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Is there a maximum distance from a planet that a moon can orbit?
What is the relationship between mass, speed and distance of a planet orbiting the sun?Two suns, one moon, and one planet?Rocky Planet in the center of SystemBased on measurements of a moon's orbit with respect to the planet, what can one calculate?Where does gravity get it's energy to cause tides?Would it be possible to pull Mars and Earth closer each other?Homework question: calculating radius and mass of planet from velocity and time“Global” orientation of an inclined orbital planeCan a planet have multiple significant sized moons?help: calculating surface temp for rotating planet from solar constant 1366 w/m2
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$begingroup$
Given a planet that orbits a star, and a moon that orbits that planet, is it possible to define a maximum orbital radius of that moon, beyond which the moon would no longer orbit the planet, but the star instead?
I initially (naively) thought this point would be where the star's gravity outweighed that of the planet:
$d_max = d_p - d_px$
$x = frac1sqrtfracm_pm_s+1$
Where:
$d_max = $ maximum orbital radius of the moon (around the planet), $d_p =$ orbital radius of the planet (around the sun), $m_p =$ mass of the planet, $m_s = $ mass of the star.
But I quickly realised this assumption was wrong (unless my shoddy maths is wrong, which is very possible), because this gives a value of 258,772 km using values of the Sun, Moon, and Earth. 125,627 km closer to the Earth than the Moon's actual orbital radius (values from Wikipedia).
Is there a maximum orbital distance? How can it be calculated?
orbital-motion planets solar-system celestial-mechanics moon
edited 6 hours ago
Qmechanic♦
110k122101292
asked 8 hours ago
leemanleeman
284
New contributor
leeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
Given a planet that orbits a star, and a moon that orbits that planet, is it possible to define a maximum orbital radius of that moon, beyond which the moon would no longer orbit the planet, but the star instead?
I initially (naively) thought this point would be where the star's gravity outweighed that of the planet:
$d_max = d_p - d_px$
$x = frac1sqrtfracm_pm_s+1$
Where:
$d_max = $ maximum orbital radius of the moon (around the planet), $d_p =$ orbital radius of the planet (around the sun), $m_p =$ mass of the planet, $m_s = $ mass of the star.
But I quickly realised this assumption was wrong (unless my shoddy maths is wrong, which is very possible), because this gives a value of 258,772 km using values of the Sun, Moon, and Earth. 125,627 km closer to the Earth than the Moon's actual orbital radius (values from Wikipedia).
Is there a maximum orbital distance? How can it be calculated?
orbital-motion planets solar-system celestial-mechanics moon
edited 6 hours ago
Qmechanic♦
110k122101292
asked 8 hours ago
leemanleeman
284
New contributor
leeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Given a planet that orbits a star, and a moon that orbits that planet, is it possible to define a maximum orbital radius of that moon, beyond which the moon would no longer orbit the planet, but the star instead?
I initially (naively) thought this point would be where the star's gravity outweighed that of the planet:
$d_max = d_p - d_px$
$x = frac1sqrtfracm_pm_s+1$
Where:
$d_max = $ maximum orbital radius of the moon (around the planet), $d_p =$ orbital radius of the planet (around the sun), $m_p =$ mass of the planet, $m_s = $ mass of the star.
But I quickly realised this assumption was wrong (unless my shoddy maths is wrong, which is very possible), because this gives a value of 258,772 km using values of the Sun, Moon, and Earth. 125,627 km closer to the Earth than the Moon's actual orbital radius (values from Wikipedia).
Is there a maximum orbital distance? How can it be calculated?
orbital-motion planets solar-system celestial-mechanics moon
edited 6 hours ago
Qmechanic♦
110k122101292
asked 8 hours ago
leemanleeman
284
New contributor
leeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Given a planet that orbits a star, and a moon that orbits that planet, is it possible to define a maximum orbital radius of that moon, beyond which the moon would no longer orbit the planet, but the star instead?
