# 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!

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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."

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Thank you! Very in depth and specific answer. Just one question: what is 'G'? Is it just a mathematical constant that we calculated to make the equation work? –  CoderTheTyler Dec 19 '12 at 2:32
Not so much calculated as measured. Think of it this way - people realized that gravitational acceleration was proportional to mass, and inversely proportional to distance squared (and not proportional to anything else). They wanted to measure mass in kg and distance in m, so to get the accelerations to come out to the right number in m/s^2, they measured G. And by measured G I mean measued the other 3 quantities and worked out what G had to be. –  Kyle Oman Dec 19 '12 at 3:04
@Kyle Couldn't have said it better myself. –  Chris White Dec 19 '12 at 4:19
@MrDoctorProfessorTyler For more info, it is called the gravitational constant. Actually, it has proven notoriously difficult to measure precisely, compared with other physical constants. –  Chris White Dec 19 '12 at 4:21
This is all rather interesting. Plus, I can actually calculate these values without some strange computational work around. Thanks everyone for the help! –  CoderTheTyler Dec 19 '12 at 23:16