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Rotation period of a planet around a star (sun)
How can I safely brighten my secondary star?Earth-like planet orbiting neutron star?Orbital period of a tidally-locked Earth-like planet around a red dwarfTwo planets in a stable horseshoe orbit?The (Alternative) Reason for the Seasons, Part 2: Variable StarIs this circumbinary planet set up in a stable manner?Iron-rich rogue planet that is 12x Earth's mass collides with the Sun; now what?Two planets with the same orbital periodHabitable zone around a Blue Supergiant160 Year Solar Orbital Period & 1.9 Year Moon Orbital Period - Is it possible?
$begingroup$
Is it possible for a planet to be much further away from its star than earth is from the sun (because for example the star is much brighter/Hotter) yet still have a comparable (as in not tens if not hundreds or thousands of times longer) year and day length to earth?
P.S. To clarify this question somewhat.
This question comes sorta as a follow up to a question i once asked regarding a planet orbiting a Blue Supergiant star.
But with the distance required to give the planet an earth like climate it would have such a huge orbit that 1 year on this world would take a little more than 13 ''Earth'' centuries wich would end with me having to make the race on this world basically immortal compared to humans or give them a livespan that (using this world's timespans and still using earth like 12 months a year) would mean that the average lifespan of a person would be less than 1 month.
TLDR. can i speed the rotation time of this planet around the star up to get to ''earth like'' (does not have to be completely exact, 1 year on this world could be 3 earth years for all i care just not earth centuries or millenia) length or would this mess up the entire world?
science-based planets orbital-mechanics
asked May 26 at 16:51
Blue DevilBlue Devil
1307
$endgroup$
add a comment |
$begingroup$
Is it possible for a planet to be much further away from its star than earth is from the sun (because for example the star is much brighter/Hotter) yet still have a comparable (as in not tens if not hundreds or thousands of times longer) year and day length to earth?
P.S. To clarify this question somewhat.
This question comes sorta as a follow up to a question i once asked regarding a planet orbiting a Blue Supergiant star.
But with the distance required to give the planet an earth like climate it would have such a huge orbit that 1 year on this world would take a little more than 13 ''Earth'' centuries wich would end with me having to make the race on this world basically immortal compared to humans or give them a livespan that (using this world's timespans and still using earth like 12 months a year) would mean that the average lifespan of a person would be less than 1 month.
TLDR. can i speed the rotation time of this planet around the star up to get to ''earth like'' (does not have to be completely exact, 1 year on this world could be 3 earth years for all i care just not earth centuries or millenia) length or would this mess up the entire world?
science-based planets orbital-mechanics
asked May 26 at 16:51
Blue DevilBlue Devil
1307
$endgroup$
7
$begingroup$
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." (Johannes Kepler, 1609.) Larger radius means longer years. The duration of days has no necessary relationship with the radius of the orbit.
$endgroup$
– AlexP
May 26 at 17:27
add a comment |
$begingroup$
Is it possible for a planet to be much further away from its star than earth is from the sun (because for example the star is much brighter/Hotter) yet still have a comparable (as in not tens if not hundreds or thousands of times longer) year and day length to earth?
P.S. To clarify this question somewhat.
This question comes sorta as a follow up to a question i once asked regarding a planet orbiting a Blue Supergiant star.
But with the distance required to give the planet an earth like climate it would have such a huge orbit that 1 year on this world would take a little more than 13 ''Earth'' centuries wich would end with me having to make the race on this world basically immortal compared to humans or give them a livespan that (using this world's timespans and still using earth like 12 months a year) would mean that the average lifespan of a person would be less than 1 month.
TLDR. can i speed the rotation time of this planet around the star up to get to ''earth like'' (does not have to be completely exact, 1 year on this world could be 3 earth years for all i care just not earth centuries or millenia) length or would this mess up the entire world?
science-based planets orbital-mechanics
asked May 26 at 16:51
Blue DevilBlue Devil
1307
$endgroup$
Is it possible for a planet to be much further away from its star than earth is from the sun (because for example the star is much brighter/Hotter) yet still have a comparable (as in not tens if not hundreds or thousands of times longer) year and day length to earth?
P.S. To clarify this question somewhat.
This question comes sorta as a follow up to a question i once asked regarding a planet orbiting a Blue Supergiant star.
But with the distance required to give the planet an earth like climate it would have such a huge orbit that 1 year on this world would take a little more than 13 ''Earth'' centuries wich would end with me having to make the race on this world basically immortal compared to humans or give them a livespan that (using this world's timespans and still using earth like 12 months a year) would mean that the average lifespan of a person would be less than 1 month.
TLDR. can i speed the rotation time of this planet around the star up to get to ''earth like'' (does not have to be completely exact, 1 year on this world could be 3 earth years for all i care just not earth centuries or millenia) length or would this mess up the entire world?
science-based planets orbital-mechanics
science-based planets orbital-mechanics
asked May 26 at 16:51
Blue DevilBlue Devil
1307
asked May 26 at 16:51
Blue DevilBlue Devil
1307
edited May 26 at 22:18
Blue Devil
asked May 26 at 16:51
Blue DevilBlue Devil
1307
asked May 26 at 16:51
Blue DevilBlue Devil
1307
asked May 26 at 16:51
Blue DevilBlue Devil
1307
1307
7
$begingroup$
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." (Johannes Kepler, 1609.) Larger radius means longer years. The duration of days has no necessary relationship with the radius of the orbit.
