Can this planet in a binary star system exist?Can a planet have a figure-8 type of orbit around two separate...
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Can this planet in a binary star system exist?
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In a certain star system, Star B orbits Star A. Planet C lies in between the two stars and also orbits Star A. Planet C and Star B share an orbital period, and are at roughly the same places in that period. Consequently, there is no night on Planet C. As Star A sets, Star B rises.
Is such a configuration possible (and if so, plausible)? Could such a planet support life, or would the constant exposure on both sides to radiation make the planet too hot?
science-based planets
New contributor
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add a comment |
$begingroup$
In a certain star system, Star B orbits Star A. Planet C lies in between the two stars and also orbits Star A. Planet C and Star B share an orbital period, and are at roughly the same places in that period. Consequently, there is no night on Planet C. As Star A sets, Star B rises.
Is such a configuration possible (and if so, plausible)? Could such a planet support life, or would the constant exposure on both sides to radiation make the planet too hot?
science-based planets
New contributor
$endgroup$
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Obligatory xkcd: what-if.xkcd.com/150. (No)
$endgroup$
– MSalters
4 hours ago
add a comment |
$begingroup$
In a certain star system, Star B orbits Star A. Planet C lies in between the two stars and also orbits Star A. Planet C and Star B share an orbital period, and are at roughly the same places in that period. Consequently, there is no night on Planet C. As Star A sets, Star B rises.
Is such a configuration possible (and if so, plausible)? Could such a planet support life, or would the constant exposure on both sides to radiation make the planet too hot?
science-based planets
New contributor
$endgroup$
In a certain star system, Star B orbits Star A. Planet C lies in between the two stars and also orbits Star A. Planet C and Star B share an orbital period, and are at roughly the same places in that period. Consequently, there is no night on Planet C. As Star A sets, Star B rises.
Is such a configuration possible (and if so, plausible)? Could such a planet support life, or would the constant exposure on both sides to radiation make the planet too hot?
science-based planets
science-based planets
New contributor
New contributor
New contributor
asked 9 hours ago
AetherfoxAetherfox
1233 bronze badges
1233 bronze badges
New contributor
New contributor
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Obligatory xkcd: what-if.xkcd.com/150. (No)
$endgroup$
– MSalters
4 hours ago
add a comment |
$begingroup$
Obligatory xkcd: what-if.xkcd.com/150. (No)
$endgroup$
– MSalters
4 hours ago
$begingroup$
Obligatory xkcd: what-if.xkcd.com/150. (No)
$endgroup$
– MSalters
4 hours ago
$begingroup$
Obligatory xkcd: what-if.xkcd.com/150. (No)
$endgroup$
– MSalters
4 hours ago
add a comment |
3 Answers
3
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oldest
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$begingroup$
The short answer is: No.
To have the same orbital period, they would need to be in the same orbit.
If the planet was in L-4 or L-5 (see Wiki: Lagrange Points)of Star B which is orbiting Star A, there would be very little night but there would still be night.
the only way for this to work is for the planet to be in L-1 between the two stars. the problem is that L-1 is not a stable location and the planet would have drifted off of that position and begun a wild ride orbit long before life could have developed.
Also, I can't imagine that L-1 would be in the habitable zone in any case. There may be a way to jigger the star masses to make that work but the planet still wouldn't stay where it belongs.
The only way for it to work is to find a planet that just happens to be in the proper position now but will soon leave it. Not much chance of life there though.
You could try an Earth-like moon orbiting a gas giant. The gas giant may be reflective enough that it will be almost as bright as the direct sunlight.
Edit:
One possible solution is that the two stars are roughly the same mass and the planet orbits the center of mass of the two stars. I still don't think that would work but someone might be able to math that out. It seems to be too much like balancing on a pin.
$endgroup$
1
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I'm curious to see someone try to tackle the "balancing on a pin" idea. My guess, like yours, is that it won't work, but I was thinking about it for a while and decided it is sufficiently complicated that my intuition could easily be wrong here.
$endgroup$
– Rob Watts
8 hours ago
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@RobWatts, Me too.
