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Can a system of three stars exist?
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Just like binary stars can a system of three stars mutually equidistant from each other?
What I mean is three stars at the vertices of a equilateral
If yes than what will be the orbit of a planet out there
If no what will be the problem for it's nonexistence?
star orbit
asked 8 hours ago
Heth GhetiyaHeth Ghetiya
112 bronze badges
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Heth Ghetiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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Just like binary stars can a system of three stars mutually equidistant from each other?
What I mean is three stars at the vertices of a equilateral
If yes than what will be the orbit of a planet out there
If no what will be the problem for it's nonexistence?
star orbit
asked 8 hours ago
Heth GhetiyaHeth Ghetiya
112 bronze badges
New contributor
Heth Ghetiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Just like binary stars can a system of three stars mutually equidistant from each other?
What I mean is three stars at the vertices of a equilateral
If yes than what will be the orbit of a planet out there
If no what will be the problem for it's nonexistence?
star orbit
asked 8 hours ago
Heth GhetiyaHeth Ghetiya
112 bronze badges
New contributor
Heth Ghetiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Just like binary stars can a system of three stars mutually equidistant from each other?
What I mean is three stars at the vertices of a equilateral
If yes than what will be the orbit of a planet out there
If no what will be the problem for it's nonexistence?
star orbit
star orbit
asked 8 hours ago
Heth GhetiyaHeth Ghetiya
112 bronze badges
New contributor
Heth Ghetiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Heth GhetiyaHeth Ghetiya
112 bronze badges
New contributor
Heth Ghetiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Heth GhetiyaHeth Ghetiya
112 bronze badges
New contributor
Heth Ghetiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Heth GhetiyaHeth Ghetiya
112 bronze badges
asked 8 hours ago
Heth GhetiyaHeth Ghetiya
112 bronze badges
112 bronze badges
New contributor
Heth Ghetiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Heth Ghetiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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Systems of three stars can exist, but a system of three stars in a triangle is unstable and won't exist in reality. There are configurations of three stars that are stable, for example, two stars in a close orbit about their common centre of gravity, and a third star in a distant orbit.
Planets can exist in such a system, they could orbit around the distant third star (like a moon orbits a planet), or they could be circumbinary, around the two close stars. Such complex systems are more likely to be unstable on the scale of billions of years. The key to stability is having each body in approximately an inverse-square gravitational field, so its orbit can be approximated by a Keplerian ellipse. That is not the case if three equal mass bodies are in an equilateral triangle.
answered 7 hours ago
James KJames K
38k2 gold badges66 silver badges128 bronze badges
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Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans.
$endgroup$
– peterh
6 hours ago
$begingroup$
There are stable equilateral systems, but not with equal masses. There's some info here & anims here. For some interesting stable n-body systems of equal mass see Cris Moore's gallery. Eg, the figure-8 orbit
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
It is possible in a Trojan configuration:
In the place of the "Planet" on the image, also a small star could exist. The third star would be at $L_4$ or at $L_5$. This configuration could be made stable.
However, as this link shows,
In unnormalized units, this criterion becomes
$$fracm_2m_1+ m_2 < 0.0385$$
We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable
equilibrium points, in the co-rotating frame, provided that mass $m_2$
is less than about $4%$ of mass $m_1$.
Thus, the mass of the second star should be at most 3.85% of the central star.
As far I know, no such known star system exists, but if it would, it would be stable.
answered 6 hours ago
peterhpeterh
1,8483 gold badges13 silver badges32 bronze badges
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
Systems of three stars can exist, but a system of three stars in a triangle is unstable and won't exist in reality. There are configurations of three stars that are stable, for example, two stars in a close orbit about their common centre of gravity, and a third star in a distant orbit.
