Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Is it possible for a moon to stay on the same side of its planet relative to the sun?

+0
−0

I'm building an earthlike world that has a moon orbiting a gas giant. Is it possible for the moon to always be between the planet and the sun? Also, is it possible for a moon to rotate around its own axis?

History
Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

This post was sourced from https://worldbuilding.stackexchange.com/q/175057. It is licensed under CC BY-SA 4.0.

0 comment threads

1 answer

+0
−0

This answer is meant as a supplement to notovny's. I agree with their conclusions (the scenario is impossible because of the instability of this Lagrange point, and the fact that the Hill sphere is too small), and I just want to derive the "curious relation" they came up with.

We start with Kepler's third law. $T_M$ and $T_p$ are the periods of the planet and the moon; $a_M$ and $a_p$ are their semi-major axes; $M_p$ and $M_S$ are the masses of the planet and the start. Let's write out Kepler's third law for both the orbit of the moon and the orbit of the planet: $$T_M^2=\frac{4\pi^2}{GM_p}a_M^3,\quad T_p^2=\frac{4\pi^2}{GM_S}a_p^3$$ If we assume the moon is in its outermost orbit, we have $$a_M=a_p\sqrt[3]{\frac{M_p}{3M_S}}$$ Now we substitute and our first equation is $$T_M^2=\frac{4\pi^2}{GM_p}a_p^3\frac{M_p}{3M_S}$$ Finally, we divide by the equation for the planet's period: $$\frac{T_M}{T_p}=\frac{M_S}{M_p}\frac{M_p}{3M_S}$$ and so $T_M\approx0.58T_p$, which is the result notovny found. It's interesting to think about this in the case of a binary planet ($M_p\approx M_M$) or a binary star ($M_S\approx M_p$). Kepler's third law is easy to modify for both of those cases. However, the derivation of the Hill radius requires that $M_p\ll M_S$, and that the Hill radius $R_H\ll a_p$. If we get rid of that requirement, then I believe a general solution would require finding the roots of a fifth-order polynomial in $x\equiv R_H/a_p$, which unfortunately has no general solution. For particular values of $M_p$ and $M_S$, we may be able to find solutions, but we'd need to look at them on a case-by-case basis.

History
Why does this post require moderator attention?
You might want to add some details to your flag.

0 comment threads

Sign up to answer this question »