Can a three body problem involving gravitational attraction be solved without the use of numeric methods?
216744
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.
561727
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
11528
just draw a free body diagram. you're solving for F parallel. that's just FgsinX which is just mgsin(30_ then you take this and divide it by mass to get the acceleration. so actually, you can cancel out mass and its just (2.15)sin(30) which is 1.075\n\nfor the frictional force, either you need more info, or it's zero.
416477
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
644170
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.
394234
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
662203
There is one, that if you were to do it and found a way to publish the results, a lot of common people would be very interested.  It is the formation and collapse of a gravitational field.\n\nWere you to go to http://timebones.blogspot.com   and then to the November Archive, toward the bottom of that month you shall find "How to Form and Collapse a Gravitational Field".  What I have suggested has been done by myself, but I have never found anyone interested in the results.
441710
The question is easy enough if you think about it properly.  What you want is the speed at which the centrifugal force (given by mv^2/r) equals the gravitational force given by the bank, and that will be mg sin(2.5 degrees).  Note that m cancels out in this, and you are given r, so v is the only unknown and you can solve for it with a scientific calculator.
391295
Split these two vaiables apart.\n\nDoubling the mass would double the gravitational force, hence the weight also would double. (purely proportional)\n\nIncreasing the distance by 4x would decrease the grav force by 16x. (Inverse square law)\n\nso...put the two factors together...\n\nIf you have 2x, and then 1/16x, then it will decrease by 1/8 of the original weight.\n\n\nm1 * m1 / d^2 = F
570881
Using 9.8 m/s^2 for gravity, the answer is 3,234 J.
384731
The motion is deccelerated by the action of gravitational and frictional force. Hence, \nm*a=Fg+Ff\nm*a=w*sinx+f*w*cosx\na=w*sinx/m+f*w*cosx/m\na=g*sinx+f*g*cosx\na=g(sinx+f*cosx)----(1)\n0.5*m*v^2=m*a*s\ns=v^2/(2*a)---(2)\nInsert (1) in (2) and solve for s\ns=v^2/(2*g(sinx+f*cosx))\ns=3.1^2/(2*9.81(sin26+0.5*cos26))\ns=0.55 meter
517016
Kepler had something to say about this type of problem.\nGMT^2=4*pi^2*a^3
89131
They used log tables.\nSimple problems like this could also be solved by using Newton's method.
118667
OK, Law of universal gravitation can be given as:\n\nF = G(m*n/r^2)\n\nWhere F = gravitational force, G = Gravitational Constant, m = Mass 1, n = Mass 2, r = distance of seperation between centers of gravity.\n\nSo, in your example, m=20kg, n = 120kg, and r = 0.5m. Plug them in and get\nF= G(2400 kg^2)/0.25 m^2\n\nSolving for F, you get F= 9600 (kg/m)^2 * G\nSince G = 6.67300 × 10^-11 m3/kg*s^2\nYou get\n6.40608 x 10^(-7) Kilogram-meters per second squared.
43767
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.
347974
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
382980
Nicole I think that to understand what is to be done here I suggest to assume that a coordinate system be attached to one of the objects. I-don’t-remember-who says, but << any reference system at rest or moving with linear constant speed has no privilege with others the like>>.\n(x) m2=4kg is reference system #2, where v2=0, and m1=2kg is moving with speed v=8+12=20 ms, impact by m1 on m2 M1=m1*v=40kg*m/s\n(y) m1=2kg is reference system #1, where v1=0, and m2=4kg is moving with speed v=8+12=20 ms, impact by m2 on m1 M2=m2*v=80kg*m/s, so R=40/80=0.5