# What is the critical mass of a planet to have an atmosphere like Earth's? [closed]

Small planets/orbits like Moon cannot have atmosphere because of their masses. They don't have enough gravity to hold an atmosphere. Then what is the critical mass that makes enough gravity to keep an atmosphere?

Please explain by Mathematical equations with the data given below:
Density of the planet is uniform everywhere; 5 g/cm3.
The atmosphere consists of pure O2 gas.
Temperature of the the atmosphere is uniform everywhere and fixed to 300K.
What is the critical size of radius or mass of this planet that makes it possible to have an atmosphere?

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## closed as off-topic by Jim, Brandon Enright, Kyle Kanos, JamalS, Colin McFaulOct 17 '14 at 21:54

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Is this homework? –  Jaime Nov 30 '12 at 17:33
@Jaime: No, it is not. –  hkBattousai Nov 30 '12 at 21:32
You can calculate the mass of the planet, therefore the gravitational attraction. You can also calculate thermal escape from the planet. An atmosphere can be held if escape is not too large. In reality you also need to take into account the stellar wind but from the questioning it looks like it's negligible. It does look like an exercise from a textbook. –  gerrit Dec 1 '12 at 16:01

All celestial bodies lose atmosphere due to a portion of the gas "near space" exceeding escape velocity. The velocity distribution of an ideal gas can be found using the Maxwell-Boltzmann distribution. So an easy approximation for this problem is to say we only want $10^{-6}$ of the molecules to have escape velocity. Using oxygen at 300K, results in an escape velocity of around 2.2 km/s. If we wanted only $10^{-16}$ of our particles to escape, that bumps the escape velocity up to 3.5 km/s

$$KE+GPE=0$$

$$\frac12mV^2 + -\frac{GMm}{r}=0$$

$$V^2-\frac{2GM}r=0$$

Let's first assume the atmosphere is thin compared to the dimensions of the planet. This allows us to use the same radius for the gravitational potential energy and the planet radius, and it allows us to neglect the mass of the atmosphere.

$$M=\rho\frac43\pi r^3$$

$$V^2-G\rho\pi\frac83 r^2=0$$

$$r=\frac{V}{\sqrt{G \rho \pi \frac83}}$$

$$M=\frac18V^3G^{-3/2}\sqrt{\frac6{\pi\rho}}$$

For escape velocities of 2.2 and 3.5 km/s the masses of the planet would be $4.7\cdot10^{22}$ and $1.9\cdot10^{23}$ kg respectively. This latter number is just a bit larger than the mass of Titan, the only known natural satellite with a dense atmosphere.

Note that the density is the denominator of this final equation indicating that a purely gaseous planet would have to be more massive to keep its atmosphere.

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