$endgroup$
– ShadoCat
8 hours ago
$begingroup$
I tried using three body problem simulator (desmos.com/calculator/icaqw49qeq), to check the posibility. And indeed, as expected for three body problem, it really is balancing on the pin. Literally. Even minor imbalance in system leads to planet in between two suns to be eventually ejected out of the central position, and often completely fired out of the system, never to return. So, not a good solution, because a meteor size of a sand of grain could destabilize the balance. Edit: Actually, lowering mass of the planet, it no longer gets fired into space, only starts orbiting stars.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
$begingroup$
"Planet C and Star B share an orbital period" - unfortunately, this won't work. The orbital period of an object is proportional to how far away it is from the object it orbits (more specifically, the oribtal period squared is proportional to the cube of the distance). That means if star B is farther away from A than C is, B will orbit more slowly.
Habitability is much more plausible - the habitable zone of a star depends on how bright the star is. If you somehow had a star on each side of the planet, you'd just need the planet to be at the outer edge (or maybe a little outside it) of the what the habitable zone would be for each star by itself. There is some point at which the stars will be far enough away that the planet is kept at an acceptable temperature only because there is no night.
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$begingroup$
Actually, it very technically, is possible, but in order to achieve that, you have to achieve such a balance that makes it so that it's equivalent to your planet being in L-1 lagrange point of a smaller star. Which makes the situation of planet unstable, and will means any disturbance of the balance, no matter how small, will lead to this scenario falling apart.
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– Failus Maximus
8 hours ago
add a comment |
$begingroup$
I think yes.
Start with Earth (planet C) and sun (star A). You might make Star A a little more massive and move Planet C's orbit out some.
Star B is a little star - equal to 75 Jupiters. It is far away, in approximately the orbit of Neptune. It is more massive than a planet and so it can (must) move much faster than Neptune does. Also you need it to move fast to complete its orbit in the time planet C completes its much smaller circle. At the same time you need it to stay far away from Planet C or its proximity is going to perturb the planet's orbit.
A thing to remember about orbits: X does not orbit Y. X and Y orbit their common center of gravity. When X (e.g. star) is of hugely greater mass than Y (e.g. planet) then their center of gravity is usually still within X. But if you have 2 stellar mass objects then the center of mass might not be within one of them.
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I don't see how Star B can be bright enough to give equal light and not be massive enough to kick the planet out.
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– ShadoCat
9 hours ago
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"It is more massive than a planet and so it can (must) move much faster than Neptune does." - this is incorrect. Orbital period is independent of mass. The only way in which the mass matters is when you get to the point where it has enough mass to significantly move whatever it's orbiting around.
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– Rob Watts
9 hours ago
$begingroup$
@RobWatts - Here is Kepler's third law. en.wikipedia.org/wiki/… Orbital period is independent of mass only when the mass of one is trivial compared to the mass of the other. But here were are discussing 2 stars so I think we are at the point you note. That point is also why I address the "common center of gravity" thing in my answer.
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– Willk
7 hours ago
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@Shadowcat - I think what you say is right, but I did not see that OP required 2 stars to give equal light. The distant star would not be very bright.
$endgroup$
– Willk
7 hours ago
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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$begingroup$
The short answer is: No.
To have the same orbital period, they would need to be in the same orbit.
If the planet was in L-4 or L-5 (see Wiki: Lagrange Points)of Star B which is orbiting Star A, there would be very little night but there would still be night.
the only way for this to work is for the planet to be in L-1 between the two stars. the problem is that L-1 is not a stable location and the planet would have drifted off of that position and begun a wild ride orbit long before life could have developed.
Also, I can't imagine that L-1 would be in the habitable zone in any case. There may be a way to jigger the star masses to make that work but the planet still wouldn't stay where it belongs.
The only way for it to work is to find a planet that just happens to be in the proper position now but will soon leave it. Not much chance of life there though.
You could try an Earth-like moon orbiting a gas giant. The gas giant may be reflective enough that it will be almost as bright as the direct sunlight.
Edit:
One possible solution is that the two stars are roughly the same mass and the planet orbits the center of mass of the two stars. I still don't think that would work but someone might be able to math that out. It seems to be too much like balancing on a pin.
$endgroup$
1
$begingroup$
I'm curious to see someone try to tackle the "balancing on a pin" idea. My guess, like yours, is that it won't work, but I was thinking about it for a while and decided it is sufficiently complicated that my intuition could easily be wrong here.
$endgroup$
– Rob Watts
8 hours ago
$begingroup$
@RobWatts, Me too.
