Timeline for Tidal forces in the early solar system
Current License: CC BY-SA 4.0
3 events
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Apr 22 at 9:34 | comment | added | gandalf61 | @user3327311 I think $r$ is always the distance from the Sun to the cloud i.e. the radius of the cloud's orbit, not the radius of the cloud itself. I don't think the tidal effect of the cloud on itself is discussed - probably assumed to be not significant. The overall point is that the tidal effect of the sun on the cloud is always greater than the cloud's self-gravitation (as long as the density of the cloud is less than the density of the Sun). | |
Apr 22 at 9:27 | comment | added | user3327311 | Yes, I agree. So it's ok to treat the effect of the Sun on the gas cloud as a tidal force proportional to $M/r^3$, where $M$ is the mass of the Sun and $r$ is the distance from the Sun to the cloud, because the size of the cloud is small compared to $r$. But why is it ok to treat the effect of the central portion of the cloud on its outer portion as a tidal force proportional to $M/r^3$, where $M$ is the mass of the cloud and $r$ is the radius of the cloud? I think that's what he's doing, anyway. | |
Apr 22 at 5:59 | history | answered | gandalf61 | CC BY-SA 4.0 |