8
$\begingroup$

Is there an easy way to calculate the Hill radius of a star in a binary system at different orbital radii where both stars are of the same mass and in circular orbit around one another’s centre of mass?

I have seen a formula for calculating Hill radii but it’s meant to address the stability of orbits around a small body which itself orbits a large one, an example of which would be the effect of the Sun's gravity on the orbit of an object around the Earth.

Can the same formula be used if both major bodies are of similar or the same size, for example to address the stability of planetary orbits around one star in a binary system?

Improve this question
$\endgroup$
6
  • $\begingroup$ Related question: astronomy.stackexchange.com/q/18844/24157 - unfortunately the answer there giving the derivation of the Hill radius (including the point where you make the approximation that the Roche lobe is small) is not ideal because it relies on an external link for the actual content. $\endgroup$
    – user24157
    Commented Dec 10, 2019 at 19:32
  • $\begingroup$ This is an interesting question, I wonder if it might be "Does anything like the concept of a Hill surface surface exist for a pair of objects of similar mass? If so, how useful is it?" $\endgroup$
    – uhoh
    Commented Dec 11, 2019 at 7:32
  • 2
    $\begingroup$ Just to clarify, are you talking about Hill radius of a planet orbiting one or both stars or hill radius of one of the stars? Each one is a kind of interesting mathematical question, but it would be 3 answers, the the S vs P type planet's hill radius might be close enough for one answer. en.wikipedia.org/wiki/Habitability_of_binary_star_systems $\endgroup$
    – userLTK
    Commented Dec 11, 2019 at 8:34
  • 1
    $\begingroup$ I'm ultimately trying to establish roughly how close two Sun / Earth analogues could be (so S type) if the stars were in a loose binary. I thought that asking about the hill radius of one of the stars would clarify the matter? I think the link might provide a clue - one fifth of the orbital radius... maybe one tenth to be on the safe side $\endgroup$
    – Slarty
    Commented Dec 11, 2019 at 9:28
  • 1
    $\begingroup$ I added some of that back into your question post (some people may not read through the comments). $\endgroup$
    – uhoh
    Commented Dec 11, 2019 at 13:36

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.