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Is there a maximum distance from a planet that a moon can orbit?

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5

$begingroup$

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

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edited 6 hours ago

Qmechanic♦

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asked 8 hours ago

leemanleeman

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5

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

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.

$endgroup$

share|cite|improve this question

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.

share|cite|improve this question

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

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

2 Answers
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9

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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.

share|cite|improve this answer

answered 7 hours ago

Michael SeifertMichael Seifert

16.4k23059

$endgroup$

add a comment | 

3

$begingroup$

What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.

share|cite|improve this answer

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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9

$begingroup$

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.

share|cite|improve this answer

answered 7 hours ago

Michael SeifertMichael Seifert

16.4k23059

$endgroup$

add a comment | 

9

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered 7 hours ago

Michael SeifertMichael Seifert

16.4k23059

$endgroup$

share|cite|improve this answer

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

3

$begingroup$

What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.

share|cite|improve this answer

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 | 

3

$begingroup$

What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.

share|cite|improve this answer

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.

share|cite|improve this answer

answered 7 hours ago

Maury MarkowitzMaury Markowitz

5,0211629

$endgroup$

share|cite|improve this answer

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