$endgroup$
– AlexP
May 26 at 17:27
add a comment |
7
$begingroup$
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." (Johannes Kepler, 1609.) Larger radius means longer years. The duration of days has no necessary relationship with the radius of the orbit.
$endgroup$
– AlexP
May 26 at 17:27
7
7
$begingroup$
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." (Johannes Kepler, 1609.) Larger radius means longer years. The duration of days has no necessary relationship with the radius of the orbit.
$endgroup$
– AlexP
May 26 at 17:27
$begingroup$
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." (Johannes Kepler, 1609.) Larger radius means longer years. The duration of days has no necessary relationship with the radius of the orbit.
$endgroup$
– AlexP
May 26 at 17:27
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Yes, it is possible. Day length is only effected by how fast the planet rotates. This is unrelated to the orbit.
Year length increases as your orbit gets bigger, and decreases as your star gets bigger. The heavier the star is, the faster your have to orbit to stay a certain distance away. The farther away the planet is, the slower it orbits. You can have an orbit of practically any period given the correct stellar mass and orbital altitude.
I don't know how this alters the heating issues though. It's possible that the habitable zone around this larger star would not be at the correct altitude.
answered May 26 at 17:34
Ryan_LRyan_L
5,467929
$endgroup$
add a comment |
$begingroup$
Yes
It's already been stated but length of day just depends on the period of rotation, which has no relation to the celestial mechanics. Well, there are some special cases where that isn't entirely true, but they are fringe cases and not pertinent to your question. So choose the length of day at your leisure.
For length of year we need a wee bit of math. Kepler's laws are the foundation on which celestial mechanics is done. Kepler's third law can be stated mathematically as
$$ a=sqrt[3]fracT^2GM4pi^2 $$ (assuming planet mass is negligible compared to the star)
where
$$ a=semispace majorspace axis $$
$$ T=orbitalspace period $$
$$ M=massspace ofspace centralspace body $$
$$ G=gravitationalspace constantspace (6.674times 10^-11fracm^3kgspace s^2) $$
If we add your constraint that the length of the year is the same as the length of an Earth year, we can rearrange to solve for mass
$$ M=frac4pi^2 a^3G*(3.154times 10^7s)^2 $$
If we plot this, we can see the relationship as semi major axis increases:
*Note that the lower limit on mass for stars is something like 0.1 solar mass.
**Also note that semi major axis is identical to orbital radius if it is a circular orbit.
If you want to also examine the amount of energy that would be received at these specified distances and star masses you would need to use some relations for how star energy output varies with mass and how solar flux varies with distance. But that seems like a whole other question.
answered May 26 at 21:45
benben
1,0469
$endgroup$
add a comment |
$begingroup$
First of all, the day lenght has nothing to do with the distance from the star. It only depends on the time it takes for the star to cross the same sky meridian.
Then, by changing the eccentricity of the orbit, you can have the planet be further away from the star for a longer time than it is closer, while keeping the year length the same.
answered May 26 at 17:25
L.Dutch♦L.Dutch
98.7k31233477
$endgroup$
1
$begingroup$
The question does not ask for a planet which is sometimes farther away from its primary than Earth is from the Sun, and sometimes closer.
$endgroup$
– AlexP
May 26 at 17:29
$begingroup$
Seems like a valid (and clever) answer to me. The winters at apihelion, though - whew.
$endgroup$
– user535733
May 26 at 17:31
$begingroup$
@AlexP, distance from the star is a function of time. Even for Earth is never constant.
$endgroup$
– L.Dutch♦
May 26 at 17:50
add a comment |
$begingroup$
Day length is no problem. Orbital period is.
Day length mainly depends on how quickly a planet rotates around its own axis, modified slightly by its orbital period. There is nothing preventing a planet far from its sun to rotate at this speed. (The opposite is not true; a planet very close to its sun is likely to be tidally locked or in spin-orbit resonance).
Having the same year length when much farther out is quite a lot more difficult. As a general rule for stars of masses near that of the sun, the luminosity of a star is proportional to its mass to the fourth power, meaning that a planet's distance from its star to achieve the same amount of heat would be proportional to the square of the mass. For stars between 2 and twenty solar masses, luminosity is proportional to the mass to the power of 3.5; not that different. For stars above 55 solar masses, the relationship between mass and luminosity becomes linear, but such stars don't last very long because of strong solar winds and are unlikely to develop planets with life.
In return, the distance the planet needs to be from its sun to have the same orbital period (year) is proportional to the cube root of the star's mass. As a star grows bigger, planets with the same orbital period will be increasingly roasted by heat and radiation.
To achieve a similar year at a far greater distance (meaning a far heavier sun), you would need a stellar object with very little radiation compared to its mass. A possible solution would be a sun in close orbit around an even more masssive black hole, with your planet orbiting far from this pair.
Say that this star has twenty times the mass of our sun. It would have roughly 50,000 times the luminosity of our sun, so your planet would have to be roughly 225 times the distance of Earth from the sun to receive the same amount of sunlight. The mass of the black hole would hence have to be 11.4 million solar masses for the planet to have the same orbital period. This is a supermassive black hole, about three times more massive than the one at the center of our galaxy (but smaller than some in other galaxies).
Say that the star instead is only 5 solar masses, with a luminosity ca. 400 times that of our sun. The orbital distance of your planet would then have to be 20 times that of Earth's to receive the same heat; about the same as Uranus' orbit. The mass of the black hole would then 'only' have to be 8,000 solar masses; much more manageable. Whether black holes of this mass exist; between supermassive and stellar black holes; is unknown.
answered May 27 at 9:07
Klaus Æ. MogensenKlaus Æ. Mogensen
3,9061718
$endgroup$
$begingroup$
If i remember correctly the star i had in mind would have been roughly comparable to Rigel so according to you would there be a way to get a planet (whether or not a planet could even exist around such a star does not matter for my story) orbiting it at a distance that would produce a simillar climate as earth while still rotating around the star at a speed that the year would not be longer than let's say max. around 1095 earth days (3 earth years)?
