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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?How to calculate linar velocity of planet orbit?Is Feynman's explanation of how the moon stays in orbit wrong?How many celestial bodies could be in stable orbit at roughly the same distance from a star?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?Homework question: calculating radius and mass of planet from velocity and time“Global” orientation of an inclined orbital planehelp: calculating surface temp for rotating planet from solar constant 1366 w/m2
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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:
$$d_textmax = d_mathrm p - d_mathrm px$$
$$x = frac1sqrtfracm_mathrm pm_mathrm s+1$$
Where:
$d_textmax = $ maximum orbital radius of the moon (around the planet), $d_mathrm p =$ orbital radius of the planet (around the sun), $m_mathrm p =$ mass of the planet, $m_mathrm 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 $258,772 mathrmkm$ using values of the Sun, Moon, and Earth. $125,627 mathrmkm$ 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 Jun 22 at 21:46
Loong
1,56112 silver badges21 bronze badges
asked Jun 19 at 20:15
leemanleeman
1682 silver badges6 bronze badges
$endgroup$
|
show 1 more 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:
$$d_textmax = d_mathrm p - d_mathrm px$$
$$x = frac1sqrtfracm_mathrm pm_mathrm s+1$$
Where:
$d_textmax = $ maximum orbital radius of the moon (around the planet), $d_mathrm p =$ orbital radius of the planet (around the sun), $m_mathrm p =$ mass of the planet, $m_mathrm 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 $258,772 mathrmkm$ using values of the Sun, Moon, and Earth. $125,627 mathrmkm$ 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 Jun 22 at 21:46
Loong
1,56112 silver badges21 bronze badges
asked Jun 19 at 20:15
leemanleeman
1682 silver badges6 bronze badges
$endgroup$
2
$begingroup$
OK, I'm not a physicist or a mathematician, but isn't the question flawed? Although we think of moons orbiting planets and planets orbiting stars, don't they actually orbit around their common centres of gravity. Therefore, don't moons already orbit the star (because they are all orbiting each other)?
$endgroup$
– Gareth
Jun 20 at 8:20
5
$begingroup$
@Gareth It's not that simple. The Moon doesn't orbit the Sun, but the Earth-Moon system does. As long as the Earth and the Moon are gravitationally bound, neither is individually orbiting the Sun. We tend to say that the Earth orbits the Sun, but that can be taken two ways - 1) As massive as the Moon is, it's still just a tiny fraction of the mass of Earth, so it can usually be ignored, just like all the other junk orbiting Earth, 2) We say "Earth", but really mean the "Earth-Moon system" (including all the other tiny bits of stuff gravitationally bound to the Earth).
$endgroup$
– Luaan
Jun 20 at 8:56
3
$begingroup$
@Gareth If you look at trajectories of Earth and the Moon relative to the Sun, they're very similar - the distance between the Earth and the Moon is tiny compared to their distance to the Sun. Both wobble, with the Earth's wobble being much smaller than the Moon's, but both of those wobble's amplitudes are too small to have much impact relative to the distance to the Sun. So in this sense, both the Earth and the Moon orbit the Sun, with some perturbations. However, all this is OT, since the OP obviously cares about when the moon stops orbiting the planet, not starts orbiting the star :P
$endgroup$
– Luaan
Jun 20 at 9:17
$begingroup$
@Luann Fair enough. And perhaps this needs a question of it's own, but if the moon were to become a trojan or greek would it be considered to be orbiting the star or the planet? Or neither?
$endgroup$
– Gareth
Jun 20 at 9:29
1
$begingroup$
That all seems like a matter definitions and choices what is a moon and planet etc. If you don't care about technicalities and astronomical vocabulary exactness (like is Pluto a planet) really you just have to look at the constraints necessary to get definitive answers. Like say given a velocity, masses, then it is determined if it is a stable orbit, hyperbole (escape velocity) or parabolic trajectory. Or basically different eccentricity cases, which there are three possibilities for any body in question.
