Why does the Sun have different day lengths, but not the gas giants?
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The Sun's rotation period varies from about 25 days at the equator to about 38 days at the poles. As I understand it, this is because the Sun is not solid, and because of the way centripetal force works, the equator must move faster than the poles.
Question: if this works, why do Jupiter/Saturn/Uranus/Neptune have well-defined days? Why don't the equators of these planets rotate faster than the poles as well? For example, Wikipedia's article on Jupiter gives the length of a Jovian day as 9h 55m 30s, which is so precise that it implies Jupiter does not have a rotational period which varies with latitude.
the-sun rotation gas-giants
asked Mar 22 at 2:38
AllureAllure
582311
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add a comment |
$begingroup$
The Sun's rotation period varies from about 25 days at the equator to about 38 days at the poles. As I understand it, this is because the Sun is not solid, and because of the way centripetal force works, the equator must move faster than the poles.
Question: if this works, why do Jupiter/Saturn/Uranus/Neptune have well-defined days? Why don't the equators of these planets rotate faster than the poles as well? For example, Wikipedia's article on Jupiter gives the length of a Jovian day as 9h 55m 30s, which is so precise that it implies Jupiter does not have a rotational period which varies with latitude.
the-sun rotation gas-giants
asked Mar 22 at 2:38
AllureAllure
582311
$endgroup$
add a comment |
$begingroup$
The Sun's rotation period varies from about 25 days at the equator to about 38 days at the poles. As I understand it, this is because the Sun is not solid, and because of the way centripetal force works, the equator must move faster than the poles.
Question: if this works, why do Jupiter/Saturn/Uranus/Neptune have well-defined days? Why don't the equators of these planets rotate faster than the poles as well? For example, Wikipedia's article on Jupiter gives the length of a Jovian day as 9h 55m 30s, which is so precise that it implies Jupiter does not have a rotational period which varies with latitude.
the-sun rotation gas-giants
asked Mar 22 at 2:38
AllureAllure
582311
$endgroup$
The Sun's rotation period varies from about 25 days at the equator to about 38 days at the poles. As I understand it, this is because the Sun is not solid, and because of the way centripetal force works, the equator must move faster than the poles.
Question: if this works, why do Jupiter/Saturn/Uranus/Neptune have well-defined days? Why don't the equators of these planets rotate faster than the poles as well? For example, Wikipedia's article on Jupiter gives the length of a Jovian day as 9h 55m 30s, which is so precise that it implies Jupiter does not have a rotational period which varies with latitude.
the-sun rotation gas-giants
the-sun rotation gas-giants
asked Mar 22 at 2:38
AllureAllure
582311
asked Mar 22 at 2:38
AllureAllure
582311
asked Mar 22 at 2:38
AllureAllure
582311
asked Mar 22 at 2:38
AllureAllure
582311
asked Mar 22 at 2:38
AllureAllure
582311
582311
add a comment |
add a comment |
1 Answer
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It's a matter of how "day" is defined.
Wikipedia's article on Jupiter cites this IAU/IAG paper for the length of a Jupiter day. In it, footnote (e) of table I has the following:
The equations for W for Jupiter, Saturn, Uranus and Neptune refer to the rotation of their magnetic fields (System III)
The radio emissions of the gas giants have well-defined periodic variations. These variations are caused by the rotation of the magnetic fields of those planets, and are evidence that they have a reasonably coherent core of some sort that's rotating at a uniform speed. The periodic variations then represent the rotation speed of that object, which is taken as the rotation speed of the planet.
We're reasonably certain the Sun doesn't have a coherent core. Measuring the variation of the magnetic field doesn't show a well-defined period, and doesn't provide a useful definition of the Sun's rotation speed.
answered Mar 22 at 3:14
MarkMark
1,874920
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4
$begingroup$
I just add that for Jupiter there is definitively no a fix day based on std rotation. At least no for the outer layer. It does even rotate in stripes and a simulation as in Celestia package is a kind of spectacle.
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– Alchimista
Mar 22 at 8:08
6
$begingroup$
“The assumption is that whatever's generating the magnetic field forms a reasonably coherent mass that's rotating at a uniform speed.” — I’d strengthen this by pointing out that it’s not just an assumption, it’s based closely on empirical facts: the magnetic field has a measurable uniform periodic behaviour, and based our understanding of planetary magnetic fields, we’re confident this corresponds to rotation of a somewhat coherent core. With the sun, as I understand it, we don’t see any uniformly periodic behaviour that we would expect to correspond to some kind of rotating mass.
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– Peter LeFanu Lumsdaine
Mar 22 at 13:37
1
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The Sun's magnetic field don't have a period as @PeterLeFanuLumsdaine stated.
$endgroup$
– Mindwin
Mar 22 at 14:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It's a matter of how "day" is defined.
