# Placing a planet on the famous Escape Velocity Graph

If I'm understanding the chart correctly, in order for a planet with lower gravity than Earth (smaller than Earth), but with decent magnetosphere, to hold on to an atmosphere with oxygen, and water on the surface, that planet would have to be in the blue but can be close to the green (between Venus and Mars)?

And if I'm way off, please don't laugh at the science noob.

But you CAN laugh at the fact that I wrote magentosphere in the graph, instead of magnetosphere.

• Ha ha. You wrote "magentosphere". The closest thing to "magenta" on that chart is the hydrogen band.
– Mark
Apr 29 at 2:02
• @Mark ...I DID say you could laugh at that...
– Len
Apr 29 at 2:06

# Correct!

Atmospheric particles are constantly moving around at enormous speed. For instance, the typical air particle (the kind you're breathing right now) at room temperature moves at around 1,000 mph. Since there are trillions of them, all moving in random directions, there is no net motion.

## Escaping Gasses

Earth's escape velocity is around 7 miles per second. If a gas particle is at the top of the atmosphere, and it suddenly achieves this velocity (as trillions are doing at this second), it will leave Earth's orbit, never to be seen again. However, in order for a gas particle to achieve this velocity, it typically has to be smacked by some very high-speed radiation.

## Keeping Gasses

As you have surmised, Earth's magnetosphere stops atmospheric gas particles from escaping in large number. It does this by deflecting high-speed cosmic radiation before it can smack into atmospheric gas particles. If Earth's magnetosphere were more powerful, even fewer cosmic radiation particles would make it through. If Earth's atmospheric temperature were lower, the baseline speed of atmospheric particles would be lower, meaning it would require even faster cosmic radiation particles to make an atmospheric particle reach escape velocity. A similar effect would be achieved if Earth's escape velocity were higher.

• While all of this is true, the need for a strong magnetosphere has been exaggerated a bit too much. If you are very close to the star, such as Mercury is, then yes, solar wind will strip your atmosphere away. If you are the distance that Venus or the Earth are, not so much. Escape velocity is far more important than magnetosphere. Mars atmosphere could be blown away because Mars gravity is weak, just like the Moon. In Venus, lack of magnetoshpere doesn't prevent a very thick atmosphere, although it may be the cause for the lack of water, if cosmic rays splat water molecules into H and O. Apr 29 at 8:58

You set the temperature to be around 250K. That means -23C or -9F.

I would say that's a tad too cold to have liquid water on the planet. If there is water it will be frozen.

Remember that at 1 bar water is liquid above 273K, so if you want liquid water, you need a temperature above that (or higher pressure).

• So horizontally I'm ok, but vertically it should be just under Earth? How's the neighborhood of 280s K?
– Len
Apr 28 at 15:31
• Salt water maybe? Apr 28 at 18:08
• @Willk, oh yeah, definitely salt water! That's helpful?
– Len
Apr 28 at 19:14
• @Len Having very little surface water could also work. That would prevent the planet from spiraling into an iceage. Think of a mars like desert planet with huge ice caps, a ark surface and a number of very salty inland seas. Apr 28 at 20:33
• @TheDyingOfLight Oh that's good, because my next question was going to be if this and (Reef World) worldbuilding.stackexchange.com/questions/201259/… could be the same world! Sounds like maybe yes(?) A reef world is not a LOT of water, is it?
– Len
Apr 28 at 20:41

Strictly speaking, the escape velocity would decrease, but the surface gravity wouldn't necessarily change, depending on your planet's parameters. The two are determined by the mass $$M$$ and radius $$R$$ by $$v_e=\sqrt{\frac{2GM}{R}},\qquad g=\frac{GM}{R^2}$$ with $$G$$ the gravitational constant. You've picked the escape velocity, but we need to know another parameter to determine the surface gravity. Picking a mass or radius would work, but I think it's better to specify a density. After all, you don't want your terrestrial planet to have the density of gas! This gives us a third formula, for the density $$\rho$$: $$\rho=\frac{M}{\frac{4\pi}{3}R^3}$$ If we specify $$v_e=7\;\text{km/s}$$ and $$\rho=5.5\;\text{g/cm}^3$$, the density of Earth, we find that the radius of your planet is $$R\approx0.56R_{\oplus}$$ and the mass is $$M\approx0.18M_{\oplus}$$, leading to a surface gravity of $$g=5.6\;\text{m/s}$$, which is indeed lower than Earth's but still fairly reasonable! So it turns out that we didn't have to be concerned after all.

In fact, if you do out the algebra, it turns out that $$g=v_e\sqrt{\frac{2}{3}\pi\rho G}\propto v_e\sqrt{\rho}$$ This means that you do have to change the density significantly to get a significant effect on the surface gravity, unless you're willing to change the escape velocity a bit.

Looking at some exoplanet models (Seager et al. 2008), it seems that this would be reasonable for a planet with more iron than is found in the terrestrial planets in our Solar System. There would still be quite a lot of silicates, as is the case on Earth, but you would need iron to reach the requisite density. After all, your planet is roughly the same radius as Mars but 80% more massive.

The temperature also looks fairly reasonable. You could honestly increase it a bit to make it easier for liquid water to exist, and all that would require is making its orbit slightly smaller or making its star slightly brighter. If your current temperature is about 250 K, you could raise it to 280 K (closer to Earth's temperature) by decreasing its orbital radius by about 20%.

• I assume the density will help in figuring out the gravity? I'm basically trying for a planet with much lower gravity than Earth that still retains a breathable atmosphere and oceans. I know that seems difficult, but there has to be a sweet spot that gives me a decent amount of all the necessary elements...
– Len
Apr 28 at 19:12
• @Len Yeah, if you specify density and escape velocity, that will determine the surface gravity. If my algebra is correct, it should be $g=v_e\sqrt{\frac{2}{3}\pi\rho G}$. Apr 28 at 20:58