$endgroup$
– Blue Devil
May 28 at 10:32
$begingroup$
@BlueDevil: Rigel A has a luminosity 120,000 times that of our Sun, so for your planet to receive the same amount of heat and light, it would need to be sqrt(120,000) ~ 350 times as far away from Rigel A as the Earth is from the sun. The mass of Rigel A is 21 times that of our sun, meaning that the orbital period of your planet at its distance is more than 1,400 years. Much closer than that, and it will be roasted. Also note that Rigel is a multiple star system, which complicates things.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 10:48
$begingroup$
So in short unless i change the star type or just completely ignore reality theres no way to give this world a year that is not uncomprehensibly long without roasting it? Would it help if i changed the blue supergiant to the lowest luminosity such a star can have? (about 10.000 times our sun)
$endgroup$
– Blue Devil
May 28 at 11:27
$begingroup$
Not really. As I write above, luminosity is roughly proportional with solar mass to the fourth power, so even a star a little larger than ours is much hotter. A star twice the mass of our sun is e.g. 16 times hotter, requiring 4 times the distance for a planet to receive the same sun and light, while the distance to have the same year is only 26% greater. You need to pair the star with a heavy, dim object to make it work.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 12:15
add a comment |
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4 Answers
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4 Answers
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$begingroup$
Yes, it is possible. Day length is only effected by how fast the planet rotates. This is unrelated to the orbit.
Year length increases as your orbit gets bigger, and decreases as your star gets bigger. The heavier the star is, the faster your have to orbit to stay a certain distance away. The farther away the planet is, the slower it orbits. You can have an orbit of practically any period given the correct stellar mass and orbital altitude.
I don't know how this alters the heating issues though. It's possible that the habitable zone around this larger star would not be at the correct altitude.
answered May 26 at 17:34
Ryan_LRyan_L
5,467929
$endgroup$
add a comment |
$begingroup$
Yes, it is possible. Day length is only effected by how fast the planet rotates. This is unrelated to the orbit.
Year length increases as your orbit gets bigger, and decreases as your star gets bigger. The heavier the star is, the faster your have to orbit to stay a certain distance away. The farther away the planet is, the slower it orbits. You can have an orbit of practically any period given the correct stellar mass and orbital altitude.
I don't know how this alters the heating issues though. It's possible that the habitable zone around this larger star would not be at the correct altitude.
answered May 26 at 17:34
Ryan_LRyan_L
5,467929
$endgroup$
add a comment |
$begingroup$
Yes, it is possible. Day length is only effected by how fast the planet rotates. This is unrelated to the orbit.
Year length increases as your orbit gets bigger, and decreases as your star gets bigger. The heavier the star is, the faster your have to orbit to stay a certain distance away. The farther away the planet is, the slower it orbits. You can have an orbit of practically any period given the correct stellar mass and orbital altitude.
I don't know how this alters the heating issues though. It's possible that the habitable zone around this larger star would not be at the correct altitude.
answered May 26 at 17:34
Ryan_LRyan_L
5,467929
$endgroup$
Yes, it is possible. Day length is only effected by how fast the planet rotates. This is unrelated to the orbit.
Year length increases as your orbit gets bigger, and decreases as your star gets bigger. The heavier the star is, the faster your have to orbit to stay a certain distance away. The farther away the planet is, the slower it orbits. You can have an orbit of practically any period given the correct stellar mass and orbital altitude.
I don't know how this alters the heating issues though. It's possible that the habitable zone around this larger star would not be at the correct altitude.
answered May 26 at 17:34
Ryan_LRyan_L
5,467929
answered May 26 at 17:34
Ryan_LRyan_L
5,467929
answered May 26 at 17:34
Ryan_LRyan_L
5,467929
answered May 26 at 17:34
Ryan_LRyan_L
5,467929
5,467929
add a comment |
add a comment |
$begingroup$
Yes
It's already been stated but length of day just depends on the period of rotation, which has no relation to the celestial mechanics. Well, there are some special cases where that isn't entirely true, but they are fringe cases and not pertinent to your question. So choose the length of day at your leisure.
For length of year we need a wee bit of math. Kepler's laws are the foundation on which celestial mechanics is done. Kepler's third law can be stated mathematically as
$$ a=sqrt[3]fracT^2GM4pi^2 $$ (assuming planet mass is negligible compared to the star)
where
$$ a=semispace majorspace axis $$
$$ T=orbitalspace period $$
$$ M=massspace ofspace centralspace body $$
$$ G=gravitationalspace constantspace (6.674times 10^-11fracm^3kgspace s^2) $$
If we add your constraint that the length of the year is the same as the length of an Earth year, we can rearrange to solve for mass
$$ M=frac4pi^2 a^3G*(3.154times 10^7s)^2 $$
If we plot this, we can see the relationship as semi major axis increases:
*Note that the lower limit on mass for stars is something like 0.1 solar mass.
