<p>You can account for this friction, but it is totally negligible. The basic reason is that the Sun and planets are very compact, while the distributed gas is very sparse.</p>
<p>The density of the interplanetary medium is of order <span class="math-container">$10^7$</span> particles per cubic metre. Let's say they are hydrogen atoms, so the mass density is of order <span class="math-container">$\rho \sim 10^{-20}$</span> kg/m<span class="math-container">$^3$</span>.</p>
<p>The drag force on something moving through a fluid is
<span class="math-container">$$F_d \sim \rho A v^2\ ,$$</span>
where <span class="math-container">$A$</span> is the effective area presented by the object and <span class="math-container">$v$</span> is the relative velocity of the object and fluid.</p>
<p>Let's assume <span class="math-container">$A= \pi R_E^2$</span> for the Earth, with <span class="math-container">$R_E \sim 6400$</span> km, and that <span class="math-container">$v \sim 30$</span> km/s, the orbital speed around the Sun. Putting everything in SI units we find <span class="math-container">$F_d \sim 10^3$</span> Newtons. Maybe that sounds a lot, but given the mass of the Earth it only produces an acceleration of <span class="math-container">$10^{-22}$</span> m/s<span class="math-container">$^2$</span> and thus is utterly negligible.</p>
<p>You can do a similar analysis for the Sun in orbit at 250 km/s around the Galaxy, with an interstellar density of about <span class="math-container">$10^6$</span> particles per cubic metre. Again, totally negligible.</p>

CC BY-SA 4.0
2023-04-20T10:39:12.363Z
Small comment: For very sparse gases, the appropriate drag law is the Epstein drag, not the Stokes drag invoked here. Doesn't change the main conclusion though, as the density is many orders of magnitude away from any drag law being relevant.
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-04-20T11:23:11.833Z
@AtmosphericPrisonEscape for the Earth the sound speed in interplanetary space is roughly equal to the orbital speed so it makes little difference whether you adopt Epstein ($\propto c_s v$)  or aerodynamic drag ($\propto v^2$).
ProfRob
CC BY-SA 4.0
2023-04-20T18:15:33.033Z
Plus the solar wind is clearing out the space the sun is moving through.
Jason Goemaat
CC BY-SA 4.0
2023-04-20T21:27:17.690Z
Answer is spot on, but drag of any sort is the wrong concept. Bodies in space will accrete any dust they intercept
thegreatemu
CC BY-SA 4.0
2023-04-20T06:38:08.950Z
2023-04-20T06:38:08.950Z
ProfRob
<p>If 10% of the galaxy is filled with gasses, and stars contribute to only 1%–2% of the total matter, then shouldn't the gas create friction for the movement of stars around the galactic center and also for planets around a star?</p>
<p>For example, if earth is revolving around the sun through a gaseous medium, then the friction due to gas should keep decelerating the earth. Why is this not happening?</p>

CC BY-SA 4.0
2023-04-20T05:16:32.467Z
Why would the gas rotate at a different speed around the Galactic center than the Sun?
planetmaker
CC BY-SA 4.0
2023-04-20T09:26:27.833Z
@planetmaker In practice, gas tends to move in orbits very close to circular, while stars can have very non-circular orbits (or even retrograde orbits). So it's perfectly possible for stars to move relative to the gas (even if the effect is, as ProfRob shows in his answer, negligible).
Peter Erwin
CC BY-SA 4.0
2023-04-20T09:27:34.067Z
Note that in the Milky Way, the stars make up about five to ten times as much of the (normal) matter as the gas does.
Peter Erwin
CC BY-SA 4.0
2023-04-20T21:24:48.267Z
@planetmaker Our sun (and most stars) have some *peculiar motion* (technical term) relative to the average galactic rotation at their point on the disk. This is important e.g. when calculating the expected signal from galactic dark matter
thegreatemu
<observational-astronomy><galaxy><planet>
Planet and star rotation through gas in the galaxy
Angela
# Planet and star rotation through gas in the galaxy

(Asked by: username_0 on 2023-04-20)

If 10% of the galaxy is filled with gasses, and stars contribute to only 1%–2% of the total matter, then shouldn't the gas create friction for the movement of stars around the galactic center and also for planets around a star?

For example, if earth is revolving around the sun through a gaseous medium, then the friction due to gas should keep decelerating the earth. Why is this not happening?

**Tags:** observational-astronomy, galaxy, planet

> **Comments:**
> **username_1**: Why would the gas rotate at a different speed around the Galactic center than the Sun?  
> **username_2**: @username_1 In practice, gas tends to move in orbits very close to circular, while stars can have very non-circular orbits (or even retrograde orbits). So it's perfectly possible for stars to move relative to the gas (even if the effect is, as username_4 shows in his answer, negligible).  
> **username_2**: Note that in the Milky Way, the stars make up about five to ten times as much of the (normal) matter as the gas does.  
> **username_3**: @username_1 Our sun (and most stars) have some *peculiar motion* (technical term) relative to the average galactic rotation at their point on the disk. This is important e.g. when calculating the expected signal from galactic dark matter  

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## 1 answers

### Answer by username_4 on 2023-04-20. Score: 10 | Accepted answer

You can account for this friction, but it is totally negligible. The basic reason is that the Sun and planets are very compact, while the distributed gas is very sparse.

The density of the interplanetary medium is of order $10^7$ particles per cubic metre. Let's say they are hydrogen atoms, so the mass density is of order $\rho \sim 10^{-20}$ kg/m$^3$.

The drag force on something moving through a fluid is
$$F\_d \sim \rho A v^2\ ,$$
where $A$ is the effective area presented by the object and $v$ is the relative velocity of the object and fluid.

Let's assume $A= \pi R\_E^2$ for the Earth, with $R\_E \sim 6400$ km, and that $v \sim 30$ km/s, the orbital speed around the Sun. Putting everything in SI units we find $F\_d \sim 10^3$ Newtons. Maybe that sounds a lot, but given the mass of the Earth it only produces an acceleration of $10^{-22}$ m/s$^2$ and thus is utterly negligible.

You can do a similar analysis for the Sun in orbit at 250 km/s around the Galaxy, with an interstellar density of about $10^6$ particles per cubic metre. Again, totally negligible.

> **Comments:**
> **username_5**: Small comment: For very sparse gases, the appropriate drag law is the Epstein drag, not the Stokes drag invoked here. Doesn't change the main conclusion though, as the density is many orders of magnitude away from any drag law being relevant.  
> **username_4**: @username_5 for the Earth the sound speed in interplanetary space is roughly equal to the orbital speed so it makes little difference whether you adopt Epstein ($\propto c_s v$)  or aerodynamic drag ($\propto v^2$).  
> **username_6**: Plus the solar wind is clearing out the space the sun is moving through.  
> **username_3**: Answer is spot on, but drag of any sort is the wrong concept. Bodies in space will accrete any dust they intercept  

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