$endgroup$
– ShadoCat
8 hours ago
$begingroup$
I tried using three body problem simulator (desmos.com/calculator/icaqw49qeq), to check the posibility. And indeed, as expected for three body problem, it really is balancing on the pin. Literally. Even minor imbalance in system leads to planet in between two suns to be eventually ejected out of the central position, and often completely fired out of the system, never to return. So, not a good solution, because a meteor size of a sand of grain could destabilize the balance. Edit: Actually, lowering mass of the planet, it no longer gets fired into space, only starts orbiting stars.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
$begingroup$
The short answer is: No.
To have the same orbital period, they would need to be in the same orbit.
If the planet was in L-4 or L-5 (see Wiki: Lagrange Points)of Star B which is orbiting Star A, there would be very little night but there would still be night.
the only way for this to work is for the planet to be in L-1 between the two stars. the problem is that L-1 is not a stable location and the planet would have drifted off of that position and begun a wild ride orbit long before life could have developed.
Also, I can't imagine that L-1 would be in the habitable zone in any case. There may be a way to jigger the star masses to make that work but the planet still wouldn't stay where it belongs.
The only way for it to work is to find a planet that just happens to be in the proper position now but will soon leave it. Not much chance of life there though.
You could try an Earth-like moon orbiting a gas giant. The gas giant may be reflective enough that it will be almost as bright as the direct sunlight.
Edit:
One possible solution is that the two stars are roughly the same mass and the planet orbits the center of mass of the two stars. I still don't think that would work but someone might be able to math that out. It seems to be too much like balancing on a pin.
$endgroup$
1
$begingroup$
I'm curious to see someone try to tackle the "balancing on a pin" idea. My guess, like yours, is that it won't work, but I was thinking about it for a while and decided it is sufficiently complicated that my intuition could easily be wrong here.
$endgroup$
– Rob Watts
8 hours ago
$begingroup$
@RobWatts, Me too.
$endgroup$
– ShadoCat
8 hours ago
$begingroup$
I tried using three body problem simulator (desmos.com/calculator/icaqw49qeq), to check the posibility. And indeed, as expected for three body problem, it really is balancing on the pin. Literally. Even minor imbalance in system leads to planet in between two suns to be eventually ejected out of the central position, and often completely fired out of the system, never to return. So, not a good solution, because a meteor size of a sand of grain could destabilize the balance. Edit: Actually, lowering mass of the planet, it no longer gets fired into space, only starts orbiting stars.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
$begingroup$
The short answer is: No.
To have the same orbital period, they would need to be in the same orbit.
If the planet was in L-4 or L-5 (see Wiki: Lagrange Points)of Star B which is orbiting Star A, there would be very little night but there would still be night.
the only way for this to work is for the planet to be in L-1 between the two stars. the problem is that L-1 is not a stable location and the planet would have drifted off of that position and begun a wild ride orbit long before life could have developed.
Also, I can't imagine that L-1 would be in the habitable zone in any case. There may be a way to jigger the star masses to make that work but the planet still wouldn't stay where it belongs.
The only way for it to work is to find a planet that just happens to be in the proper position now but will soon leave it. Not much chance of life there though.
You could try an Earth-like moon orbiting a gas giant. The gas giant may be reflective enough that it will be almost as bright as the direct sunlight.
Edit:
One possible solution is that the two stars are roughly the same mass and the planet orbits the center of mass of the two stars. I still don't think that would work but someone might be able to math that out. It seems to be too much like balancing on a pin.
$endgroup$
The short answer is: No.
To have the same orbital period, they would need to be in the same orbit.
If the planet was in L-4 or L-5 (see Wiki: Lagrange Points)of Star B which is orbiting Star A, there would be very little night but there would still be night.
the only way for this to work is for the planet to be in L-1 between the two stars. the problem is that L-1 is not a stable location and the planet would have drifted off of that position and begun a wild ride orbit long before life could have developed.
Also, I can't imagine that L-1 would be in the habitable zone in any case. There may be a way to jigger the star masses to make that work but the planet still wouldn't stay where it belongs.
The only way for it to work is to find a planet that just happens to be in the proper position now but will soon leave it. Not much chance of life there though.
You could try an Earth-like moon orbiting a gas giant. The gas giant may be reflective enough that it will be almost as bright as the direct sunlight.
