memory loss. g=(re2/r2)g. equation for inside or outside?

Answer 1

What you wrote is incomplete and/or incorrect. The g's in the equation would cancel out; so that you'd have 1 =(re2/r2), which could be true but only in the special case where re = r. In other words, 1 = 1 is what you wrote; I don't think you meant to do that.

What I think you are trying to write is g/G = (r/R)^2; where g and G are accelerations due to gravity at two different radii r < R from the center of mass M. Then you'd have g = G (r/R)^2 as the correct equation. What this shows is that the acceleration due to gravity varies inversely as the square of the distance from the center of the mass M.

For example, let R = 2r and r = 1 Earth radius, and G = 9.81 m/sec^2 on Earth's surface (i.e., at r). What's g, the acceleration at R. For this we solve g = G (r/R)^2 = 9.81(1/2)^2 = 9.81(1/4) ~ 2.2 m/sec^2 roughly or about 1/4 of the acceleration due to gravity on the Earth's surface.

For remembering this relationship, remember the physics of any force field as it spreads out from a point. For gravity, the physics is best known as Newton's gravity law F = GmM/r^2 = mg = W the weight of a mass m under the gravity acceleration g. The two m's, the weighed object, cancel...leaving us with g = GM/r^2 where G in this case is a constant, M is the source of the gravity (Earth), and r is the radius from the source.

And there you have it...g ~ 1/r^2, it is proportional to 1/r^2, because G and M are constants (usually) and, combined, they are the constant of proportionality. So if we have two radii (r and R), we'd have two g's (g and g'). Then the ratio g/g' = (1/r^2)//((1/R^2) = (R/r)^2 such that g = g'(R/r)^2 which is the equation you were unclear on.