**Also note that semi major axis is identical to orbital radius if it is a circular orbit.
If you want to also examine the amount of energy that would be received at these specified distances and star masses you would need to use some relations for how star energy output varies with mass and how solar flux varies with distance. But that seems like a whole other question.
answered May 26 at 21:45
benben
1,0469
$endgroup$
add a comment |
$begingroup$
Yes
It's already been stated but length of day just depends on the period of rotation, which has no relation to the celestial mechanics. Well, there are some special cases where that isn't entirely true, but they are fringe cases and not pertinent to your question. So choose the length of day at your leisure.
For length of year we need a wee bit of math. Kepler's laws are the foundation on which celestial mechanics is done. Kepler's third law can be stated mathematically as
$$ a=sqrt[3]fracT^2GM4pi^2 $$ (assuming planet mass is negligible compared to the star)
where
$$ a=semispace majorspace axis $$
$$ T=orbitalspace period $$
$$ M=massspace ofspace centralspace body $$
$$ G=gravitationalspace constantspace (6.674times 10^-11fracm^3kgspace s^2) $$
If we add your constraint that the length of the year is the same as the length of an Earth year, we can rearrange to solve for mass
$$ M=frac4pi^2 a^3G*(3.154times 10^7s)^2 $$
If we plot this, we can see the relationship as semi major axis increases:
*Note that the lower limit on mass for stars is something like 0.1 solar mass.
**Also note that semi major axis is identical to orbital radius if it is a circular orbit.
If you want to also examine the amount of energy that would be received at these specified distances and star masses you would need to use some relations for how star energy output varies with mass and how solar flux varies with distance. But that seems like a whole other question.
answered May 26 at 21:45
benben
1,0469
$endgroup$
add a comment |
$begingroup$
Yes
It's already been stated but length of day just depends on the period of rotation, which has no relation to the celestial mechanics. Well, there are some special cases where that isn't entirely true, but they are fringe cases and not pertinent to your question. So choose the length of day at your leisure.
For length of year we need a wee bit of math. Kepler's laws are the foundation on which celestial mechanics is done. Kepler's third law can be stated mathematically as
$$ a=sqrt[3]fracT^2GM4pi^2 $$ (assuming planet mass is negligible compared to the star)
where
$$ a=semispace majorspace axis $$
$$ T=orbitalspace period $$
$$ M=massspace ofspace centralspace body $$
$$ G=gravitationalspace constantspace (6.674times 10^-11fracm^3kgspace s^2) $$
If we add your constraint that the length of the year is the same as the length of an Earth year, we can rearrange to solve for mass
$$ M=frac4pi^2 a^3G*(3.154times 10^7s)^2 $$
If we plot this, we can see the relationship as semi major axis increases:
*Note that the lower limit on mass for stars is something like 0.1 solar mass.
**Also note that semi major axis is identical to orbital radius if it is a circular orbit.
If you want to also examine the amount of energy that would be received at these specified distances and star masses you would need to use some relations for how star energy output varies with mass and how solar flux varies with distance. But that seems like a whole other question.
answered May 26 at 21:45
benben
1,0469
$endgroup$
Yes
It's already been stated but length of day just depends on the period of rotation, which has no relation to the celestial mechanics. Well, there are some special cases where that isn't entirely true, but they are fringe cases and not pertinent to your question. So choose the length of day at your leisure.
For length of year we need a wee bit of math. Kepler's laws are the foundation on which celestial mechanics is done. Kepler's third law can be stated mathematically as
$$ a=sqrt[3]fracT^2GM4pi^2 $$ (assuming planet mass is negligible compared to the star)
where
$$ a=semispace majorspace axis $$
$$ T=orbitalspace period $$
$$ M=massspace ofspace centralspace body $$
$$ G=gravitationalspace constantspace (6.674times 10^-11fracm^3kgspace s^2) $$
If we add your constraint that the length of the year is the same as the length of an Earth year, we can rearrange to solve for mass
$$ M=frac4pi^2 a^3G*(3.154times 10^7s)^2 $$
If we plot this, we can see the relationship as semi major axis increases:
*Note that the lower limit on mass for stars is something like 0.1 solar mass.
**Also note that semi major axis is identical to orbital radius if it is a circular orbit.
If you want to also examine the amount of energy that would be received at these specified distances and star masses you would need to use some relations for how star energy output varies with mass and how solar flux varies with distance. But that seems like a whole other question.
answered May 26 at 21:45
benben
1,0469
answered May 26 at 21:45
benben
1,0469
answered May 26 at 21:45
benben
1,0469
answered May 26 at 21:45
benben
1,0469
1,0469
add a comment |
add a comment |
$begingroup$
First of all, the day lenght has nothing to do with the distance from the star. It only depends on the time it takes for the star to cross the same sky meridian.
Then, by changing the eccentricity of the orbit, you can have the planet be further away from the star for a longer time than it is closer, while keeping the year length the same.
answered May 26 at 17:25
L.Dutch♦L.Dutch
98.7k31233477
$endgroup$
1
$begingroup$
The question does not ask for a planet which is sometimes farther away from its primary than Earth is from the Sun, and sometimes closer.
$endgroup$
– AlexP
May 26 at 17:29
$begingroup$
Seems like a valid (and clever) answer to me. The winters at apihelion, though - whew.
$endgroup$
– user535733
May 26 at 17:31
$begingroup$
@AlexP, distance from the star is a function of time. Even for Earth is never constant.
$endgroup$
– L.Dutch♦
May 26 at 17:50
add a comment |
$begingroup$
First of all, the day lenght has nothing to do with the distance from the star. It only depends on the time it takes for the star to cross the same sky meridian.
