Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

put on hold as off-topic by Jim, Brandon Enright, Kyle Kanos, JamalS, Colin McFaul Oct 17 at 21:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Jim, Brandon Enright, Kyle Kanos, JamalS, Colin McFaul
If this question can be reworded to fit the rules in the help center, please edit the question.

1  
Is this homework? –  Jaime Nov 30 '12 at 17:33
    
@Jaime: No, it is not. –  hkBattousai Nov 30 '12 at 21:32
1  
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

1 Answer 1

up vote 1 down vote accepted

All celestial bodies loose 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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.