Edit:
One possible solution is that the two stars are roughly the same mass and the planet orbits the center of mass of the two stars. I still don't think that would work but someone might be able to math that out. It seems to be too much like balancing on a pin.
answered 9 hours ago
ShadoCatShadoCat
16k21 silver badges55 bronze badges
16k21 silver badges55 bronze badges
1
$begingroup$
I'm curious to see someone try to tackle the "balancing on a pin" idea. My guess, like yours, is that it won't work, but I was thinking about it for a while and decided it is sufficiently complicated that my intuition could easily be wrong here.
$endgroup$
– Rob Watts
8 hours ago
$begingroup$
@RobWatts, Me too.
$endgroup$
– ShadoCat
8 hours ago
$begingroup$
I tried using three body problem simulator (desmos.com/calculator/icaqw49qeq), to check the posibility. And indeed, as expected for three body problem, it really is balancing on the pin. Literally. Even minor imbalance in system leads to planet in between two suns to be eventually ejected out of the central position, and often completely fired out of the system, never to return. So, not a good solution, because a meteor size of a sand of grain could destabilize the balance. Edit: Actually, lowering mass of the planet, it no longer gets fired into space, only starts orbiting stars.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
1
$begingroup$
I'm curious to see someone try to tackle the "balancing on a pin" idea. My guess, like yours, is that it won't work, but I was thinking about it for a while and decided it is sufficiently complicated that my intuition could easily be wrong here.
$endgroup$
– Rob Watts
8 hours ago
$begingroup$
@RobWatts, Me too.
$endgroup$
– ShadoCat
8 hours ago
$begingroup$
I tried using three body problem simulator (desmos.com/calculator/icaqw49qeq), to check the posibility. And indeed, as expected for three body problem, it really is balancing on the pin. Literally. Even minor imbalance in system leads to planet in between two suns to be eventually ejected out of the central position, and often completely fired out of the system, never to return. So, not a good solution, because a meteor size of a sand of grain could destabilize the balance. Edit: Actually, lowering mass of the planet, it no longer gets fired into space, only starts orbiting stars.
$endgroup$
– Failus Maximus
8 hours ago
1
1
$begingroup$
I'm curious to see someone try to tackle the "balancing on a pin" idea. My guess, like yours, is that it won't work, but I was thinking about it for a while and decided it is sufficiently complicated that my intuition could easily be wrong here.
$endgroup$
– Rob Watts
8 hours ago
$begingroup$
I'm curious to see someone try to tackle the "balancing on a pin" idea. My guess, like yours, is that it won't work, but I was thinking about it for a while and decided it is sufficiently complicated that my intuition could easily be wrong here.
$endgroup$
– Rob Watts
8 hours ago
$begingroup$
@RobWatts, Me too.
$endgroup$
– ShadoCat
8 hours ago
$begingroup$
@RobWatts, Me too.
$endgroup$
– ShadoCat
8 hours ago
$begingroup$
I tried using three body problem simulator (desmos.com/calculator/icaqw49qeq), to check the posibility. And indeed, as expected for three body problem, it really is balancing on the pin. Literally. Even minor imbalance in system leads to planet in between two suns to be eventually ejected out of the central position, and often completely fired out of the system, never to return. So, not a good solution, because a meteor size of a sand of grain could destabilize the balance. Edit: Actually, lowering mass of the planet, it no longer gets fired into space, only starts orbiting stars.
$endgroup$
– Failus Maximus
8 hours ago
$begingroup$
I tried using three body problem simulator (desmos.com/calculator/icaqw49qeq), to check the posibility. And indeed, as expected for three body problem, it really is balancing on the pin. Literally. Even minor imbalance in system leads to planet in between two suns to be eventually ejected out of the central position, and often completely fired out of the system, never to return. So, not a good solution, because a meteor size of a sand of grain could destabilize the balance. Edit: Actually, lowering mass of the planet, it no longer gets fired into space, only starts orbiting stars.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
$begingroup$
"Planet C and Star B share an orbital period" - unfortunately, this won't work. The orbital period of an object is proportional to how far away it is from the object it orbits (more specifically, the oribtal period squared is proportional to the cube of the distance). That means if star B is farther away from A than C is, B will orbit more slowly.
Habitability is much more plausible - the habitable zone of a star depends on how bright the star is. If you somehow had a star on each side of the planet, you'd just need the planet to be at the outer edge (or maybe a little outside it) of the what the habitable zone would be for each star by itself. There is some point at which the stars will be far enough away that the planet is kept at an acceptable temperature only because there is no night.