$endgroup$
– marshal craft
Jun 20 at 12:56
|
show 1 more 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:
$$d_textmax = d_mathrm p - d_mathrm px$$
$$x = frac1sqrtfracm_mathrm pm_mathrm s+1$$
Where:
$d_textmax = $ maximum orbital radius of the moon (around the planet), $d_mathrm p =$ orbital radius of the planet (around the sun), $m_mathrm p =$ mass of the planet, $m_mathrm 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 $258,772 mathrmkm$ using values of the Sun, Moon, and Earth. $125,627 mathrmkm$ 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 Jun 22 at 21:46
Loong
1,56112 silver badges21 bronze badges
asked Jun 19 at 20:15
leemanleeman
1682 silver badges6 bronze badges
$endgroup$
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:
$$d_textmax = d_mathrm p - d_mathrm px$$
$$x = frac1sqrtfracm_mathrm pm_mathrm s+1$$
Where:
$d_textmax = $ maximum orbital radius of the moon (around the planet), $d_mathrm p =$ orbital radius of the planet (around the sun), $m_mathrm p =$ mass of the planet, $m_mathrm 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 $258,772 mathrmkm$ using values of the Sun, Moon, and Earth. $125,627 mathrmkm$ 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 Jun 22 at 21:46
Loong
1,56112 silver badges21 bronze badges
asked Jun 19 at 20:15
leemanleeman
1682 silver badges6 bronze badges
edited Jun 22 at 21:46
Loong
1,56112 silver badges21 bronze badges
asked Jun 19 at 20:15
leemanleeman
1682 silver badges6 bronze badges
edited Jun 22 at 21:46
Loong
1,56112 silver badges21 bronze badges
edited Jun 22 at 21:46
Loong
1,56112 silver badges21 bronze badges
edited Jun 22 at 21:46
Loong
1,56112 silver badges21 bronze badges
1,56112 silver badges21 bronze badges
asked Jun 19 at 20:15
leemanleeman
1682 silver badges6 bronze badges
asked Jun 19 at 20:15
leemanleeman
1682 silver badges6 bronze badges
asked Jun 19 at 20:15
leemanleeman
1682 silver badges6 bronze badges
1682 silver badges6 bronze badges
2
$begingroup$
OK, I'm not a physicist or a mathematician, but isn't the question flawed? Although we think of moons orbiting planets and planets orbiting stars, don't they actually orbit around their common centres of gravity. Therefore, don't moons already orbit the star (because they are all orbiting each other)?
$endgroup$
– Gareth
Jun 20 at 8:20
5
$begingroup$
@Gareth It's not that simple. The Moon doesn't orbit the Sun, but the Earth-Moon system does. As long as the Earth and the Moon are gravitationally bound, neither is individually orbiting the Sun. We tend to say that the Earth orbits the Sun, but that can be taken two ways - 1) As massive as the Moon is, it's still just a tiny fraction of the mass of Earth, so it can usually be ignored, just like all the other junk orbiting Earth, 2) We say "Earth", but really mean the "Earth-Moon system" (including all the other tiny bits of stuff gravitationally bound to the Earth).
$endgroup$
– Luaan
Jun 20 at 8:56
3
$begingroup$
@Gareth If you look at trajectories of Earth and the Moon relative to the Sun, they're very similar - the distance between the Earth and the Moon is tiny compared to their distance to the Sun. Both wobble, with the Earth's wobble being much smaller than the Moon's, but both of those wobble's amplitudes are too small to have much impact relative to the distance to the Sun. So in this sense, both the Earth and the Moon orbit the Sun, with some perturbations. However, all this is OT, since the OP obviously cares about when the moon stops orbiting the planet, not starts orbiting the star :P
$endgroup$
– Luaan
Jun 20 at 9:17
$begingroup$
@Luann Fair enough. And perhaps this needs a question of it's own, but if the moon were to become a trojan or greek would it be considered to be orbiting the star or the planet? Or neither?
$endgroup$
– Gareth
Jun 20 at 9:29
1
$begingroup$
That all seems like a matter definitions and choices what is a moon and planet etc. If you don't care about technicalities and astronomical vocabulary exactness (like is Pluto a planet) really you just have to look at the constraints necessary to get definitive answers. Like say given a velocity, masses, then it is determined if it is a stable orbit, hyperbole (escape velocity) or parabolic trajectory. Or basically different eccentricity cases, which there are three possibilities for any body in question.
