Planetary Gravity and its effects This is my first question on the Physics portion of Stack Exchange.  I was hoping to get some light on the topic of gravity.  I don't have much background knowledge of physics so I might as well start here.
I have a few planet and gravity related questions:
1). Does planetary gravity depend on size or density? Or both?
2). Is it plausible for a large (very large), terrestrial planet to have tall mountains like Mount Everest or even like on Mars?
3). Is there a way, given an average density per cubic meter along with the average radius of the planet, to calculate the pull of gravity?
I've been scouring the internet for these things with no luck so far.  The idea behind this is to make a planet rendering program which takes all of this into effect.  If this forum doesn't answer this sort of thing, feel free to point me in the right direction.  Thanks!
 A: 1/3) As Newton pointed out way back in the Principia, the gravitational attraction due to a spherically symmetric mass distribution is, assuming you are outside of it, the same as if all the mass were at a point at the center. Thus the gravitational acceleration at the surface of a sphere is determined solely by the total mass $M$ and the radius $R$. What is that acceleration? Well, Newton helps us again and says it is
$$ a = \frac{GM}{R^2}, $$
where $G = 6.67\times10^{-11}\ \mathrm{m}^3/(\mathrm{kg}\cdot\mathrm{s}^2)$. (Historical point - Newton had no way to measure $G$ with the instruments of his day; that came later.)
Now you bring up density, which allows us to write this in another way. The average density of our sphere is
$$ \rho = \frac{M}{(4\pi/3)R^3}, $$
so we can rearrange things to say
$$ a = \frac{4\pi}{3} GR\rho. $$
Thus if your two variables of interest are $R$ and $\rho$, the "surface gravity" (i.e. acceleration at the surface) is directly proportional to both. Note that in our solar system, average densities vary from a little under $1\ \mathrm{g}/\mathrm{cm}^3$ to a little more than $5\ \mathrm{g}/\mathrm{cm}$. The change in $R$ is much more extreme.
However you calculate the acceleration, the force on a mass $m$ at the surface is $F = ma$, again as per Newton.
2) In general, the bigger something is, the less it can deviate from having a smooth surface. The reason is tall mountains will be crushed under their own weight, and valleys will not be able to support their walls. The specific limit depends on what you mean by "very large."
