is of "zero" mass can be solved, but there is no general analytic solution to this problem.
1133022
It theoreticaly should not be too dificult to solve using vectors.  I am not going to solve it because I would prefer having exact values to work with, but basicaly, what you would need to do is find the complete distance, also make a graph where you find the time intersect of the torpedo and boat, and work with the x/y components to figure out what would be the ideal inclination.  It is a complex problem where you must work with more than one vector system, but it is doable if you work in parts.
273787
It has been shown that there is an answer to the great question of life, the universe and everything. It was computed by Deep Thought, but really didn't seem to provide , well... an answer.\n\nThe great computer kindly pointed out that what the problem really was that no-one knew the question. Accordingly, the computer designed its successor, the Earth, to find the question to the ultimate answer.\n\nHowever, due to a slight Golgafrincham problem, and a small hooha with some Vogons the program went a bit wrong. The whole problem remains just as mysterious as it always has been, but a little more frustrating.
1157446
The reason nothing is sent out from earth up or down is rather simple, Nothing is within reach in those directions. For the most part the planetary system; the only thing we are able to explore at current technology level, is on a flat planer system due to the suns gravity force. Certainly the universe is tri-dimensional but it is A: very hard to depict in that way. and B: Impossible for us to study in terms of anything besides telescope observation.
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
363951
theorically yes.   The problem is discovering and proving the right theory.   It maybe 500 years but we will eventually solve the cold fusion problem.
816836
You need one more datum, either the third side or any angle.  If it were a right triangle, the right angle would comprise the third datum.
917347
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
188376
I was a rock star in college physics!   Top of the class!  But I can't remember any of it now.  My teacher always said to start each solution by drawing a picture, then making a list of the varriable you know, and writing down the equations you thik you might need to solve.  It worked for me!\n\nPost your problem, maybe I can help.  Traditional physics problems have not changed much in the last 5 years, maybe I can remember.
799446
electrons can't line up in a plane they wouldn't all fit... unless it was like a big 747... or one of those military airbus things... and by 3d do you mean 2 spacial dimentions + time or 3 spacial dimensions? cause we already live with 3 spacial dmension (4 including time)
116787
Euclid! Come and help us please!!
296792
Your experimental model is corrupt in terms of current known information about matter and energy in a space-time continuum.  No prediction or calculation can be applied.\nWith two point masses, the only force in the system would be do to gravity.  This is an assumption based upon multiple moving masses in what we would call the normal physical universe.\nIf the distance between their centers (d) was sufficiently small, the points would be touching one another and no force would be exerted on the masses.  If the distance was sufficiently large, then it is possible that the force would be so small that the masses would not move.  Only in a finite range of distances would any gravity possibly exist.\nOne can make no assumptions about G as a gravitational constant, since G assumes that one of the masses is the earth.  One would need additional information about the masses to solve the problem.
700100
Many other materials can be in three states. It depends on the temperature!\nNear the 'triple point' the three phases exist together.\n\nTh
874570
NO!!  If people would stop (mis)quoting half-remembered precepts here, and do a spot of THINKING, instead, we might get somewhere.\n\n1.  If a moving spherical ball directly hits an identical free, stationary ball head on in an elastic collision, the moving ball STOPS, and the previously stationary one heads off with the original velocity of the other.  So ALL of the moving ball's K.E. has been transferred to the other ball.  \n\nHere are two ways to see this:\n\n1a.  The device known somewhat indelicately to British schoolboys as "Newton's balls" (but in squeamish America as "The Swinging Wonder") provides an excellent table-top demonstration of this.  Most undergraduate physics instructors must haul this out from time to time.\n\n1b.  You can easily solve it analytically, but a thought experiment will also demonstrate it.  In the originally stationary ball's frame, let the other ball approach with speed v.  Now shift (with an appropriately directed speed of v/2) to the frame in which the two equal balls are simply approaching each other, each with a speed in this frame of that same v/2 (oppositely directed, of course)!  Clearly, from the symmetry of the situation, they'll bounce off each other with the same (oppositely directed) speeds in this frame, still of v/2.  Now go back to the original frame: the previously moving ball is now completely stationary, while the originally stationary one is haring off with speed v.  (You may need to make little sketches of what is happening as viewed successively from each of these frames, to see this; I admit that that's what I did.)\n\n2a.  A similar approach can be used to solve the general case for an arbitrary ratio of masses, m (initially moving)/M (initially stationary).  Hint: have the balls approach one another in their center of mass frame with equal but opposite momenta of mv and  MV.  They'll clearly bounce off one another in such a frame, retaining the same values of their respective (but now oppositely directed) speeds.  Then you can do the other frame-jugglings from there. \n\nHowever, this analytical approach, done accurately, is a tedious mess.  So, to appreciate the general problem more physically, consider just the following:\n\n2b.  The case of m << M.\n\nConsider the "equal but opposite momentum" frame for these two balls, in which MV = mv.  This means that if ball M is VERY MASSIVE, the speed it will have in such a frame, V = (m/M)v; that is, it's VERY SMALL compared to little puny m's high speed.\n\nSo, transforming to the frame in which M itself (the big guy) is initially stationary takes only a speed shift of O(m/M)v.  (In fact identically so.)  This only affects the speed of puny m minutely, so m's initial K.E. is still of order O(m v^2).  (There's no need to do picky constant book-keeping for these purposes.)  \n\nNow after the collision: In the original frame, Big M is now moving (and in fact at 2V if you think about it.)  But, being loose about the constants, M's K.E is only O(MV^2).  \n\nSo, the ratio (m's K.E.)/(M's K.E.) ~ (mv^2)/(MV^2)\n\n= (m/M)(M/m)^2 (since mv = MV)\n\n= M/m.\n\nBINGO!  The little bugger still carries by far the larger amount of the kinetic energy.  The massive ball (i) moves sluggishly, and (ii) because  K.E. involves both mass but also the SQUARE of the speed, the TWO POWERS of (small) speed more than cancel out the effect of its (large) mass.\n\nThis is a well-known fact in particle collision theory, that little particles basically bounce off big ones.  \n\nYou could also say that it's well-known in "space billiards": when spacecraft head close to Jupiter to receive a "gravitational sling-shot effect" that sends them off to the outermost planets.  \n\nIt's more complicated because of 3-D effects there, but analogous frame-shifting methods can be used to show that the spacecraft can pick up considerable speed AND change
402813
No, I think it'll take too long for them to join together.  They're generally so far apart that the gravitational pull on each other is very very small.  Besides, the universe is constantly expanding, so instead we'll end up in a position where everything is just cold and the density of matter is very small.  That's just one theory, though.
283486
to imagine it, it would be very difficult. I suggest you experiment on it. Try using 3 ball, one for the sun, one for the earth, and another for the moon. Now do hte things you think. \nBut for me, the earth spins on its axis east to west and revolves the sun at east to west direction. EWhile the moon also revolves the earth at an east to west direction.
1082511
I think your question have a typo.  There are three variables and I am not sure which one are you trying to solve for.