I initially (naively) thought this point would be where the star's gravity outweighed that of the planet:
$d_max = d_p - d_px$
$x = frac1sqrtfracm_pm_s+1$
Where:
$d_max = $ maximum orbital radius of the moon (around the planet), $d_p =$ orbital radius of the planet (around the sun), $m_p =$ mass of the planet, $m_s = $ mass of the star.
But I quickly realised this assumption was wrong (unless my shoddy maths is wrong, which is very possible), because this gives a value of 258,772 km using values of the Sun, Moon, and Earth. 125,627 km closer to the Earth than the Moon's actual orbital radius (values from Wikipedia).
Is there a maximum orbital distance? How can it be calculated?
orbital-motion planets solar-system celestial-mechanics moon
orbital-motion planets solar-system celestial-mechanics moon
edited 6 hours ago
Qmechanic♦
110k122101292
asked 8 hours ago
leemanleeman
284
New contributor
leeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Qmechanic♦
110k122101292
asked 8 hours ago
leemanleeman
284
New contributor
leeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Qmechanic♦
110k122101292
edited 6 hours ago
Qmechanic♦
110k122101292
edited 6 hours ago
Qmechanic♦
110k122101292
110k122101292
asked 8 hours ago
leemanleeman
284
New contributor
leeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
leemanleeman
284
asked 8 hours ago
leemanleeman
284
284
New contributor
leeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
leeman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The concept you're looking for is that of a planet's Hill sphere. If a planet of mass $m$ is in a roughly circular orbit of radius $a$ about a star of mass $M$, then the radius of this "sphere" is given by
$$
r_H = a sqrt[3]fracm3M.
$$
For the Earth-Moon system, this yields $r_H approx 0.01 text AU$, or about 1.5 million kilometers.
The calculation given in the Wikipedia article shows how to derive this in terms of rotating reference frames. But for a qualitative explanation of why your reasoning didn't work, you have to remember that the moon and the planet are not stationary; both of them are accelerating towards the star. This means that it's not the entire weight of the moon that matters, but rather the tidal force on the moon as measured in the planet's frame. This effect, along with the fact that the centripetal force needed for the star to "steal" the planet is a bit less when the moon is between the star and the planet, leads to the expression given above.
answered 7 hours ago
Michael SeifertMichael Seifert
16.4k23059
$endgroup$
add a comment |
$begingroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered 7 hours ago
Maury MarkowitzMaury Markowitz
5,0211629
$endgroup$
1
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
7 hours ago
1
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
7 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The concept you're looking for is that of a planet's Hill sphere. If a planet of mass $m$ is in a roughly circular orbit of radius $a$ about a star of mass $M$, then the radius of this "sphere" is given by
$$
r_H = a sqrt[3]fracm3M.
$$
For the Earth-Moon system, this yields $r_H approx 0.01 text AU$, or about 1.5 million kilometers.
The calculation given in the Wikipedia article shows how to derive this in terms of rotating reference frames. But for a qualitative explanation of why your reasoning didn't work, you have to remember that the moon and the planet are not stationary; both of them are accelerating towards the star. This means that it's not the entire weight of the moon that matters, but rather the tidal force on the moon as measured in the planet's frame. This effect, along with the fact that the centripetal force needed for the star to "steal" the planet is a bit less when the moon is between the star and the planet, leads to the expression given above.
answered 7 hours ago
Michael SeifertMichael Seifert
16.4k23059
$endgroup$
add a comment |
$begingroup$
The concept you're looking for is that of a planet's Hill sphere. If a planet of mass $m$ is in a roughly circular orbit of radius $a$ about a star of mass $M$, then the radius of this "sphere" is given by
$$
r_H = a sqrt[3]fracm3M.
$$
For the Earth-Moon system, this yields $r_H approx 0.01 text AU$, or about 1.5 million kilometers.