Then, by changing the eccentricity of the orbit, you can have the planet be further away from the star for a longer time than it is closer, while keeping the year length the same.
answered May 26 at 17:25
L.Dutch♦L.Dutch
98.7k31233477
$endgroup$
1
$begingroup$
The question does not ask for a planet which is sometimes farther away from its primary than Earth is from the Sun, and sometimes closer.
$endgroup$
– AlexP
May 26 at 17:29
$begingroup$
Seems like a valid (and clever) answer to me. The winters at apihelion, though - whew.
$endgroup$
– user535733
May 26 at 17:31
$begingroup$
@AlexP, distance from the star is a function of time. Even for Earth is never constant.
$endgroup$
– L.Dutch♦
May 26 at 17:50
add a comment |
$begingroup$
First of all, the day lenght has nothing to do with the distance from the star. It only depends on the time it takes for the star to cross the same sky meridian.
Then, by changing the eccentricity of the orbit, you can have the planet be further away from the star for a longer time than it is closer, while keeping the year length the same.
answered May 26 at 17:25
L.Dutch♦L.Dutch
98.7k31233477
$endgroup$
First of all, the day lenght has nothing to do with the distance from the star. It only depends on the time it takes for the star to cross the same sky meridian.
Then, by changing the eccentricity of the orbit, you can have the planet be further away from the star for a longer time than it is closer, while keeping the year length the same.
answered May 26 at 17:25
L.Dutch♦L.Dutch
98.7k31233477
answered May 26 at 17:25
L.Dutch♦L.Dutch
98.7k31233477
answered May 26 at 17:25
L.Dutch♦L.Dutch
98.7k31233477
answered May 26 at 17:25
L.Dutch♦L.Dutch
98.7k31233477
98.7k31233477
1
$begingroup$
The question does not ask for a planet which is sometimes farther away from its primary than Earth is from the Sun, and sometimes closer.
$endgroup$
– AlexP
May 26 at 17:29
$begingroup$
Seems like a valid (and clever) answer to me. The winters at apihelion, though - whew.
$endgroup$
– user535733
May 26 at 17:31
$begingroup$
@AlexP, distance from the star is a function of time. Even for Earth is never constant.
$endgroup$
– L.Dutch♦
May 26 at 17:50
add a comment |
1
$begingroup$
The question does not ask for a planet which is sometimes farther away from its primary than Earth is from the Sun, and sometimes closer.
$endgroup$
– AlexP
May 26 at 17:29
$begingroup$
Seems like a valid (and clever) answer to me. The winters at apihelion, though - whew.
$endgroup$
– user535733
May 26 at 17:31
$begingroup$
@AlexP, distance from the star is a function of time. Even for Earth is never constant.
$endgroup$
– L.Dutch♦
May 26 at 17:50
1
1
$begingroup$
The question does not ask for a planet which is sometimes farther away from its primary than Earth is from the Sun, and sometimes closer.
$endgroup$
– AlexP
May 26 at 17:29
$begingroup$
The question does not ask for a planet which is sometimes farther away from its primary than Earth is from the Sun, and sometimes closer.
$endgroup$
– AlexP
May 26 at 17:29
$begingroup$
Seems like a valid (and clever) answer to me. The winters at apihelion, though - whew.
$endgroup$
– user535733
May 26 at 17:31
$begingroup$
Seems like a valid (and clever) answer to me. The winters at apihelion, though - whew.
$endgroup$
– user535733
May 26 at 17:31
$begingroup$
@AlexP, distance from the star is a function of time. Even for Earth is never constant.
$endgroup$
– L.Dutch♦
May 26 at 17:50
$begingroup$
@AlexP, distance from the star is a function of time. Even for Earth is never constant.
$endgroup$
– L.Dutch♦
May 26 at 17:50
add a comment |
$begingroup$
Day length is no problem. Orbital period is.
Day length mainly depends on how quickly a planet rotates around its own axis, modified slightly by its orbital period. There is nothing preventing a planet far from its sun to rotate at this speed. (The opposite is not true; a planet very close to its sun is likely to be tidally locked or in spin-orbit resonance).
Having the same year length when much farther out is quite a lot more difficult. As a general rule for stars of masses near that of the sun, the luminosity of a star is proportional to its mass to the fourth power, meaning that a planet's distance from its star to achieve the same amount of heat would be proportional to the square of the mass. For stars between 2 and twenty solar masses, luminosity is proportional to the mass to the power of 3.5; not that different. For stars above 55 solar masses, the relationship between mass and luminosity becomes linear, but such stars don't last very long because of strong solar winds and are unlikely to develop planets with life.
In return, the distance the planet needs to be from its sun to have the same orbital period (year) is proportional to the cube root of the star's mass. As a star grows bigger, planets with the same orbital period will be increasingly roasted by heat and radiation.
To achieve a similar year at a far greater distance (meaning a far heavier sun), you would need a stellar object with very little radiation compared to its mass. A possible solution would be a sun in close orbit around an even more masssive black hole, with your planet orbiting far from this pair.
Say that this star has twenty times the mass of our sun. It would have roughly 50,000 times the luminosity of our sun, so your planet would have to be roughly 225 times the distance of Earth from the sun to receive the same amount of sunlight. The mass of the black hole would hence have to be 11.4 million solar masses for the planet to have the same orbital period. This is a supermassive black hole, about three times more massive than the one at the center of our galaxy (but smaller than some in other galaxies).