Wikipedia's article on Jupiter cites this IAU/IAG paper for the length of a Jupiter day. In it, footnote (e) of table I has the following:
The equations for W for Jupiter, Saturn, Uranus and Neptune refer to the rotation of their magnetic fields (System III)
The radio emissions of the gas giants have well-defined periodic variations. These variations are caused by the rotation of the magnetic fields of those planets, and are evidence that they have a reasonably coherent core of some sort that's rotating at a uniform speed. The periodic variations then represent the rotation speed of that object, which is taken as the rotation speed of the planet.
We're reasonably certain the Sun doesn't have a coherent core. Measuring the variation of the magnetic field doesn't show a well-defined period, and doesn't provide a useful definition of the Sun's rotation speed.
answered Mar 22 at 3:14
MarkMark
1,874920
$endgroup$
4
$begingroup$
I just add that for Jupiter there is definitively no a fix day based on std rotation. At least no for the outer layer. It does even rotate in stripes and a simulation as in Celestia package is a kind of spectacle.
$endgroup$
– Alchimista
Mar 22 at 8:08
6
$begingroup$
“The assumption is that whatever's generating the magnetic field forms a reasonably coherent mass that's rotating at a uniform speed.” — I’d strengthen this by pointing out that it’s not just an assumption, it’s based closely on empirical facts: the magnetic field has a measurable uniform periodic behaviour, and based our understanding of planetary magnetic fields, we’re confident this corresponds to rotation of a somewhat coherent core. With the sun, as I understand it, we don’t see any uniformly periodic behaviour that we would expect to correspond to some kind of rotating mass.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 22 at 13:37
1
$begingroup$
The Sun's magnetic field don't have a period as @PeterLeFanuLumsdaine stated.
$endgroup$
– Mindwin
Mar 22 at 14:06
add a comment |
$begingroup$
It's a matter of how "day" is defined.
Wikipedia's article on Jupiter cites this IAU/IAG paper for the length of a Jupiter day. In it, footnote (e) of table I has the following:
The equations for W for Jupiter, Saturn, Uranus and Neptune refer to the rotation of their magnetic fields (System III)
The radio emissions of the gas giants have well-defined periodic variations. These variations are caused by the rotation of the magnetic fields of those planets, and are evidence that they have a reasonably coherent core of some sort that's rotating at a uniform speed. The periodic variations then represent the rotation speed of that object, which is taken as the rotation speed of the planet.
We're reasonably certain the Sun doesn't have a coherent core. Measuring the variation of the magnetic field doesn't show a well-defined period, and doesn't provide a useful definition of the Sun's rotation speed.
answered Mar 22 at 3:14
MarkMark
1,874920
$endgroup$
4
$begingroup$
I just add that for Jupiter there is definitively no a fix day based on std rotation. At least no for the outer layer. It does even rotate in stripes and a simulation as in Celestia package is a kind of spectacle.
$endgroup$
– Alchimista
Mar 22 at 8:08
6
$begingroup$
“The assumption is that whatever's generating the magnetic field forms a reasonably coherent mass that's rotating at a uniform speed.” — I’d strengthen this by pointing out that it’s not just an assumption, it’s based closely on empirical facts: the magnetic field has a measurable uniform periodic behaviour, and based our understanding of planetary magnetic fields, we’re confident this corresponds to rotation of a somewhat coherent core. With the sun, as I understand it, we don’t see any uniformly periodic behaviour that we would expect to correspond to some kind of rotating mass.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 22 at 13:37
1
$begingroup$
The Sun's magnetic field don't have a period as @PeterLeFanuLumsdaine stated.
$endgroup$
– Mindwin
Mar 22 at 14:06
add a comment |
$begingroup$
It's a matter of how "day" is defined.
Wikipedia's article on Jupiter cites this IAU/IAG paper for the length of a Jupiter day. In it, footnote (e) of table I has the following:
The equations for W for Jupiter, Saturn, Uranus and Neptune refer to the rotation of their magnetic fields (System III)
The radio emissions of the gas giants have well-defined periodic variations. These variations are caused by the rotation of the magnetic fields of those planets, and are evidence that they have a reasonably coherent core of some sort that's rotating at a uniform speed. The periodic variations then represent the rotation speed of that object, which is taken as the rotation speed of the planet.
We're reasonably certain the Sun doesn't have a coherent core. Measuring the variation of the magnetic field doesn't show a well-defined period, and doesn't provide a useful definition of the Sun's rotation speed.
answered Mar 22 at 3:14
MarkMark
1,874920
$endgroup$
It's a matter of how "day" is defined.