The calculation given in the Wikipedia article shows how to derive this in terms of rotating reference frames. But for a qualitative explanation of why your reasoning didn't work, you have to remember that the moon and the planet are not stationary; both of them are accelerating towards the star. This means that it's not the entire weight of the moon that matters, but rather the tidal force on the moon as measured in the planet's frame. This effect, along with the fact that the centripetal force needed for the star to "steal" the planet is a bit less when the moon is between the star and the planet, leads to the expression given above.
answered 7 hours ago
Michael SeifertMichael Seifert
16.4k23059
$endgroup$
add a comment |
$begingroup$
The concept you're looking for is that of a planet's Hill sphere. If a planet of mass $m$ is in a roughly circular orbit of radius $a$ about a star of mass $M$, then the radius of this "sphere" is given by
$$
r_H = a sqrt[3]fracm3M.
$$
For the Earth-Moon system, this yields $r_H approx 0.01 text AU$, or about 1.5 million kilometers.
The calculation given in the Wikipedia article shows how to derive this in terms of rotating reference frames. But for a qualitative explanation of why your reasoning didn't work, you have to remember that the moon and the planet are not stationary; both of them are accelerating towards the star. This means that it's not the entire weight of the moon that matters, but rather the tidal force on the moon as measured in the planet's frame. This effect, along with the fact that the centripetal force needed for the star to "steal" the planet is a bit less when the moon is between the star and the planet, leads to the expression given above.
answered 7 hours ago
Michael SeifertMichael Seifert
16.4k23059
$endgroup$
The concept you're looking for is that of a planet's Hill sphere. If a planet of mass $m$ is in a roughly circular orbit of radius $a$ about a star of mass $M$, then the radius of this "sphere" is given by
$$
r_H = a sqrt[3]fracm3M.
$$
For the Earth-Moon system, this yields $r_H approx 0.01 text AU$, or about 1.5 million kilometers.
The calculation given in the Wikipedia article shows how to derive this in terms of rotating reference frames. But for a qualitative explanation of why your reasoning didn't work, you have to remember that the moon and the planet are not stationary; both of them are accelerating towards the star. This means that it's not the entire weight of the moon that matters, but rather the tidal force on the moon as measured in the planet's frame. This effect, along with the fact that the centripetal force needed for the star to "steal" the planet is a bit less when the moon is between the star and the planet, leads to the expression given above.
answered 7 hours ago
Michael SeifertMichael Seifert
16.4k23059
answered 7 hours ago
Michael SeifertMichael Seifert
16.4k23059
answered 7 hours ago
Michael SeifertMichael Seifert
16.4k23059
answered 7 hours ago
Michael SeifertMichael Seifert
16.4k23059
16.4k23059
add a comment |
add a comment |
$begingroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered 7 hours ago
Maury MarkowitzMaury Markowitz
5,0211629
$endgroup$
1
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
7 hours ago
1
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
7 hours ago
add a comment |
$begingroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered 7 hours ago
Maury MarkowitzMaury Markowitz
5,0211629
$endgroup$
1
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
7 hours ago
1
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
7 hours ago
add a comment |
$begingroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered 7 hours ago
Maury MarkowitzMaury Markowitz
5,0211629
$endgroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered 7 hours ago
Maury MarkowitzMaury Markowitz
5,0211629
answered 7 hours ago
Maury MarkowitzMaury Markowitz
5,0211629
answered 7 hours ago
Maury MarkowitzMaury Markowitz
5,0211629
answered 7 hours ago
Maury MarkowitzMaury Markowitz
5,0211629
5,0211629
1
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
7 hours ago
1
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
7 hours ago
add a comment |
1
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
7 hours ago
1
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
7 hours ago
1
1
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
7 hours ago
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
7 hours ago
1
1
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
7 hours ago
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
7 hours ago
add a comment |
leeman is a new contributor. Be nice, and check out our Code of Conduct.
leeman is a new contributor. Be nice, and check out our Code of Conduct.
leeman is a new contributor. Be nice, and check out our Code of Conduct.
leeman is a new contributor. Be nice, and check out our Code of Conduct.
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