Can Newton gravitational constant be equal to Space curvature/stress energy momentum tensor?
814715
no. \nthe stress-energy tensor tells us how the energy is spread, and moves, across spacetime.\nthe gravitational constant is a parameter that determines to what degree 1 kg of mass (for instance) will "warp" the spacetime. \n\nso, you can't derive it from the einstein equation.\n\nin fact, relativists usually set the gravitational constant to = 1, so that they don't have to deal with it. although it's essential when you're trying to determine exactly what your physical predictions will be, it's not really an active element of einstein's theory. it just kind of sits there.
643671
The gravitational constant is constant by definition and is independent of all other physical values however there is some evidence to suggest that it is not a universal constant and indeed it's value may change over time.
611971
Mass is a constant; weight varies by gravitational pull.
1109075
Einstein's law of gravitation. It certainly looks nicer than Newton's :P
733729
http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
184873
Yes.\ngravity is inversely proportional to radius squared, and radius is inversely proportional to the curvature of the same surface, so your statement is true.
253430
Well if gravity is what Einstein said it is=Curvature of space time.\nThan you are going to find some space time and uncurve it!\nThis would be anti-Gravity.
546224
Energy = force x distance\n=>  Energy = (mass x acceleration) x Distance\n=>  Energy = mass x acceleration x Distance\n=>  Energy = mass x (speed / time) x Distance\n=>  Energy = mass x speed x (Distance/time)\n=>  Energy = mass x speed x speed\n=>  Energy = mass x (speed)²
565790
g=m*a\nGravitational Force=Mass * Acceleration
773114
F=F1-F2\nF=qm(M1/R1^2-M2/R2^2)\nWhere \nq- universal gravitational constant (6.672 x 10^-11 N m^2/kg^2) \nM1- mass of  407kg\nM2 – mass of  102kg\nM - 21.7 kg\nR1-distance between M1 and the object\nR1-distance between M2 and the object\n\nIn the first case R1=R2=Rso we have \n\nF=qm(M1 -M2)/R^2\nF=6.672 x 10^-11 x 21.7(407 + 102kg) / (0.475)^2=3.266e-6 N\n\n\n\nFor the second question \n\nF=F1-F2=0\n\nSo \n\nM1/R1^2=M2/R2^2\n\nR1/R2 = (M2/M1)^.5\n\nAnd we know that R1+R2=0.475 m\nSo R2=0.475 –R1\n\nThen we have \n\nR1 = (M2/M1)^.5 (0.475 –R1)\nR1=((M2/M1)^.5  x 0.475)/(1+(M2/M1)^.5)\nR1=(1.997 x.475)/(1 + 1.997)= 0.317 m\nAnd R2 must me then 0.475-.317=0.158  m
945562
depends on which g you are talking about:\nG, the gravitational constant was discovered by Cavendish.  Upon this discovery, he was able to find the mass of the earth.\ng, the acceleration due to gravity has been explained by 2 ppl already.
951543
It is not mass that curves space time. Remember that mass is transformed between frames under relativity. Instead it is stress energy. This means that frame invariant combination of mass and energy, including the effect of the movement of the mass.\n\nAnd there is no prospect that these are mutual.
67071
In Newtonian space, F = ma or m = F / a (mass is equal to the force it has divided by the acceleration applied to it).\nIn an space-time continuum (relativity aspects as per Einstein) the mass is relative to its location and velocity with respect to the observer.
1082685
Simply equate gravity function with centripetal acceleration.\n\nGMe/(r+h)^2 = V^2/(r+h)....where r is earth radius and h is height\n\nsolve for v
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.
783835
gravitational potential energy\nm*g*h where m=the mass g= the acceleration of gravity which is(9.8m/s^2). h= the height.
727214
http://en.wikipedia.org/wiki/Strain_%28materials_science%29\nhttp://en.wikipedia.org/wiki/Specific_gravity\nhttp://en.wikipedia.org/wiki/Stress_%28physics%29\nhttp://en.wikipedia.org/wiki/Reynold%27s_number