Say that the star instead is only 5 solar masses, with a luminosity ca. 400 times that of our sun. The orbital distance of your planet would then have to be 20 times that of Earth's to receive the same heat; about the same as Uranus' orbit. The mass of the black hole would then 'only' have to be 8,000 solar masses; much more manageable. Whether black holes of this mass exist; between supermassive and stellar black holes; is unknown.
answered May 27 at 9:07
Klaus Æ. MogensenKlaus Æ. Mogensen
3,9061718
$endgroup$
$begingroup$
If i remember correctly the star i had in mind would have been roughly comparable to Rigel so according to you would there be a way to get a planet (whether or not a planet could even exist around such a star does not matter for my story) orbiting it at a distance that would produce a simillar climate as earth while still rotating around the star at a speed that the year would not be longer than let's say max. around 1095 earth days (3 earth years)?
$endgroup$
– Blue Devil
May 28 at 10:32
$begingroup$
@BlueDevil: Rigel A has a luminosity 120,000 times that of our Sun, so for your planet to receive the same amount of heat and light, it would need to be sqrt(120,000) ~ 350 times as far away from Rigel A as the Earth is from the sun. The mass of Rigel A is 21 times that of our sun, meaning that the orbital period of your planet at its distance is more than 1,400 years. Much closer than that, and it will be roasted. Also note that Rigel is a multiple star system, which complicates things.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 10:48
$begingroup$
So in short unless i change the star type or just completely ignore reality theres no way to give this world a year that is not uncomprehensibly long without roasting it? Would it help if i changed the blue supergiant to the lowest luminosity such a star can have? (about 10.000 times our sun)
$endgroup$
– Blue Devil
May 28 at 11:27
$begingroup$
Not really. As I write above, luminosity is roughly proportional with solar mass to the fourth power, so even a star a little larger than ours is much hotter. A star twice the mass of our sun is e.g. 16 times hotter, requiring 4 times the distance for a planet to receive the same sun and light, while the distance to have the same year is only 26% greater. You need to pair the star with a heavy, dim object to make it work.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 12:15
add a comment |
$begingroup$
Day length is no problem. Orbital period is.
Day length mainly depends on how quickly a planet rotates around its own axis, modified slightly by its orbital period. There is nothing preventing a planet far from its sun to rotate at this speed. (The opposite is not true; a planet very close to its sun is likely to be tidally locked or in spin-orbit resonance).
Having the same year length when much farther out is quite a lot more difficult. As a general rule for stars of masses near that of the sun, the luminosity of a star is proportional to its mass to the fourth power, meaning that a planet's distance from its star to achieve the same amount of heat would be proportional to the square of the mass. For stars between 2 and twenty solar masses, luminosity is proportional to the mass to the power of 3.5; not that different. For stars above 55 solar masses, the relationship between mass and luminosity becomes linear, but such stars don't last very long because of strong solar winds and are unlikely to develop planets with life.
In return, the distance the planet needs to be from its sun to have the same orbital period (year) is proportional to the cube root of the star's mass. As a star grows bigger, planets with the same orbital period will be increasingly roasted by heat and radiation.
To achieve a similar year at a far greater distance (meaning a far heavier sun), you would need a stellar object with very little radiation compared to its mass. A possible solution would be a sun in close orbit around an even more masssive black hole, with your planet orbiting far from this pair.
Say that this star has twenty times the mass of our sun. It would have roughly 50,000 times the luminosity of our sun, so your planet would have to be roughly 225 times the distance of Earth from the sun to receive the same amount of sunlight. The mass of the black hole would hence have to be 11.4 million solar masses for the planet to have the same orbital period. This is a supermassive black hole, about three times more massive than the one at the center of our galaxy (but smaller than some in other galaxies).
Say that the star instead is only 5 solar masses, with a luminosity ca. 400 times that of our sun. The orbital distance of your planet would then have to be 20 times that of Earth's to receive the same heat; about the same as Uranus' orbit. The mass of the black hole would then 'only' have to be 8,000 solar masses; much more manageable. Whether black holes of this mass exist; between supermassive and stellar black holes; is unknown.
answered May 27 at 9:07
Klaus Æ. MogensenKlaus Æ. Mogensen
3,9061718
$endgroup$
$begingroup$
If i remember correctly the star i had in mind would have been roughly comparable to Rigel so according to you would there be a way to get a planet (whether or not a planet could even exist around such a star does not matter for my story) orbiting it at a distance that would produce a simillar climate as earth while still rotating around the star at a speed that the year would not be longer than let's say max. around 1095 earth days (3 earth years)?
$endgroup$
– Blue Devil
May 28 at 10:32
$begingroup$
@BlueDevil: Rigel A has a luminosity 120,000 times that of our Sun, so for your planet to receive the same amount of heat and light, it would need to be sqrt(120,000) ~ 350 times as far away from Rigel A as the Earth is from the sun. The mass of Rigel A is 21 times that of our sun, meaning that the orbital period of your planet at its distance is more than 1,400 years. Much closer than that, and it will be roasted. Also note that Rigel is a multiple star system, which complicates things.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 10:48
$begingroup$
So in short unless i change the star type or just completely ignore reality theres no way to give this world a year that is not uncomprehensibly long without roasting it? Would it help if i changed the blue supergiant to the lowest luminosity such a star can have? (about 10.000 times our sun)
$endgroup$
– Blue Devil
May 28 at 11:27
$begingroup$
Not really. As I write above, luminosity is roughly proportional with solar mass to the fourth power, so even a star a little larger than ours is much hotter. A star twice the mass of our sun is e.g. 16 times hotter, requiring 4 times the distance for a planet to receive the same sun and light, while the distance to have the same year is only 26% greater. You need to pair the star with a heavy, dim object to make it work.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 12:15
add a comment |
$begingroup$
Day length is no problem. Orbital period is.