$endgroup$
$begingroup$
Actually, it very technically, is possible, but in order to achieve that, you have to achieve such a balance that makes it so that it's equivalent to your planet being in L-1 lagrange point of a smaller star. Which makes the situation of planet unstable, and will means any disturbance of the balance, no matter how small, will lead to this scenario falling apart.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
$begingroup$
"Planet C and Star B share an orbital period" - unfortunately, this won't work. The orbital period of an object is proportional to how far away it is from the object it orbits (more specifically, the oribtal period squared is proportional to the cube of the distance). That means if star B is farther away from A than C is, B will orbit more slowly.
Habitability is much more plausible - the habitable zone of a star depends on how bright the star is. If you somehow had a star on each side of the planet, you'd just need the planet to be at the outer edge (or maybe a little outside it) of the what the habitable zone would be for each star by itself. There is some point at which the stars will be far enough away that the planet is kept at an acceptable temperature only because there is no night.
$endgroup$
$begingroup$
Actually, it very technically, is possible, but in order to achieve that, you have to achieve such a balance that makes it so that it's equivalent to your planet being in L-1 lagrange point of a smaller star. Which makes the situation of planet unstable, and will means any disturbance of the balance, no matter how small, will lead to this scenario falling apart.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
$begingroup$
"Planet C and Star B share an orbital period" - unfortunately, this won't work. The orbital period of an object is proportional to how far away it is from the object it orbits (more specifically, the oribtal period squared is proportional to the cube of the distance). That means if star B is farther away from A than C is, B will orbit more slowly.
Habitability is much more plausible - the habitable zone of a star depends on how bright the star is. If you somehow had a star on each side of the planet, you'd just need the planet to be at the outer edge (or maybe a little outside it) of the what the habitable zone would be for each star by itself. There is some point at which the stars will be far enough away that the planet is kept at an acceptable temperature only because there is no night.
$endgroup$
"Planet C and Star B share an orbital period" - unfortunately, this won't work. The orbital period of an object is proportional to how far away it is from the object it orbits (more specifically, the oribtal period squared is proportional to the cube of the distance). That means if star B is farther away from A than C is, B will orbit more slowly.
Habitability is much more plausible - the habitable zone of a star depends on how bright the star is. If you somehow had a star on each side of the planet, you'd just need the planet to be at the outer edge (or maybe a little outside it) of the what the habitable zone would be for each star by itself. There is some point at which the stars will be far enough away that the planet is kept at an acceptable temperature only because there is no night.
edited 9 hours ago
answered 9 hours ago
Rob WattsRob Watts
16.5k5 gold badges38 silver badges78 bronze badges
16.5k5 gold badges38 silver badges78 bronze badges
$begingroup$
Actually, it very technically, is possible, but in order to achieve that, you have to achieve such a balance that makes it so that it's equivalent to your planet being in L-1 lagrange point of a smaller star. Which makes the situation of planet unstable, and will means any disturbance of the balance, no matter how small, will lead to this scenario falling apart.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
$begingroup$
Actually, it very technically, is possible, but in order to achieve that, you have to achieve such a balance that makes it so that it's equivalent to your planet being in L-1 lagrange point of a smaller star. Which makes the situation of planet unstable, and will means any disturbance of the balance, no matter how small, will lead to this scenario falling apart.
$endgroup$
– Failus Maximus
8 hours ago
$begingroup$
Actually, it very technically, is possible, but in order to achieve that, you have to achieve such a balance that makes it so that it's equivalent to your planet being in L-1 lagrange point of a smaller star. Which makes the situation of planet unstable, and will means any disturbance of the balance, no matter how small, will lead to this scenario falling apart.
$endgroup$
– Failus Maximus
8 hours ago
$begingroup$
Actually, it very technically, is possible, but in order to achieve that, you have to achieve such a balance that makes it so that it's equivalent to your planet being in L-1 lagrange point of a smaller star. Which makes the situation of planet unstable, and will means any disturbance of the balance, no matter how small, will lead to this scenario falling apart.
$endgroup$
– Failus Maximus
8 hours ago
add a comment |
$begingroup$
I think yes.
Start with Earth (planet C) and sun (star A). You might make Star A a little more massive and move Planet C's orbit out some.
Star B is a little star - equal to 75 Jupiters. It is far away, in approximately the orbit of Neptune. It is more massive than a planet and so it can (must) move much faster than Neptune does. Also you need it to move fast to complete its orbit in the time planet C completes its much smaller circle. At the same time you need it to stay far away from Planet C or its proximity is going to perturb the planet's orbit.
