Let's imagine we have an algorithm that produce an elevation-map for a sphere. I wonder if the ratio between the planet radius and the delta between the highest and lowest altitude is a constant or can be guessed depending a few factors (main chemical components of the planet, atmosphere thickness, ...). Of course, I speak about telluric planets.

For example, Earth has a delta of, approximately, 20 km (Mount Everest in Nepal is 8,848 m hight and Mariana trench is 10,911 m deep in Pacific Ocean). See Wikipedia for more details. And, radius is about 6300 km. So, the final ratio is 0,003 (radius/delta).

On Mars (see here), the highest point is the peak of Olympus Mons at 21,229 m, and the deepest is in the Hellas Impact Crater which is 8,200 m deep. So, the total delta is about 29 km. Then, Mars radius is about 3400 km, which makes a ratio of 0.008.

As you can see, the variation of this ratio between these two planets are quite different.

So, I would like to have some way of "guessing" this ratio (maybe I am missing a few factors that I did not take into account, the radius is probably not enough). My point is to be able to make a map-making algorithm that will stay within realistic elevations when computing the points.

It can also be that I am totally wrong and such ratio do not exist (or has absolutely no sense at all), but, then, I would like to have a few arguments about it.


Just to make it clear, what I am looking for is an equation providing the delta of the elevation map (highest and deepest points) according to several parameters such as planet density and planet size (radius) and others...

Something like:

$$\Delta \text{(meter)} = \text{constant(m}^3\text{/kg)} \times \text{planet radius(meters)} \times \text{planet density(kg/m}^3\text{)}$$


I have collected a few samples to illustrate the formula that I am looking for. I recall that I am looking for the elevation delta based on various physical parameters which are only linked to the physics and NOT evolution of the landscape (no tectonic activity, no erosion, ...).

         delta     radius    density        surface gravity
Earth    20 km     6300 km   5.51 g/cm^3    1g
Mars     29 km     3400 km   3.93 g/cm^3    .376 g
Mercury  30 km     2439 km   5.43 g/cm^3    .38 g
Moon     18 km     1700 km   3.34 g/cm^3    .16 g

Somehow, I suspect that the planet radius and the surface gravity are involved in the formula but I don't see quite well how they interact right now. And, I suspect that I am still missing one parameter.

  • 6
    $\begingroup$ If I have time later, I'll find the references, but the highest peak a planet can support is based on the mass of the planet, that's why Mars has such higher peaks, less gravity pulling them down. $\endgroup$
    – bowlturner
    Sep 21, 2014 at 15:19
  • $\begingroup$ Gravity (or Mass)... Of course ! This was my missing parameter... I guess that computing this delta will involve: planet average density, radius and some magic constants. If you can find the precise equation it would be extremely useful ! Thanks ! $\endgroup$
    – perror
    Sep 21, 2014 at 15:21
  • $\begingroup$ You state the ratio is radius/delta, but the figures you have calculated appear to be delta/radius. Either is fine as long as they are consistent, so I don't know which one to correct... $\endgroup$ Sep 21, 2014 at 15:46
  • 1
    $\begingroup$ There are many factors to consider. An equation can only give rough upper and lower bounds, but I imagine such bounds would be useful. The exact figure will depend on the depth and density of atmosphere, the rate of incident space debris, the level of tectonic activity, the presence or absence of weathering and water erosion, and lots of things I can't think of... Also the answer will vary widely depending on whether you measure altitude from an idealised spherical planet surface, or from "sea level" (the oblate spheroid associated with rotation, regardless of whether there is a sea). $\endgroup$ Sep 21, 2014 at 15:57
  • $\begingroup$ What I am looking for is a way to compute this delta (in order to give limits to the algorithm building the elevation map). As I am looking for an equation, parameters can move from left to right and reverse... $\endgroup$
    – perror
    Sep 21, 2014 at 17:02

2 Answers 2


The main factors that would be involved are:

  1. Surface gravity - an effect of diameter and density
  2. Tectonic activity levels
  3. Erosion rate.

Mars has lower surface gravity, and due to its thinner atmosphere, a lower erosion rate. This means that there is less gravity to prevent taller mountains and less weather to wear them down.

So, unfortunately there isn't a single factor.

  • 1
    $\begingroup$ Those factors are connected, lower gravity tends to lead to less atmosphere. $\endgroup$
    – Tim B
    Sep 21, 2014 at 16:17
  • $\begingroup$ In fact, I would like to leave the factors such as tectonic activity and erosion rate to the algorithm computing the elevation map. On the other hand, the effect of gravity and/or planet mass cannot be workaround by the algorithm and are constants... and this is precisely what I would like to know (all the effects that cannot be changed in the algorithm). $\endgroup$
    – perror
    Sep 21, 2014 at 17:08
  • 1
    $\begingroup$ I think this is a good answer, but I would add vulcanism as a separate item, as it can be caused by forces not related to tectonics. The Hawaiian islands were caused by an interior hot spot. My recollection is that Mars has exhibited very little, if any tectonic activity, so Olympus Mons was probably created the same way. $\endgroup$ Sep 25, 2014 at 13:15
  • $\begingroup$ There is evidence that suggests that Mars once had a significant amount of surface water, and there is evidence of plate tectonics there: news.discovery.com/space/… $\endgroup$
    – Monty Wild
    Sep 25, 2014 at 14:51
  • $\begingroup$ @Donald.McLean: In fact, the reason Olympus Mons is so tall is because Mars has no tectonic activity. Where Hawaii was created as the plate moved across the hotspot, Olympus Mons just stayed in one place, so all the eruptions built on the previous ones, creating one super-mountain instead of a series of small mountains. $\endgroup$
    – mao47
    Oct 16, 2014 at 18:11

The tallest mountains possible depend on the strength of the rocks, which are fluid under high pressures and geologic time. They will sag even as their being pushed up. Mountains on the moon are very tall because the rocks are dry. Mountains on Venus are stubby because of the heat.

So it depends greatly on the composition and gravity. A larger planet will have stronger surface gravity and thus shorter mountains, which is scaling the opposite direction you were supposing.

You could look up the details for various rocky worlds and check for yourself.


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