Day length mainly depends on how quickly a planet rotates around its own axis, modified slightly by its orbital period. There is nothing preventing a planet far from its sun to rotate at this speed. (The opposite is not true; a planet very close to its sun is likely to be tidally locked or in spin-orbit resonance).
Having the same year length when much farther out is quite a lot more difficult. As a general rule for stars of masses near that of the sun, the luminosity of a star is proportional to its mass to the fourth power, meaning that a planet's distance from its star to achieve the same amount of heat would be proportional to the square of the mass. For stars between 2 and twenty solar masses, luminosity is proportional to the mass to the power of 3.5; not that different. For stars above 55 solar masses, the relationship between mass and luminosity becomes linear, but such stars don't last very long because of strong solar winds and are unlikely to develop planets with life.
In return, the distance the planet needs to be from its sun to have the same orbital period (year) is proportional to the cube root of the star's mass. As a star grows bigger, planets with the same orbital period will be increasingly roasted by heat and radiation.
To achieve a similar year at a far greater distance (meaning a far heavier sun), you would need a stellar object with very little radiation compared to its mass. A possible solution would be a sun in close orbit around an even more masssive black hole, with your planet orbiting far from this pair.
Say that this star has twenty times the mass of our sun. It would have roughly 50,000 times the luminosity of our sun, so your planet would have to be roughly 225 times the distance of Earth from the sun to receive the same amount of sunlight. The mass of the black hole would hence have to be 11.4 million solar masses for the planet to have the same orbital period. This is a supermassive black hole, about three times more massive than the one at the center of our galaxy (but smaller than some in other galaxies).
Say that the star instead is only 5 solar masses, with a luminosity ca. 400 times that of our sun. The orbital distance of your planet would then have to be 20 times that of Earth's to receive the same heat; about the same as Uranus' orbit. The mass of the black hole would then 'only' have to be 8,000 solar masses; much more manageable. Whether black holes of this mass exist; between supermassive and stellar black holes; is unknown.
answered May 27 at 9:07
Klaus Æ. MogensenKlaus Æ. Mogensen
3,9061718
$endgroup$
Day length is no problem. Orbital period is.
Day length mainly depends on how quickly a planet rotates around its own axis, modified slightly by its orbital period. There is nothing preventing a planet far from its sun to rotate at this speed. (The opposite is not true; a planet very close to its sun is likely to be tidally locked or in spin-orbit resonance).
Having the same year length when much farther out is quite a lot more difficult. As a general rule for stars of masses near that of the sun, the luminosity of a star is proportional to its mass to the fourth power, meaning that a planet's distance from its star to achieve the same amount of heat would be proportional to the square of the mass. For stars between 2 and twenty solar masses, luminosity is proportional to the mass to the power of 3.5; not that different. For stars above 55 solar masses, the relationship between mass and luminosity becomes linear, but such stars don't last very long because of strong solar winds and are unlikely to develop planets with life.
In return, the distance the planet needs to be from its sun to have the same orbital period (year) is proportional to the cube root of the star's mass. As a star grows bigger, planets with the same orbital period will be increasingly roasted by heat and radiation.
To achieve a similar year at a far greater distance (meaning a far heavier sun), you would need a stellar object with very little radiation compared to its mass. A possible solution would be a sun in close orbit around an even more masssive black hole, with your planet orbiting far from this pair.
Say that this star has twenty times the mass of our sun. It would have roughly 50,000 times the luminosity of our sun, so your planet would have to be roughly 225 times the distance of Earth from the sun to receive the same amount of sunlight. The mass of the black hole would hence have to be 11.4 million solar masses for the planet to have the same orbital period. This is a supermassive black hole, about three times more massive than the one at the center of our galaxy (but smaller than some in other galaxies).
Say that the star instead is only 5 solar masses, with a luminosity ca. 400 times that of our sun. The orbital distance of your planet would then have to be 20 times that of Earth's to receive the same heat; about the same as Uranus' orbit. The mass of the black hole would then 'only' have to be 8,000 solar masses; much more manageable. Whether black holes of this mass exist; between supermassive and stellar black holes; is unknown.
answered May 27 at 9:07
Klaus Æ. MogensenKlaus Æ. Mogensen
3,9061718
answered May 27 at 9:07
Klaus Æ. MogensenKlaus Æ. Mogensen
3,9061718
answered May 27 at 9:07
Klaus Æ. MogensenKlaus Æ. Mogensen
3,9061718
answered May 27 at 9:07
Klaus Æ. MogensenKlaus Æ. Mogensen
3,9061718
3,9061718
$begingroup$
If i remember correctly the star i had in mind would have been roughly comparable to Rigel so according to you would there be a way to get a planet (whether or not a planet could even exist around such a star does not matter for my story) orbiting it at a distance that would produce a simillar climate as earth while still rotating around the star at a speed that the year would not be longer than let's say max. around 1095 earth days (3 earth years)?
