# What is the smallest possible planet with same sea level atmospheric pressure as Earth? [closed]

Suppose there is a planet made of pure tungsten (a quite common and heavy element), with perfect spherical shape.

Which should be its radius so that the air pressure and ground/sea level is the same as on earth ?

I don't know if it is sufficient to have a 1G gravity, because the smaller radius could affect the density of the atmosphere as the gravity will decrease differently that from earth.

For sake of simplificy, say that this planet is on the same orbit as Earth, orbiting around a twin of our Sun. Also magnetic field and atmosphere composition should be comparable to the Earth's one (altough this is not realistic, due to the lack of body of waters).

• I'm voting to close this question as off-topic because it is purely hypothetical (astronomy.meta.stackexchange.com/a/298/1569) – user1569 Oct 29 '18 at 15:25
• Gravity will not decrease differently above the ground level. Gravitational force depends on the distance from the center and on the total mass between you and the center. How the mass is distributed doesn't matter. – Carl Witthoft Oct 29 '18 at 17:53
• @CarlWitthoft The gravitational field profile would of course be different. $dg/dr = 2GM/r^3$. $M$ is the same but $r$ would be smaller. – ProfRob Oct 29 '18 at 22:36
• @RobJeffries Yeah, true that. $r$ is smaller at the surface, which changes the atmospheric profile until you get up to the equivalent of Earth's radius. Sorry about that. – Carl Witthoft Oct 30 '18 at 12:58
• There's not enough information in the question to constrain things. Is the planet's mass supposed to be same as the Earth's? Is the mass of the atmosphere supposed to be the same? You could have a planet with the same mass and radius as the Earth, with a much higher air pressure (see Venus), or with zero air pressure, depending on how much of an atmosphere you give it. – Peter Erwin Oct 30 '18 at 13:26

In the textbook treatment the pressure decreases exponentially with height as $$P(z)=P(0)\exp(-z/H)$$ where $$H=kT/mg$$ is the scale height of the atmosphere. This assumes that the gravity $$g$$ is constant with altitude, which is a decent approximation for planets with an atmospheric height much smaller than their radius. Since $$H$$ is about 8 km for Earth, this makes sense to use.
For an atmosphere with Earth-like temperature $$T$$ and mean molecular mass $$m$$ it would hence seem that for the tungsten planet that if we want $$P(0)$$ to be like on Earth $$g$$ must be equal to Earth's 9.82 m/s$$^2$$. So if we have a planet with $$g=G\rho (4\pi r^3/3)/r^2$$, we get $$r=3g/4\pi G\rho$$. For $$\rho=19300$$ kg/m$$^3$$ this gives $$r=1820$$ km.
The limit is met somewhere before white dwarf density. They have a density of about a billion kg/m$$^3$$ and a 35 meter radius white dwarf would have terrestrial surface gravity. At this point the above approximation has broken down (the scale height is bigger than the radius), and besides the low gravity would not be enough to hold it together. The real determinant of the limit will be the highest density material you can get that is stable at (near) zero pressure. For molecular matter tungsten is likely close to the limit, and we have reason to doubt degenerate white dwarf matter is up to the job.