$endgroup$
– marshal craft
Jun 20 at 12:56
|
show 1 more comment
2
$begingroup$
OK, I'm not a physicist or a mathematician, but isn't the question flawed? Although we think of moons orbiting planets and planets orbiting stars, don't they actually orbit around their common centres of gravity. Therefore, don't moons already orbit the star (because they are all orbiting each other)?
$endgroup$
– Gareth
Jun 20 at 8:20
5
$begingroup$
@Gareth It's not that simple. The Moon doesn't orbit the Sun, but the Earth-Moon system does. As long as the Earth and the Moon are gravitationally bound, neither is individually orbiting the Sun. We tend to say that the Earth orbits the Sun, but that can be taken two ways - 1) As massive as the Moon is, it's still just a tiny fraction of the mass of Earth, so it can usually be ignored, just like all the other junk orbiting Earth, 2) We say "Earth", but really mean the "Earth-Moon system" (including all the other tiny bits of stuff gravitationally bound to the Earth).
$endgroup$
– Luaan
Jun 20 at 8:56
3
$begingroup$
@Gareth If you look at trajectories of Earth and the Moon relative to the Sun, they're very similar - the distance between the Earth and the Moon is tiny compared to their distance to the Sun. Both wobble, with the Earth's wobble being much smaller than the Moon's, but both of those wobble's amplitudes are too small to have much impact relative to the distance to the Sun. So in this sense, both the Earth and the Moon orbit the Sun, with some perturbations. However, all this is OT, since the OP obviously cares about when the moon stops orbiting the planet, not starts orbiting the star :P
$endgroup$
– Luaan
Jun 20 at 9:17
$begingroup$
@Luann Fair enough. And perhaps this needs a question of it's own, but if the moon were to become a trojan or greek would it be considered to be orbiting the star or the planet? Or neither?
$endgroup$
– Gareth
Jun 20 at 9:29
1
$begingroup$
That all seems like a matter definitions and choices what is a moon and planet etc. If you don't care about technicalities and astronomical vocabulary exactness (like is Pluto a planet) really you just have to look at the constraints necessary to get definitive answers. Like say given a velocity, masses, then it is determined if it is a stable orbit, hyperbole (escape velocity) or parabolic trajectory. Or basically different eccentricity cases, which there are three possibilities for any body in question.
$endgroup$
– marshal craft
Jun 20 at 12:56
2
2
$begingroup$
OK, I'm not a physicist or a mathematician, but isn't the question flawed? Although we think of moons orbiting planets and planets orbiting stars, don't they actually orbit around their common centres of gravity. Therefore, don't moons already orbit the star (because they are all orbiting each other)?
$endgroup$
– Gareth
Jun 20 at 8:20
$begingroup$
OK, I'm not a physicist or a mathematician, but isn't the question flawed? Although we think of moons orbiting planets and planets orbiting stars, don't they actually orbit around their common centres of gravity. Therefore, don't moons already orbit the star (because they are all orbiting each other)?
$endgroup$
– Gareth
Jun 20 at 8:20
5
5
$begingroup$
@Gareth It's not that simple. The Moon doesn't orbit the Sun, but the Earth-Moon system does. As long as the Earth and the Moon are gravitationally bound, neither is individually orbiting the Sun. We tend to say that the Earth orbits the Sun, but that can be taken two ways - 1) As massive as the Moon is, it's still just a tiny fraction of the mass of Earth, so it can usually be ignored, just like all the other junk orbiting Earth, 2) We say "Earth", but really mean the "Earth-Moon system" (including all the other tiny bits of stuff gravitationally bound to the Earth).
$endgroup$
– Luaan
Jun 20 at 8:56
$begingroup$
@Gareth It's not that simple. The Moon doesn't orbit the Sun, but the Earth-Moon system does. As long as the Earth and the Moon are gravitationally bound, neither is individually orbiting the Sun. We tend to say that the Earth orbits the Sun, but that can be taken two ways - 1) As massive as the Moon is, it's still just a tiny fraction of the mass of Earth, so it can usually be ignored, just like all the other junk orbiting Earth, 2) We say "Earth", but really mean the "Earth-Moon system" (including all the other tiny bits of stuff gravitationally bound to the Earth).