$endgroup$
– Blue Devil
May 28 at 10:32
$begingroup$
@BlueDevil: Rigel A has a luminosity 120,000 times that of our Sun, so for your planet to receive the same amount of heat and light, it would need to be sqrt(120,000) ~ 350 times as far away from Rigel A as the Earth is from the sun. The mass of Rigel A is 21 times that of our sun, meaning that the orbital period of your planet at its distance is more than 1,400 years. Much closer than that, and it will be roasted. Also note that Rigel is a multiple star system, which complicates things.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 10:48
$begingroup$
So in short unless i change the star type or just completely ignore reality theres no way to give this world a year that is not uncomprehensibly long without roasting it? Would it help if i changed the blue supergiant to the lowest luminosity such a star can have? (about 10.000 times our sun)
$endgroup$
– Blue Devil
May 28 at 11:27
$begingroup$
Not really. As I write above, luminosity is roughly proportional with solar mass to the fourth power, so even a star a little larger than ours is much hotter. A star twice the mass of our sun is e.g. 16 times hotter, requiring 4 times the distance for a planet to receive the same sun and light, while the distance to have the same year is only 26% greater. You need to pair the star with a heavy, dim object to make it work.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 12:15
add a comment |
$begingroup$
If i remember correctly the star i had in mind would have been roughly comparable to Rigel so according to you would there be a way to get a planet (whether or not a planet could even exist around such a star does not matter for my story) orbiting it at a distance that would produce a simillar climate as earth while still rotating around the star at a speed that the year would not be longer than let's say max. around 1095 earth days (3 earth years)?
$endgroup$
– Blue Devil
May 28 at 10:32
$begingroup$
@BlueDevil: Rigel A has a luminosity 120,000 times that of our Sun, so for your planet to receive the same amount of heat and light, it would need to be sqrt(120,000) ~ 350 times as far away from Rigel A as the Earth is from the sun. The mass of Rigel A is 21 times that of our sun, meaning that the orbital period of your planet at its distance is more than 1,400 years. Much closer than that, and it will be roasted. Also note that Rigel is a multiple star system, which complicates things.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 10:48
$begingroup$
So in short unless i change the star type or just completely ignore reality theres no way to give this world a year that is not uncomprehensibly long without roasting it? Would it help if i changed the blue supergiant to the lowest luminosity such a star can have? (about 10.000 times our sun)
$endgroup$
– Blue Devil
May 28 at 11:27
$begingroup$
Not really. As I write above, luminosity is roughly proportional with solar mass to the fourth power, so even a star a little larger than ours is much hotter. A star twice the mass of our sun is e.g. 16 times hotter, requiring 4 times the distance for a planet to receive the same sun and light, while the distance to have the same year is only 26% greater. You need to pair the star with a heavy, dim object to make it work.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 12:15
$begingroup$
If i remember correctly the star i had in mind would have been roughly comparable to Rigel so according to you would there be a way to get a planet (whether or not a planet could even exist around such a star does not matter for my story) orbiting it at a distance that would produce a simillar climate as earth while still rotating around the star at a speed that the year would not be longer than let's say max. around 1095 earth days (3 earth years)?
$endgroup$
– Blue Devil
May 28 at 10:32
$begingroup$
If i remember correctly the star i had in mind would have been roughly comparable to Rigel so according to you would there be a way to get a planet (whether or not a planet could even exist around such a star does not matter for my story) orbiting it at a distance that would produce a simillar climate as earth while still rotating around the star at a speed that the year would not be longer than let's say max. around 1095 earth days (3 earth years)?
$endgroup$
– Blue Devil
May 28 at 10:32
$begingroup$
@BlueDevil: Rigel A has a luminosity 120,000 times that of our Sun, so for your planet to receive the same amount of heat and light, it would need to be sqrt(120,000) ~ 350 times as far away from Rigel A as the Earth is from the sun. The mass of Rigel A is 21 times that of our sun, meaning that the orbital period of your planet at its distance is more than 1,400 years. Much closer than that, and it will be roasted. Also note that Rigel is a multiple star system, which complicates things.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 10:48
$begingroup$
@BlueDevil: Rigel A has a luminosity 120,000 times that of our Sun, so for your planet to receive the same amount of heat and light, it would need to be sqrt(120,000) ~ 350 times as far away from Rigel A as the Earth is from the sun. The mass of Rigel A is 21 times that of our sun, meaning that the orbital period of your planet at its distance is more than 1,400 years. Much closer than that, and it will be roasted. Also note that Rigel is a multiple star system, which complicates things.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 10:48
$begingroup$
So in short unless i change the star type or just completely ignore reality theres no way to give this world a year that is not uncomprehensibly long without roasting it? Would it help if i changed the blue supergiant to the lowest luminosity such a star can have? (about 10.000 times our sun)
$endgroup$
– Blue Devil
May 28 at 11:27
$begingroup$
So in short unless i change the star type or just completely ignore reality theres no way to give this world a year that is not uncomprehensibly long without roasting it? Would it help if i changed the blue supergiant to the lowest luminosity such a star can have? (about 10.000 times our sun)
$endgroup$
– Blue Devil
May 28 at 11:27
$begingroup$
Not really. As I write above, luminosity is roughly proportional with solar mass to the fourth power, so even a star a little larger than ours is much hotter. A star twice the mass of our sun is e.g. 16 times hotter, requiring 4 times the distance for a planet to receive the same sun and light, while the distance to have the same year is only 26% greater. You need to pair the star with a heavy, dim object to make it work.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 12:15
$begingroup$
Not really. As I write above, luminosity is roughly proportional with solar mass to the fourth power, so even a star a little larger than ours is much hotter. A star twice the mass of our sun is e.g. 16 times hotter, requiring 4 times the distance for a planet to receive the same sun and light, while the distance to have the same year is only 26% greater. You need to pair the star with a heavy, dim object to make it work.
$endgroup$
– Klaus Æ. Mogensen
May 28 at 12:15
add a comment |
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"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." (Johannes Kepler, 1609.) Larger radius means longer years. The duration of days has no necessary relationship with the radius of the orbit.
$endgroup$
– AlexP
May 26 at 17:27