The height of any point $$P$$ above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point $$P$$ will be : (Given $$g$$ = acceleration due to gravity at the surface of earth).

We need to find the acceleration due to gravity at point $$P$$ which is at a height equal to the diameter of the Earth above the surface. The height of point $$P$$ above the surface is the diameter of the Earth, $$2R$$, where $$R$$ is the radius of the Earth, so the distance from the center of the Earth is $$d = R + h = R + 2R = 3R$$.

The acceleration due to gravity at a distance $$d$$ from the center (when $$d > R$$) is given by $$g' = \frac{g R^2}{d^2}$$. Substituting $$d = 3R$$ yields $$g' = \frac{g R^2}{(3R)^2} = \frac{g R^2}{9R^2} = \frac{g}{9}$$.

The correct answer is Option D: $$\frac{g}{9}$$.