$endgroup$
– Luaan
Jun 20 at 8:56
3
3
$begingroup$
@Gareth If you look at trajectories of Earth and the Moon relative to the Sun, they're very similar - the distance between the Earth and the Moon is tiny compared to their distance to the Sun. Both wobble, with the Earth's wobble being much smaller than the Moon's, but both of those wobble's amplitudes are too small to have much impact relative to the distance to the Sun. So in this sense, both the Earth and the Moon orbit the Sun, with some perturbations. However, all this is OT, since the OP obviously cares about when the moon stops orbiting the planet, not starts orbiting the star :P
$endgroup$
– Luaan
Jun 20 at 9:17
$begingroup$
@Gareth If you look at trajectories of Earth and the Moon relative to the Sun, they're very similar - the distance between the Earth and the Moon is tiny compared to their distance to the Sun. Both wobble, with the Earth's wobble being much smaller than the Moon's, but both of those wobble's amplitudes are too small to have much impact relative to the distance to the Sun. So in this sense, both the Earth and the Moon orbit the Sun, with some perturbations. However, all this is OT, since the OP obviously cares about when the moon stops orbiting the planet, not starts orbiting the star :P
$endgroup$
– Luaan
Jun 20 at 9:17
$begingroup$
@Luann Fair enough. And perhaps this needs a question of it's own, but if the moon were to become a trojan or greek would it be considered to be orbiting the star or the planet? Or neither?
$endgroup$
– Gareth
Jun 20 at 9:29
$begingroup$
@Luann Fair enough. And perhaps this needs a question of it's own, but if the moon were to become a trojan or greek would it be considered to be orbiting the star or the planet? Or neither?
$endgroup$
– Gareth
Jun 20 at 9:29
1
1
$begingroup$
That all seems like a matter definitions and choices what is a moon and planet etc. If you don't care about technicalities and astronomical vocabulary exactness (like is Pluto a planet) really you just have to look at the constraints necessary to get definitive answers. Like say given a velocity, masses, then it is determined if it is a stable orbit, hyperbole (escape velocity) or parabolic trajectory. Or basically different eccentricity cases, which there are three possibilities for any body in question.
$endgroup$
– marshal craft
Jun 20 at 12:56
$begingroup$
That all seems like a matter definitions and choices what is a moon and planet etc. If you don't care about technicalities and astronomical vocabulary exactness (like is Pluto a planet) really you just have to look at the constraints necessary to get definitive answers. Like say given a velocity, masses, then it is determined if it is a stable orbit, hyperbole (escape velocity) or parabolic trajectory. Or basically different eccentricity cases, which there are three possibilities for any body in question.
$endgroup$
– marshal craft
Jun 20 at 12:56
|
show 1 more comment
3 Answers
3
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 Sun-Earth 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.
As pointed out by @uhoh in the comments, the L1 and L2 Earth-Sun Lagrange points are precisely this distance from the Earth. These are precisely the points where the gravitational forces of the Earth and the Sun combine in such a way that an object can orbit the Sun with the same period as the Earth, but at a different radius. In a rotating reference frame, this means that the influences of the Earth, the Sun, and the centrifugal force are precisely canceling out; any closer to Earth than that, and the Earth's forces dominate. Thus, the L1 and L2 Lagrange points are on the boundary of the Hill sphere.
edited Jun 21 at 11:01
Euro Micelli
1535 bronze badges
answered Jun 19 at 20:56
Michael SeifertMichael Seifert
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4
$begingroup$
I love answers where you learn something new! I'd never heard of the Hill Sphere before... Also, it seems to be a common theme in orbital dynamics to forget that everything is moving. The slingshot maneuver makes no sense at all until you realise this.
$endgroup$
– Oscar Bravo
Jun 20 at 8:27
1
$begingroup$
Nice answer!+1I think that I would say "For the Sun-Earth system..." rather than the "Earth-Moon system"; the Earth's Hill sphere is defined and exists whether or not The Moon exists, it's an artifact of the Sun-Earth system. You might also mention that the Earth-Moon Lagrange points L1 and L2 are also at about 1.5 million kilometers. It's not the same thing but it's somewhat related.
$endgroup$
– uhoh
Jun 20 at 9:17
2
$begingroup$
Michael, note that @uhoh stated "It's not the same thing but it's somewhat related." The distances between the Sun-Earth L1 and L2 points and the Earth are the real solutions to two slightly different fifth order polynomials, neither of which has a closed form expression. The expression $asqrt[3]frac m 3M$ is a close approximation of these distances. Two additional comments: (1) The so-called Hill sphere isn't really a sphere, and (2) An orbit with a radius that is more than half the Hill sphere radius is most likely unstable.