Wikipedia's article on Jupiter cites this IAU/IAG paper for the length of a Jupiter day. In it, footnote (e) of table I has the following:
The equations for W for Jupiter, Saturn, Uranus and Neptune refer to the rotation of their magnetic fields (System III)
The radio emissions of the gas giants have well-defined periodic variations. These variations are caused by the rotation of the magnetic fields of those planets, and are evidence that they have a reasonably coherent core of some sort that's rotating at a uniform speed. The periodic variations then represent the rotation speed of that object, which is taken as the rotation speed of the planet.
We're reasonably certain the Sun doesn't have a coherent core. Measuring the variation of the magnetic field doesn't show a well-defined period, and doesn't provide a useful definition of the Sun's rotation speed.
answered Mar 22 at 3:14
MarkMark
1,874920
edited Mar 22 at 19:55
answered Mar 22 at 3:14
MarkMark
1,874920
answered Mar 22 at 3:14
MarkMark
1,874920
answered Mar 22 at 3:14
MarkMark
1,874920
1,874920
4
$begingroup$
I just add that for Jupiter there is definitively no a fix day based on std rotation. At least no for the outer layer. It does even rotate in stripes and a simulation as in Celestia package is a kind of spectacle.
$endgroup$
– Alchimista
Mar 22 at 8:08
6
$begingroup$
“The assumption is that whatever's generating the magnetic field forms a reasonably coherent mass that's rotating at a uniform speed.” — I’d strengthen this by pointing out that it’s not just an assumption, it’s based closely on empirical facts: the magnetic field has a measurable uniform periodic behaviour, and based our understanding of planetary magnetic fields, we’re confident this corresponds to rotation of a somewhat coherent core. With the sun, as I understand it, we don’t see any uniformly periodic behaviour that we would expect to correspond to some kind of rotating mass.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 22 at 13:37
1
$begingroup$
The Sun's magnetic field don't have a period as @PeterLeFanuLumsdaine stated.
$endgroup$
– Mindwin
Mar 22 at 14:06
add a comment |
4
$begingroup$
I just add that for Jupiter there is definitively no a fix day based on std rotation. At least no for the outer layer. It does even rotate in stripes and a simulation as in Celestia package is a kind of spectacle.
$endgroup$
– Alchimista
Mar 22 at 8:08
6
$begingroup$
“The assumption is that whatever's generating the magnetic field forms a reasonably coherent mass that's rotating at a uniform speed.” — I’d strengthen this by pointing out that it’s not just an assumption, it’s based closely on empirical facts: the magnetic field has a measurable uniform periodic behaviour, and based our understanding of planetary magnetic fields, we’re confident this corresponds to rotation of a somewhat coherent core. With the sun, as I understand it, we don’t see any uniformly periodic behaviour that we would expect to correspond to some kind of rotating mass.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 22 at 13:37
1
$begingroup$
The Sun's magnetic field don't have a period as @PeterLeFanuLumsdaine stated.
$endgroup$
– Mindwin
Mar 22 at 14:06
4
4
$begingroup$
I just add that for Jupiter there is definitively no a fix day based on std rotation. At least no for the outer layer. It does even rotate in stripes and a simulation as in Celestia package is a kind of spectacle.
$endgroup$
– Alchimista
Mar 22 at 8:08
$begingroup$
I just add that for Jupiter there is definitively no a fix day based on std rotation. At least no for the outer layer. It does even rotate in stripes and a simulation as in Celestia package is a kind of spectacle.
$endgroup$
– Alchimista
Mar 22 at 8:08
6
6
$begingroup$
“The assumption is that whatever's generating the magnetic field forms a reasonably coherent mass that's rotating at a uniform speed.” — I’d strengthen this by pointing out that it’s not just an assumption, it’s based closely on empirical facts: the magnetic field has a measurable uniform periodic behaviour, and based our understanding of planetary magnetic fields, we’re confident this corresponds to rotation of a somewhat coherent core. With the sun, as I understand it, we don’t see any uniformly periodic behaviour that we would expect to correspond to some kind of rotating mass.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 22 at 13:37
$begingroup$
“The assumption is that whatever's generating the magnetic field forms a reasonably coherent mass that's rotating at a uniform speed.” — I’d strengthen this by pointing out that it’s not just an assumption, it’s based closely on empirical facts: the magnetic field has a measurable uniform periodic behaviour, and based our understanding of planetary magnetic fields, we’re confident this corresponds to rotation of a somewhat coherent core. With the sun, as I understand it, we don’t see any uniformly periodic behaviour that we would expect to correspond to some kind of rotating mass.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 22 at 13:37
1
1
$begingroup$
The Sun's magnetic field don't have a period as @PeterLeFanuLumsdaine stated.
$endgroup$
– Mindwin
Mar 22 at 14:06
$begingroup$
The Sun's magnetic field don't have a period as @PeterLeFanuLumsdaine stated.
$endgroup$
– Mindwin
Mar 22 at 14:06
add a comment |
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