A thing to remember about orbits: X does not orbit Y. X and Y orbit their common center of gravity. When X (e.g. star) is of hugely greater mass than Y (e.g. planet) then their center of gravity is usually still within X. But if you have 2 stellar mass objects then the center of mass might not be within one of them.
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I don't see how Star B can be bright enough to give equal light and not be massive enough to kick the planet out.
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– ShadoCat
9 hours ago
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"It is more massive than a planet and so it can (must) move much faster than Neptune does." - this is incorrect. Orbital period is independent of mass. The only way in which the mass matters is when you get to the point where it has enough mass to significantly move whatever it's orbiting around.
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– Rob Watts
9 hours ago
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@RobWatts - Here is Kepler's third law. en.wikipedia.org/wiki/… Orbital period is independent of mass only when the mass of one is trivial compared to the mass of the other. But here were are discussing 2 stars so I think we are at the point you note. That point is also why I address the "common center of gravity" thing in my answer.
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– Willk
7 hours ago
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@Shadowcat - I think what you say is right, but I did not see that OP required 2 stars to give equal light. The distant star would not be very bright.
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– Willk
7 hours ago
add a comment |
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I think yes.
Start with Earth (planet C) and sun (star A). You might make Star A a little more massive and move Planet C's orbit out some.
Star B is a little star - equal to 75 Jupiters. It is far away, in approximately the orbit of Neptune. It is more massive than a planet and so it can (must) move much faster than Neptune does. Also you need it to move fast to complete its orbit in the time planet C completes its much smaller circle. At the same time you need it to stay far away from Planet C or its proximity is going to perturb the planet's orbit.
A thing to remember about orbits: X does not orbit Y. X and Y orbit their common center of gravity. When X (e.g. star) is of hugely greater mass than Y (e.g. planet) then their center of gravity is usually still within X. But if you have 2 stellar mass objects then the center of mass might not be within one of them.
$endgroup$
$begingroup$
I don't see how Star B can be bright enough to give equal light and not be massive enough to kick the planet out.
$endgroup$
– ShadoCat
9 hours ago
$begingroup$
"It is more massive than a planet and so it can (must) move much faster than Neptune does." - this is incorrect. Orbital period is independent of mass. The only way in which the mass matters is when you get to the point where it has enough mass to significantly move whatever it's orbiting around.
$endgroup$
– Rob Watts
9 hours ago
$begingroup$
@RobWatts - Here is Kepler's third law. en.wikipedia.org/wiki/… Orbital period is independent of mass only when the mass of one is trivial compared to the mass of the other. But here were are discussing 2 stars so I think we are at the point you note. That point is also why I address the "common center of gravity" thing in my answer.
$endgroup$
– Willk
7 hours ago
$begingroup$
@Shadowcat - I think what you say is right, but I did not see that OP required 2 stars to give equal light. The distant star would not be very bright.
$endgroup$
– Willk
7 hours ago
add a comment |
$begingroup$
I think yes.
Start with Earth (planet C) and sun (star A). You might make Star A a little more massive and move Planet C's orbit out some.
Star B is a little star - equal to 75 Jupiters. It is far away, in approximately the orbit of Neptune. It is more massive than a planet and so it can (must) move much faster than Neptune does. Also you need it to move fast to complete its orbit in the time planet C completes its much smaller circle. At the same time you need it to stay far away from Planet C or its proximity is going to perturb the planet's orbit.
A thing to remember about orbits: X does not orbit Y. X and Y orbit their common center of gravity. When X (e.g. star) is of hugely greater mass than Y (e.g. planet) then their center of gravity is usually still within X. But if you have 2 stellar mass objects then the center of mass might not be within one of them.
$endgroup$
I think yes.
Start with Earth (planet C) and sun (star A). You might make Star A a little more massive and move Planet C's orbit out some.
Star B is a little star - equal to 75 Jupiters. It is far away, in approximately the orbit of Neptune. It is more massive than a planet and so it can (must) move much faster than Neptune does. Also you need it to move fast to complete its orbit in the time planet C completes its much smaller circle. At the same time you need it to stay far away from Planet C or its proximity is going to perturb the planet's orbit.
