What's the magnitude of the gravitational force on a 32kg sphere located at the origin due to the other sphere
406596
no idea...but I think you use the formula (Gravitational constant x mass 1 x mass 2) / distance squared
393015
If a body is fixed at its center of gravity, it is in a state of neutral equilibrium.\n\nTherefore, the whole weight of the body can be assumed to act at the center of gravity.\n\nThe following method is used to find an expression for the C.G of a solid body.\n\nThe solid body is considered to be broken up into elementary particles.\n\nIf m1, m2, m3 etc are the masses of the particles, their weights are m1g,m2g, m3g etc.\n\nThe weights are acting parallel to each other in the downward direction.\n\nIf we assume that the C.G is at the point X,  and x1,x2,x3 etc are the distances of m1,m2,m3, then the  sum of MOMENTS of all these weights about the origin must be equal to the moment of the whole weight of the body about its C.G.\n\nTherefore,   Mg .X =  ( m1g .x1 + m2g .x2+ m3g .x3 + m4g .x4+ ....) where M = sum of all masses.\n\nIn short, the moments of weights of small particles in a solid body about any point is equal to the moment of the whole weight of the body about its C.G.
1037374
Every mass in the universe attracts every other mass in the universe. The force of the attraction between any two objects is proportional to their two masses and inversely proportional to the square of the distance between them.\n\nThe formula for the force is F = G x m_1 x m_2 / r^2 where \n\nG  = 6.6742 × 10^( −11) N m^2 / kg^2    is the gravitational constant.\n\nThe gravitation that you feel is from the attaraction between the mass of the earth and the mass of your body.\n\nGravity is not caused by the the earth's magnetic field, its rotation  on  its axis, or its revolving around the sun.
757859
haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
247359
Mass interacting with gravity
831377
it is the force of gravity it's simply 9.8 m/s/s or 32ft/s/s     \nacceleration of gravity is -9.8m/s/s or 32ft/s/s\ng force is just the absolute value
894961
The gravitational force (Fg) exerted by Saturn on its moon is\n\nFg = G*M*m/r^2\n\nwhere G is the universal gravitation constant, M is the mass of Saturn, m is the mass of the moon, and r is the radius of the orbit.\n\nThe gravitational force is balanced by the centripetal force (Fc) due to the motion of the moon in a 'circular' orbit around Saturn.\n\nFc = m*v^2/r \n\nwhere v is the orbital speed, which can be expressed as\n\nv = 2*pi*r/T\n\nwhere T is the period of the orbit. Inserting this into the equation for the centripetal force you have\n\nFc = 4*pi^2*m*r/T^2\n\nNow since\n\nFg = G*M*m/r^2 = 4*pi^2*m*r/T^2 = Fc\n\nThe mass of the moon, m, cancels leaving\n\nG*M/r^2 = 4*pi^2*r/T^2\n\nwhich can be rearranged to give Kepler's third law that says that the square of the orbital period is proportional to the cube of the orbit's semimajor axis (taken as the radius for this problem, which assumes a circular orbit). Solving for the mass of Saturn\n\nM = 4*pi^2*r^3/(G*T^2).
833966
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.
1081934
There are too many formulas to calculate that answer. You have to take in account for example, position of object to find potential energy, mass and the weight due to the gravitational force, also Force that is going to be exerted the object and the amount of that energy that will be lost. SO it is a lot to  find that out. But their is not one formula for it. Sorry, But if you do find out Let me know.
989214
F = -G(m1*m2)/r^2\nF = force\nG = gravitational constant = (6.6742 +/- 0.001)*10^(-11) N * m^2 / kg^2\nm1 = mass of object 1 = 47 kg\nm2 = mass of object 2 = 6.00×10^24 kg\nr = distance between objects\n\nF = -(6.6742 +/- 0.001)*10^(-11) * (47 kg * 6.00*10^24 kg) / r^2\nF = -4.225 * 10^14 N*m^2 / r^2\nSince the distanct between the Earth and the person weren't given this is all the farther the force can be calculated to.  The force will be equal on both the objects (person and Earth) only that the acceleration of the Earth will be almost nothing due to its size and the force will be in the opposite direction.\n\nBut based on the acceleration of the person, you can calculate a force, but it will only be an estimate\nF = m*a = 47 kg * 9.8*10 m / s^2 = 460.6 N\nForce on Earth is 460.6 N\n\nNow the acceleration of the Earth is\n460.6 = 6.00*10^24 * a\na = 7.68 * 10^(-23) m/s^2
864950
Gravitational Potential Energy = m*g*h joules\n        where   m - mass of the object in kg\n                     g - accelaration due to gravity in m/s^2\n                     h - height of the object from the reference in m\nHere the height from reference is 500m\ng is taken to be a constant and its value is 9.81m/s^2 though it may vary slightly with latitude, altitude and other such factors.\nMass is given as 0.400kg\nThus Potential energy = 0.400 * 9.81 * 500\n          Potential Energy = 1962 joules\n\nHope so u got wat u needed.