$endgroup$
– David Hammen
Jun 20 at 13:04
1
$begingroup$
@DavidHammen: True, but the radius of the Hill "sphere" is also only approximately given by $asqrt[3]frac m 3M$; in fact, the exact expressions for the bounds of the Hill "sphere" are the roots of the same fifth-order polynomial that gives the distances to L1 and L2; see the derivation in the Wiki article. And yes, it's not actually a sphere in the geometric sense (hence the scare quotes on "sphere" in my second sentence), though I suppose you could argue that it's a "sphere" in the sense of a "sphere of influence."
$endgroup$
– Michael Seifert
Jun 20 at 13:47
add a comment |
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What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered Jun 19 at 21:00
Maury MarkowitzMaury Markowitz
5,1041 gold badge6 silver badges29 bronze badges
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3
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Sigh, posted while I was typing.
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– Maury Markowitz
Jun 19 at 21:01
2
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
Jun 19 at 21:02
7
$begingroup$
It's always reassuring to have two concurring opinions ;-).
$endgroup$
– Peter A. Schneider
Jun 20 at 15:41
add a comment |
$begingroup$
There are several different ways of characterizing whether something is "really" in orbit around the sun or Earth. One is asking which gravitational force is larger. As you found, by that standard, the moon primarily orbits around the sun, with its orbit being next to Earth's. Another criteria would be looking at which body's gravitational potential energy dominates: if you wanted to fly a rocket from the moon to outside the solar system, which body's gravity would contribute more to the energy you would need? For that question, the moon is even more clearly in orbit around the sun rather than Earth. But you can also ask at what point would Earth's gravity be so weak that perturbations from the sun would take the moon out of a stable orbit, which gives you the Hill sphere, of which the moon is well within. There are other pairs of bodies, such as Pluto and Charon, for which these different definitions also give different answers.
answered Jun 21 at 15:57
AcccumulationAcccumulation
4,0837 silver badges16 bronze badges
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3 Answers
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3 Answers
3
active
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active
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$begingroup$
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 Sun-Earth 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.
As pointed out by @uhoh in the comments, the L1 and L2 Earth-Sun Lagrange points are precisely this distance from the Earth. These are precisely the points where the gravitational forces of the Earth and the Sun combine in such a way that an object can orbit the Sun with the same period as the Earth, but at a different radius. In a rotating reference frame, this means that the influences of the Earth, the Sun, and the centrifugal force are precisely canceling out; any closer to Earth than that, and the Earth's forces dominate. Thus, the L1 and L2 Lagrange points are on the boundary of the Hill sphere.
edited Jun 21 at 11:01
Euro Micelli
1535 bronze badges
answered Jun 19 at 20:56
Michael SeifertMichael Seifert
17.1k2 gold badges33 silver badges60 bronze badges
$endgroup$
4
$begingroup$
I love answers where you learn something new! I'd never heard of the Hill Sphere before... Also, it seems to be a common theme in orbital dynamics to forget that everything is moving. The slingshot maneuver makes no sense at all until you realise this.
$endgroup$
– Oscar Bravo
Jun 20 at 8:27
1
$begingroup$
Nice answer!+1I think that I would say "For the Sun-Earth system..." rather than the "Earth-Moon system"; the Earth's Hill sphere is defined and exists whether or not The Moon exists, it's an artifact of the Sun-Earth system. You might also mention that the Earth-Moon Lagrange points L1 and L2 are also at about 1.5 million kilometers. It's not the same thing but it's somewhat related.
$endgroup$
– uhoh
Jun 20 at 9:17
2
$begingroup$
Michael, note that @uhoh stated "It's not the same thing but it's somewhat related." The distances between the Sun-Earth L1 and L2 points and the Earth are the real solutions to two slightly different fifth order polynomials, neither of which has a closed form expression. The expression $asqrt[3]frac m 3M$ is a close approximation of these distances. Two additional comments: (1) The so-called Hill sphere isn't really a sphere, and (2) An orbit with a radius that is more than half the Hill sphere radius is most likely unstable.
