Density and Gravity Let's say we have two spheres. One is twice as dense as the other but they have the same mass. Were you to stand on the surface of each would you experience any difference in gravitational force or would they be the same?  
To clarify, I am not curious as to the sphere's gravitational effects on each other, but as to whether the gravitational force one would experience on one would differ from that of the other.  
 A: The acceleration due to gravity $g$ on a sphere of radius $r$ with mass $M$ is given by 
$$g = GM / r^2$$
where $G$ is the universal constant of gravitation.  
The volume of a sphere is given by
$$V = \frac43 \pi r^3$$
and density $\rho$ is given by 
$$\rho = M / V$$
Combining those equations and eliminating $r$ we get
$$g = G \left(\frac{4\pi}{3}\right)^{\frac23}M^{\frac13}\rho^{\frac23}$$
So if mass is constant and density is doubled, gravity is scaled by $2^{\frac23}$, or approximately 1.5874. So if you did this to the Earth $g$ would go up from $9.81ms^{-2}$ to  $15.57ms^{-2}$.
A: You'd experience more gravitation standing on the denser sphere, because you'd be closer to its center of mass.
Considering that the two spheres are the same mass but different densities, and density is mass/volume, then this implies that the more dense sphere has a smaller volume than the other.  And since the volume of a sphere is $v = \frac{4}{3} \pi r^3$, then this implies that reduced $v$ equals reduced radius ($r$).
Smaller radius means that someone standing on its surface is closer to its center of mass.
Addendum
I forgot to mention that gravity works with the inverse-square law.
A: The gravity experienced on the surface of a sphere of mass $ M$ is proportional to $M/r^2$, where  $r$ is the radius of the sphere.  If the sphere is compressed to give it a greater density without changing its mass, $r$ gets smaller, so the force gets greater.    Bottom line:  the gravitational force is greater on the surface of the denser (smaller) of two equal-mass spheres.
A: I will give laymans tearms a shot.... Imagine the sphere looks like a clock and you are standing over the 12. You have that one number holding you down. Now shrink the clock. You now cover the 11, 12, and 1. You now have three numbers holding you down. Now think of the numbers as atoms....
