<p>In the textbook treatment the pressure decreases exponentially with height as <span class="math-container">$P(z)=P(0)\exp(-z/H)$</span> where <span class="math-container">$H=kT/mg$</span> is the scale height of the atmosphere. This assumes that the gravity <span class="math-container">$g$</span> is constant with altitude, which is a decent approximation for planets with an atmospheric height much smaller than their radius. Since <span class="math-container">$H$</span> is about 8 km for Earth, this makes sense to use. </p>

<p>For an atmosphere with Earth-like temperature <span class="math-container">$T$</span> and mean molecular mass <span class="math-container">$m$</span> it would hence seem that for the tungsten planet that if we want <span class="math-container">$P(0)$</span> to be like on Earth <span class="math-container">$g$</span> must be equal to Earth's 9.82 m/s<span class="math-container">$^2$</span>. So if we have a planet with <span class="math-container">$g=G\rho (4\pi r^3/3)/r^2$</span>, we get <span class="math-container">$r=3g/4\pi G\rho$</span>. For <span class="math-container">$\rho=19300$</span> kg/m<span class="math-container">$^3$</span> this gives <span class="math-container">$r=1820$</span> km. </p>

<p>The limit is met somewhere before white dwarf density. They have a density of about a billion kg/m<span class="math-container">$^3$</span> 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. </p>

<p>Note that small but less dense bodies can have quite dense atmospheres. Titan has radius about 2550 km and a surface pressure 1.48 times Earth, despite lower surface gravity. The trick here is that the atmosphere is colder and slightly denser. </p>

CC BY-SA 4.0
2018-10-29T16:05:34.213Z
Venus has a lower mass, radius, and surface gravity than Earth, and it's much hotter, but its atmospheric pressure is huge.
PM 2Ring
CC BY-SA 4.0
2018-10-29T16:30:33.787Z
Why some planets have atmospheres and others don't, is not well understood and doesn't depend on surface gravity or temperature (which one? surface, radiative or effective temperature?) only.
AtmosphericPrisonEscape
CC BY-SA 4.0
2018-10-29T17:51:24.203Z
@PM2Ring  yeah, but its atmosphere is a heckuvalot thicker and extends higher
Carl Witthoft
CC BY-SA 4.0
2018-10-30T17:54:32.797Z
@PM2Ring - The amount of volatiles a planet gets depends on its formation and history. Venus presumably had outgassing of most of its carbon, producing a dense and high pressure atmosphere (with a big scale height due to the high temperature). If one allows adding atmosphere, then you might shrink tungsten-Earth somewhat and add more atmospheric mass to compensate for the lower gravity.
Anders Sandberg
CC BY-SA 4.0
2018-10-29T15:32:03.340Z
2018-10-29T15:32:03.340Z
Anders Sandberg
<p>Suppose there is a planet made of pure tungsten (a quite common and heavy element), with perfect spherical shape.</p>

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

<p>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.</p>

<p>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).</p>

CC BY-SA 4.0
2018-10-29T17:53:20.903Z
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
CC BY-SA 4.0
2018-10-29T22:36:32.067Z
@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
CC BY-SA 4.0
2018-10-30T12:58:41.357Z
@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
CC BY-SA 4.0
2018-10-30T13:26:21.623Z
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
CC BY-SA 4.0
2018-10-30T13:41:28.100Z
@PeterErwin: planet mass will not be the same, because the smaller radius allow smaller mass. Regarding the mass of the atmosphere, for simplicity I set the average temperature similar to Earth and also the start similar to our Sun.
Nicola Lepetit
<gravity><planetary-atmosphere><earth-like-planet>
What is the smallest possible planet with same sea level atmospheric pressure as Earth?
Nicola Lepetit
# What is the smallest possible planet with same sea level atmospheric pressure as Earth?

(Asked by: username_0 on 2018-10-29)

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).

**Tags:** gravity, planetary-atmosphere, earth-like-planet

> **Comments:**
> **username_1**: 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.  
> **username_2**: @CarlWitthoft The gravitational field profile would of course be different. $dg/dr = 2GM/r^3$. $M$ is the same but $r$ would be smaller.  
> **username_1**: @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.  
> **username_3**: 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.  
> **username_0**: @PeterErwin: planet mass will not be the same, because the smaller radius allow smaller mass. Regarding the mass of the atmosphere, for simplicity I set the average temperature similar to Earth and also the start similar to our Sun.  

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## 1 answers

### Answer by username_4 on 2018-10-29. Score: 4

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.

Note that small but less dense bodies can have quite dense atmospheres. Titan has radius about 2550 km and a surface pressure 1.48 times Earth, despite lower surface gravity. The trick here is that the atmosphere is colder and slightly denser.

> **Comments:**
> **username_5**: Venus has a lower mass, radius, and surface gravity than Earth, and it's much hotter, but its atmospheric pressure is huge.  
> **username_6**: Why some planets have atmospheres and others don't, is not well understood and doesn't depend on surface gravity or temperature (which one? surface, radiative or effective temperature?) only.  
> **username_1**: @PM2Ring  yeah, but its atmosphere is a heckuvalot thicker and extends higher  
> **username_4**: @PM2Ring - The amount of volatiles a planet gets depends on its formation and history. Venus presumably had outgassing of most of its carbon, producing a dense and high pressure atmosphere (with a big scale height due to the high temperature). If one allows adding atmosphere, then you might shrink tungsten-Earth somewhat and add more atmospheric mass to compensate for the lower gravity.  

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