$endgroup$
– David Hammen
Jun 20 at 13:04
1
$begingroup$
@DavidHammen: True, but the radius of the Hill "sphere" is also only approximately given by $asqrt[3]frac m 3M$; in fact, the exact expressions for the bounds of the Hill "sphere" are the roots of the same fifth-order polynomial that gives the distances to L1 and L2; see the derivation in the Wiki article. And yes, it's not actually a sphere in the geometric sense (hence the scare quotes on "sphere" in my second sentence), though I suppose you could argue that it's a "sphere" in the sense of a "sphere of influence."
$endgroup$
– Michael Seifert
Jun 20 at 13:47
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 Sun-Earth 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.
As pointed out by @uhoh in the comments, the L1 and L2 Earth-Sun Lagrange points are precisely this distance from the Earth. These are precisely the points where the gravitational forces of the Earth and the Sun combine in such a way that an object can orbit the Sun with the same period as the Earth, but at a different radius. In a rotating reference frame, this means that the influences of the Earth, the Sun, and the centrifugal force are precisely canceling out; any closer to Earth than that, and the Earth's forces dominate. Thus, the L1 and L2 Lagrange points are on the boundary of the Hill sphere.
edited Jun 21 at 11:01
Euro Micelli
1535 bronze badges
answered Jun 19 at 20:56
Michael SeifertMichael Seifert
17.1k2 gold badges33 silver badges60 bronze badges
$endgroup$
4
$begingroup$
I love answers where you learn something new! I'd never heard of the Hill Sphere before... Also, it seems to be a common theme in orbital dynamics to forget that everything is moving. The slingshot maneuver makes no sense at all until you realise this.
$endgroup$
– Oscar Bravo
Jun 20 at 8:27
1
$begingroup$
Nice answer!+1I think that I would say "For the Sun-Earth system..." rather than the "Earth-Moon system"; the Earth's Hill sphere is defined and exists whether or not The Moon exists, it's an artifact of the Sun-Earth system. You might also mention that the Earth-Moon Lagrange points L1 and L2 are also at about 1.5 million kilometers. It's not the same thing but it's somewhat related.
$endgroup$
– uhoh
Jun 20 at 9:17
2
$begingroup$
Michael, note that @uhoh stated "It's not the same thing but it's somewhat related." The distances between the Sun-Earth L1 and L2 points and the Earth are the real solutions to two slightly different fifth order polynomials, neither of which has a closed form expression. The expression $asqrt[3]frac m 3M$ is a close approximation of these distances. Two additional comments: (1) The so-called Hill sphere isn't really a sphere, and (2) An orbit with a radius that is more than half the Hill sphere radius is most likely unstable.
$endgroup$
– David Hammen
Jun 20 at 13:04
1
$begingroup$
@DavidHammen: True, but the radius of the Hill "sphere" is also only approximately given by $asqrt[3]frac m 3M$; in fact, the exact expressions for the bounds of the Hill "sphere" are the roots of the same fifth-order polynomial that gives the distances to L1 and L2; see the derivation in the Wiki article. And yes, it's not actually a sphere in the geometric sense (hence the scare quotes on "sphere" in my second sentence), though I suppose you could argue that it's a "sphere" in the sense of a "sphere of influence."
$endgroup$
– Michael Seifert
Jun 20 at 13:47
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 Sun-Earth 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.
As pointed out by @uhoh in the comments, the L1 and L2 Earth-Sun Lagrange points are precisely this distance from the Earth. These are precisely the points where the gravitational forces of the Earth and the Sun combine in such a way that an object can orbit the Sun with the same period as the Earth, but at a different radius. In a rotating reference frame, this means that the influences of the Earth, the Sun, and the centrifugal force are precisely canceling out; any closer to Earth than that, and the Earth's forces dominate. Thus, the L1 and L2 Lagrange points are on the boundary of the Hill sphere.
edited Jun 21 at 11:01
Euro Micelli
1535 bronze badges
answered Jun 19 at 20:56
Michael SeifertMichael Seifert
17.1k2 gold badges33 silver badges60 bronze badges
$endgroup$
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 Sun-Earth 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.