A thing to remember about orbits: X does not orbit Y. X and Y orbit their common center of gravity. When X (e.g. star) is of hugely greater mass than Y (e.g. planet) then their center of gravity is usually still within X. But if you have 2 stellar mass objects then the center of mass might not be within one of them.
answered 9 hours ago
WillkWillk
136k34 gold badges256 silver badges566 bronze badges
136k34 gold badges256 silver badges566 bronze badges
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I don't see how Star B can be bright enough to give equal light and not be massive enough to kick the planet out.
$endgroup$
– ShadoCat
9 hours ago
$begingroup$
"It is more massive than a planet and so it can (must) move much faster than Neptune does." - this is incorrect. Orbital period is independent of mass. The only way in which the mass matters is when you get to the point where it has enough mass to significantly move whatever it's orbiting around.
$endgroup$
– Rob Watts
9 hours ago
$begingroup$
@RobWatts - Here is Kepler's third law. en.wikipedia.org/wiki/… Orbital period is independent of mass only when the mass of one is trivial compared to the mass of the other. But here were are discussing 2 stars so I think we are at the point you note. That point is also why I address the "common center of gravity" thing in my answer.
$endgroup$
– Willk
7 hours ago
$begingroup$
@Shadowcat - I think what you say is right, but I did not see that OP required 2 stars to give equal light. The distant star would not be very bright.
$endgroup$
– Willk
7 hours ago
add a comment |
$begingroup$
I don't see how Star B can be bright enough to give equal light and not be massive enough to kick the planet out.
$endgroup$
– ShadoCat
9 hours ago
$begingroup$
"It is more massive than a planet and so it can (must) move much faster than Neptune does." - this is incorrect. Orbital period is independent of mass. The only way in which the mass matters is when you get to the point where it has enough mass to significantly move whatever it's orbiting around.
$endgroup$
– Rob Watts
9 hours ago
$begingroup$
@RobWatts - Here is Kepler's third law. en.wikipedia.org/wiki/… Orbital period is independent of mass only when the mass of one is trivial compared to the mass of the other. But here were are discussing 2 stars so I think we are at the point you note. That point is also why I address the "common center of gravity" thing in my answer.
$endgroup$
– Willk
7 hours ago
$begingroup$
@Shadowcat - I think what you say is right, but I did not see that OP required 2 stars to give equal light. The distant star would not be very bright.
$endgroup$
– Willk
7 hours ago
$begingroup$
I don't see how Star B can be bright enough to give equal light and not be massive enough to kick the planet out.
$endgroup$
– ShadoCat
9 hours ago
$begingroup$
I don't see how Star B can be bright enough to give equal light and not be massive enough to kick the planet out.
$endgroup$
– ShadoCat
9 hours ago
$begingroup$
"It is more massive than a planet and so it can (must) move much faster than Neptune does." - this is incorrect. Orbital period is independent of mass. The only way in which the mass matters is when you get to the point where it has enough mass to significantly move whatever it's orbiting around.
$endgroup$
– Rob Watts
9 hours ago
$begingroup$
"It is more massive than a planet and so it can (must) move much faster than Neptune does." - this is incorrect. Orbital period is independent of mass. The only way in which the mass matters is when you get to the point where it has enough mass to significantly move whatever it's orbiting around.
$endgroup$
– Rob Watts
9 hours ago
$begingroup$
@RobWatts - Here is Kepler's third law. en.wikipedia.org/wiki/… Orbital period is independent of mass only when the mass of one is trivial compared to the mass of the other. But here were are discussing 2 stars so I think we are at the point you note. That point is also why I address the "common center of gravity" thing in my answer.
$endgroup$
– Willk
7 hours ago
$begingroup$
@RobWatts - Here is Kepler's third law. en.wikipedia.org/wiki/… Orbital period is independent of mass only when the mass of one is trivial compared to the mass of the other. But here were are discussing 2 stars so I think we are at the point you note. That point is also why I address the "common center of gravity" thing in my answer.
$endgroup$
– Willk
7 hours ago
$begingroup$
@Shadowcat - I think what you say is right, but I did not see that OP required 2 stars to give equal light. The distant star would not be very bright.
$endgroup$
– Willk
7 hours ago
$begingroup$
@Shadowcat - I think what you say is right, but I did not see that OP required 2 stars to give equal light. The distant star would not be very bright.
$endgroup$
– Willk
7 hours ago
add a comment |
Aetherfox is a new contributor. Be nice, and check out our Code of Conduct.
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Obligatory xkcd: what-if.xkcd.com/150. (No)
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– MSalters
4 hours ago