Planets can exist in such a system, they could orbit around the distant third star (like a moon orbits a planet), or they could be circumbinary, around the two close stars. Such complex systems are more likely to be unstable on the scale of billions of years. The key to stability is having each body in approximately an inverse-square gravitational field, so its orbit can be approximated by a Keplerian ellipse. That is not the case if three equal mass bodies are in an equilateral triangle.
answered 7 hours ago
James KJames K
38k2 gold badges66 silver badges128 bronze badges
$endgroup$
$begingroup$
Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans.
$endgroup$
– peterh
6 hours ago
$begingroup$
There are stable equilateral systems, but not with equal masses. There's some info here & anims here. For some interesting stable n-body systems of equal mass see Cris Moore's gallery. Eg, the figure-8 orbit
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
Systems of three stars can exist, but a system of three stars in a triangle is unstable and won't exist in reality. There are configurations of three stars that are stable, for example, two stars in a close orbit about their common centre of gravity, and a third star in a distant orbit.
Planets can exist in such a system, they could orbit around the distant third star (like a moon orbits a planet), or they could be circumbinary, around the two close stars. Such complex systems are more likely to be unstable on the scale of billions of years. The key to stability is having each body in approximately an inverse-square gravitational field, so its orbit can be approximated by a Keplerian ellipse. That is not the case if three equal mass bodies are in an equilateral triangle.
answered 7 hours ago
James KJames K
38k2 gold badges66 silver badges128 bronze badges
$endgroup$
$begingroup$
Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans.
$endgroup$
– peterh
6 hours ago
$begingroup$
There are stable equilateral systems, but not with equal masses. There's some info here & anims here. For some interesting stable n-body systems of equal mass see Cris Moore's gallery. Eg, the figure-8 orbit
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
Systems of three stars can exist, but a system of three stars in a triangle is unstable and won't exist in reality. There are configurations of three stars that are stable, for example, two stars in a close orbit about their common centre of gravity, and a third star in a distant orbit.
Planets can exist in such a system, they could orbit around the distant third star (like a moon orbits a planet), or they could be circumbinary, around the two close stars. Such complex systems are more likely to be unstable on the scale of billions of years. The key to stability is having each body in approximately an inverse-square gravitational field, so its orbit can be approximated by a Keplerian ellipse. That is not the case if three equal mass bodies are in an equilateral triangle.
answered 7 hours ago
James KJames K
38k2 gold badges66 silver badges128 bronze badges
$endgroup$
Systems of three stars can exist, but a system of three stars in a triangle is unstable and won't exist in reality. There are configurations of three stars that are stable, for example, two stars in a close orbit about their common centre of gravity, and a third star in a distant orbit.
Planets can exist in such a system, they could orbit around the distant third star (like a moon orbits a planet), or they could be circumbinary, around the two close stars. Such complex systems are more likely to be unstable on the scale of billions of years. The key to stability is having each body in approximately an inverse-square gravitational field, so its orbit can be approximated by a Keplerian ellipse. That is not the case if three equal mass bodies are in an equilateral triangle.
answered 7 hours ago
James KJames K
38k2 gold badges66 silver badges128 bronze badges
answered 7 hours ago
James KJames K
38k2 gold badges66 silver badges128 bronze badges
answered 7 hours ago
James KJames K
38k2 gold badges66 silver badges128 bronze badges
answered 7 hours ago
James KJames K
38k2 gold badges66 silver badges128 bronze badges
38k2 gold badges66 silver badges128 bronze badges
$begingroup$
Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans.
$endgroup$
– peterh
6 hours ago
$begingroup$
There are stable equilateral systems, but not with equal masses. There's some info here & anims here. For some interesting stable n-body systems of equal mass see Cris Moore's gallery. Eg, the figure-8 orbit
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans.
$endgroup$
– peterh
6 hours ago
$begingroup$
There are stable equilateral systems, but not with equal masses. There's some info here & anims here. For some interesting stable n-body systems of equal mass see Cris Moore's gallery. Eg, the figure-8 orbit
$endgroup$
– PM 2Ring
6 hours ago
$begingroup$
Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans.
$endgroup$
– peterh
6 hours ago
$begingroup$
Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans.
