What will be the value of 'g' and 'G' at the center of the earth and at the poles.?
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g=9.8 m/sec square at the pole \ng=0  m/sec at the center of earth\nG=6.67 * 10 (to power -11) newton merter square/Kilogram square (universal Constant remains same)
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K so...I haven't done any Grav froces recently...but I'll try\n\nIm guessing use  Net force = ma\nThe net force is the Gravitational force, Gm1m2 / r^2 is = m2 4 (pie)^2 r/ T^2\n\nfor the 4 (pie)^2 r/ T^2    r is the distance form the planet's center, to the moon, for T, find how many second 40 days is.\n\nand the other r is the planet's radius\n\nm2 cancels\n\nsolve for m1, and yo uhave the mass of the planet.\n\nI guess plug in numbers after that, into the equaltion Fg = Gm1 m2  /  r^2\n\nm2 would be the 70 kg\n\nhope thats right
913200
Both their accelerations will be the same and equal to the acceleration due to gravity, g=9.81 m/s/s...This is coz the only force acting on them is the force due to gravity
223868
depends.....at the equator, from the surface to sea level, it is 6,378.135 km. from the centre to the north or south pole, at sea level, it is 6,356.750 km. remember, both of those are to sea level, and they differ because the world is not a perfect sphere.
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W=mg ... where g is the gravitational acceleration
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WEight =[GMo * Mass to be weighted]/[Distance from mass to be weighted to center of the earth]^2\nWhere GMo/R^2 is the acceleration.\nG= gravity constant\nMo = mass of the earth in kilograms\nR = distance ro center of the earth in meters\nIf the mass to be weighted is one kilogram , the acceleration would be numerically equal to the weight.\nThe Units of weight is the same unit as gravity force which is Units of  Newtons.
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If we assume the Earth is a perfect spehere, the gravitational accleration would, indeed, be equal at any point along the surface of the sphere.\nWhen you take the rotation of the Earth into account, yes, at the equator, an object's "weight" is less than it would be at the poles.\n\nThe gravitational acceleration acts as the centripetal acceleration which keeps us all on on the Earth instead of flying out into space in a straight line.\nAt the poles, no centripetal acceleration is needed (neglecting Earth's revolution around the sun), but at the equation, the objects inertia acts to lessen the appearent effects of gravity and reduce its weight.  This is NOT due to the fictitious centrifuga force.
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check out http://www.answers.com/center%20of%20earth
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G=uys\nF=***\nE=verthing
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Jupiter has a "g" value of approx 26. The calculation can be made possible by simply knowing the Mass and Diameter of the planet, and by bearing in mind Issac's Newton's theory on gravitation.\n\nFgrav=(G*m1*m2)/(d*d)\n\nwhere G is the universal gravitational constant = 6.67*10^ -11 Nm^2/kg^2\nm1=mass of body 1\nm2=mass of body 2\nd=distance between body 1 and 2 (centre to centre)\n\n\nNow consider ur mass to be m2. So the force exterted by u against the planet would be mg=Fgrav. So the value of g simplifies out to be,\n\ng=(G*mass of planet)/d^2\n\nFor Jupiter, \nmass=1.9*10^27 kg\nradius=6.98*10^7 m \nTherefore, g= approx 26
945010
Equatorial gravity 3.69 m/s2\n(0.376g)
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sorry i am no good at math, but this may help:\nhttp://www.fourmilab.ch/earthview/pacalc.html
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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.
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If you're not taking earth rotation into account, the value of GMe is 9.8 * (6.38E6)^2, so the square of the radius required is that value divided by .660, the root of which is 2.4584521E7.  Subtracting the earth radius from that gives\n\n1.8204521E7, or about 18 million km.\n\nIf you're considering a point that stays directly above the earth as it rotates, the nswer depends on the latitude and is the solution of a cubic equation.   I'm not going there for you!
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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
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Add v2/R.\nv=speed of rotation at the suface of the earth.\n(it depends on the geographic altitude fi)\nR = radius of the earth.
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Measurements. The acceleration of gravity is:\ng = G*m/R^2\nWhere G is the gravitational constant 6.67*10^-11 N*m^2/s^2, m is the mass of the planet, and R is the distance between yourself and the planet. For the surface of the Earth, m and R are approximately:\nMe = 5.97*10^24 kg\nRe = 6.38*10^6 m\nUsing this, you should get about 9.8 m/s^2.