882023
The first is a sphere of radius a centered at (0,0,a) and the second is a sphere of radius a/2 centered at (0,0,a/2). Both spheres are above the xy-plane and are tangent to each other at the origin. The smaller sphere is contained inside the larger.\n\nIn spherical coordinates, the equations reduce to\nrho=2a*cos(phi) and\nrho=a*cos(phi).\n\nHence the limits are\nrho: from a*cos(phi) to 2a*cos(phi)\nphi: from 0 to pi/2\ntheta: from 0 to 2pi.\n\nI am using the mathematician's convention where phi is the angle down from the z-axis and theta is the angle from the xz-plane. Physicists use the reverse convention, so be careful!\n\nDon't forget the rho^2 sin(phi) !
100886
The equations weren't visible in Ginabean's answer (at least not on my display) and the other answer dealt with Earth, not the fictional planet postulated.  So I'll add some things.\n\nWe worked out the mass of this planet using the density and the radius in your other question.  M = 8.92*10^22 kg.\n\nNow, the force of gravity is given by\n\nF = G M m /r^2  , where M is the mass of the first object, in this case the planet, m is the mass of the other object, in this case something on the surface of the planet, and r is the distance between their centers, in this case the radius of the planet.  (Because the planet is spherical, we can treat the mass of the planet as if it were all located at its center.)\n\nSo, the acceleration caused by gravity on this planet, g, is F/m where F is the force of gravity and m is the mass of the object, \n\ng =F/m = G M /r^2\n\ng = (6.67 * 10^-11 m^3/s^2kg) * 8.92*10^22 kg / (1600000 m)^2\n\ng = 2.32 m/s^2\n\nThis is about a fourth of the values on earth.  That makes sense.  The mass and radius of this imaginary planet are comparable to the Moon (M = 7.36 x 10^22 kg, r = 1738 km) where the acceleration of gravity is about 1/6 of the Earth.
965826
L = 8pi G rho/c^4 is the definition of the cosmo constant L (1/m^2); where c = light speed and rho is the energy density of the known universe in a vacuum.\n\nF = GmM/r^2 is the equation for the attraction force F between two masses m and M, r distance apart.\n\nL was created by Einstein to make his equations come out right.  By that, he meant the universe should be spherical (not flat or saddle shaped).  His original results showed a flat universe, and he just "knew" that was wrong; so on came L to fix the shape problem.\n\nF was the result of experiments by Newton.  Through observation, he found that the attractive force between two bodies was proportional to the inverse square of the distance between them and to the product of the magnitudes of the two masses.  That is, F varies as (~) (mV/r^2).  \n\nWhenever there is a proportionality, as in F ~ mM/r^2, there has to be a constant of proportionality.  In this case Newton called it G; so that F ~ mM/r^2 becomes F = GmM/r^2 when the constant is inserted.\n\nThe G in both equations is the same constant.  But Sir Newton came upon it long before Einstein found it could be used in his equations and the calculation of L.  So there is no way that Newton (1642-1727) could know about Einstein's (1879-1955) use of it much later.
285876
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.
1036294
letter b is the answer.. because the weight of the object is approx. 627.20kg multiplied by 9.8(acceleration due to gravity) so the product is the force..
967250
The gravitotional force F of two masses m1 and m2 separated by a distance r is given by the equation\n\nF = G * m1 * m2 / r^2\n\nwhere G is a physical constant with value 6.67*10^ -11 N*m^2/kg^2\n\nSInce the third object is placed midway then its distance from either m1 or m2 is 0.15 m.\n\nSubstituting we get\nF1=3.9*10^-5 N\nand\nF2=9,5*10^-5 N\n\nSince the two forces have opposite directions then \nFtotal = F1 - F2 = -5.6*10^-5 N having the direction of F2.\n\nIn order for total to be zero we have\nF1=F2 ->\nG*m1*m/x^2=G*m2*m/y^2 ->\nm1/x^2 = m2/y^2 ->\nm1*y^2=m2*x^2 (1)\n\nwhere x and y the diastances of m from m1 and m2 respectively.\n\nsince\n\nx+y=r=0.300 m\n\nit follows that y = 0.3-x (2)\nCombining (1) and (2) we have\n\nm1*(0.3-x)^2=m2*x^2 ->\n215*(0.09-0.6*x+x^2)=515x^2 ->\n300x^2+129x-19,35=0\n\nsolving this we get\n\nx1=0.12m\nand\nx2=-0.55 m which is of course rejected since it represents a point where there can be no equilibrium.