$endgroup$
– peterh
6 hours ago
$begingroup$
There are stable equilateral systems, but not with equal masses. There's some info here & anims here. For some interesting stable n-body systems of equal mass see Cris Moore's gallery. Eg, the figure-8 orbit
$endgroup$
– PM 2Ring
6 hours ago
$begingroup$
There are stable equilateral systems, but not with equal masses. There's some info here & anims here. For some interesting stable n-body systems of equal mass see Cris Moore's gallery. Eg, the figure-8 orbit
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
It is possible in a Trojan configuration:
In the place of the "Planet" on the image, also a small star could exist. The third star would be at $L_4$ or at $L_5$. This configuration could be made stable.
However, as this link shows,
In unnormalized units, this criterion becomes
$$fracm_2m_1+ m_2 < 0.0385$$
We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable
equilibrium points, in the co-rotating frame, provided that mass $m_2$
is less than about $4%$ of mass $m_1$.
Thus, the mass of the second star should be at most 3.85% of the central star.
As far I know, no such known star system exists, but if it would, it would be stable.
answered 6 hours ago
peterhpeterh
1,8483 gold badges13 silver badges32 bronze badges
$endgroup$
add a comment |
$begingroup$
It is possible in a Trojan configuration:
In the place of the "Planet" on the image, also a small star could exist. The third star would be at $L_4$ or at $L_5$. This configuration could be made stable.
However, as this link shows,
In unnormalized units, this criterion becomes
$$fracm_2m_1+ m_2 < 0.0385$$
We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable
equilibrium points, in the co-rotating frame, provided that mass $m_2$
is less than about $4%$ of mass $m_1$.
Thus, the mass of the second star should be at most 3.85% of the central star.
As far I know, no such known star system exists, but if it would, it would be stable.
answered 6 hours ago
peterhpeterh
1,8483 gold badges13 silver badges32 bronze badges
$endgroup$
add a comment |
$begingroup$
It is possible in a Trojan configuration:
In the place of the "Planet" on the image, also a small star could exist. The third star would be at $L_4$ or at $L_5$. This configuration could be made stable.
However, as this link shows,
In unnormalized units, this criterion becomes
$$fracm_2m_1+ m_2 < 0.0385$$
We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable
equilibrium points, in the co-rotating frame, provided that mass $m_2$
is less than about $4%$ of mass $m_1$.
Thus, the mass of the second star should be at most 3.85% of the central star.
As far I know, no such known star system exists, but if it would, it would be stable.
answered 6 hours ago
peterhpeterh
1,8483 gold badges13 silver badges32 bronze badges
$endgroup$
It is possible in a Trojan configuration:
In the place of the "Planet" on the image, also a small star could exist. The third star would be at $L_4$ or at $L_5$. This configuration could be made stable.
However, as this link shows,
In unnormalized units, this criterion becomes
$$fracm_2m_1+ m_2 < 0.0385$$
We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable
equilibrium points, in the co-rotating frame, provided that mass $m_2$
is less than about $4%$ of mass $m_1$.
Thus, the mass of the second star should be at most 3.85% of the central star.
As far I know, no such known star system exists, but if it would, it would be stable.
answered 6 hours ago
peterhpeterh
1,8483 gold badges13 silver badges32 bronze badges
edited 3 hours ago
answered 6 hours ago
peterhpeterh
1,8483 gold badges13 silver badges32 bronze badges
answered 6 hours ago
peterhpeterh
1,8483 gold badges13 silver badges32 bronze badges
answered 6 hours ago
peterhpeterh
1,8483 gold badges13 silver badges32 bronze badges
1,8483 gold badges13 silver badges32 bronze badges
add a comment |
add a comment |
Heth Ghetiya is a new contributor. Be nice, and check out our Code of Conduct.
Heth Ghetiya is a new contributor. Be nice, and check out our Code of Conduct.
Heth Ghetiya is a new contributor. Be nice, and check out our Code of Conduct.
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