<P> where r ^ (\ displaystyle (\ hat (\ mathbf (r)))) is a unit vector lying along the line joining M and m and pointing from m to M. Therefore, the gravitational field of M is </P> <Dl> <Dd> g (r) = F (r) m = − G M r 2 r ^ . (\ displaystyle \ mathbf (g) (\ mathbf (r)) = (\ frac (\ mathbf (F) (\ mathbf (r))) (m)) = - (\ frac (GM) (r ^ (2))) (\ hat (\ mathbf (r))).) </Dd> </Dl> <Dd> g (r) = F (r) m = − G M r 2 r ^ . (\ displaystyle \ mathbf (g) (\ mathbf (r)) = (\ frac (\ mathbf (F) (\ mathbf (r))) (m)) = - (\ frac (GM) (r ^ (2))) (\ hat (\ mathbf (r))).) </Dd> <P> The experimental observation that inertial mass and gravitational mass are equal to an unprecedented level of accuracy leads to the identity that gravitational field strength is identical to the acceleration experienced by a particle . This is the starting point of the equivalence principle, which leads to general relativity . </P>

A field of force can exist only between