As pointed out by @uhoh in the comments, the L1 and L2 Earth-Sun Lagrange points are precisely this distance from the Earth. These are precisely the points where the gravitational forces of the Earth and the Sun combine in such a way that an object can orbit the Sun with the same period as the Earth, but at a different radius. In a rotating reference frame, this means that the influences of the Earth, the Sun, and the centrifugal force are precisely canceling out; any closer to Earth than that, and the Earth's forces dominate. Thus, the L1 and L2 Lagrange points are on the boundary of the Hill sphere.
edited Jun 21 at 11:01
Euro Micelli
1535 bronze badges
answered Jun 19 at 20:56
Michael SeifertMichael Seifert
17.1k2 gold badges33 silver badges60 bronze badges
edited Jun 21 at 11:01
Euro Micelli
1535 bronze badges
edited Jun 21 at 11:01
Euro Micelli
1535 bronze badges
edited Jun 21 at 11:01
Euro Micelli
1535 bronze badges
1535 bronze badges
answered Jun 19 at 20:56
Michael SeifertMichael Seifert
17.1k2 gold badges33 silver badges60 bronze badges
answered Jun 19 at 20:56
Michael SeifertMichael Seifert
17.1k2 gold badges33 silver badges60 bronze badges
answered Jun 19 at 20:56
Michael SeifertMichael Seifert
17.1k2 gold badges33 silver badges60 bronze badges
17.1k2 gold badges33 silver badges60 bronze badges
4
$begingroup$
I love answers where you learn something new! I'd never heard of the Hill Sphere before... Also, it seems to be a common theme in orbital dynamics to forget that everything is moving. The slingshot maneuver makes no sense at all until you realise this.
$endgroup$
– Oscar Bravo
Jun 20 at 8:27
1
$begingroup$
Nice answer!+1I think that I would say "For the Sun-Earth system..." rather than the "Earth-Moon system"; the Earth's Hill sphere is defined and exists whether or not The Moon exists, it's an artifact of the Sun-Earth system. You might also mention that the Earth-Moon Lagrange points L1 and L2 are also at about 1.5 million kilometers. It's not the same thing but it's somewhat related.
$endgroup$
– uhoh
Jun 20 at 9:17
2
$begingroup$
Michael, note that @uhoh stated "It's not the same thing but it's somewhat related." The distances between the Sun-Earth L1 and L2 points and the Earth are the real solutions to two slightly different fifth order polynomials, neither of which has a closed form expression. The expression $asqrt[3]frac m 3M$ is a close approximation of these distances. Two additional comments: (1) The so-called Hill sphere isn't really a sphere, and (2) An orbit with a radius that is more than half the Hill sphere radius is most likely unstable.
$endgroup$
– David Hammen
Jun 20 at 13:04
1
$begingroup$
@DavidHammen: True, but the radius of the Hill "sphere" is also only approximately given by $asqrt[3]frac m 3M$; in fact, the exact expressions for the bounds of the Hill "sphere" are the roots of the same fifth-order polynomial that gives the distances to L1 and L2; see the derivation in the Wiki article. And yes, it's not actually a sphere in the geometric sense (hence the scare quotes on "sphere" in my second sentence), though I suppose you could argue that it's a "sphere" in the sense of a "sphere of influence."
$endgroup$
– Michael Seifert
Jun 20 at 13:47
add a comment |
4
$begingroup$
I love answers where you learn something new! I'd never heard of the Hill Sphere before... Also, it seems to be a common theme in orbital dynamics to forget that everything is moving. The slingshot maneuver makes no sense at all until you realise this.
$endgroup$
– Oscar Bravo
Jun 20 at 8:27
1
$begingroup$
Nice answer!+1I think that I would say "For the Sun-Earth system..." rather than the "Earth-Moon system"; the Earth's Hill sphere is defined and exists whether or not The Moon exists, it's an artifact of the Sun-Earth system. You might also mention that the Earth-Moon Lagrange points L1 and L2 are also at about 1.5 million kilometers. It's not the same thing but it's somewhat related.
$endgroup$
– uhoh
Jun 20 at 9:17
2
$begingroup$
Michael, note that @uhoh stated "It's not the same thing but it's somewhat related." The distances between the Sun-Earth L1 and L2 points and the Earth are the real solutions to two slightly different fifth order polynomials, neither of which has a closed form expression. The expression $asqrt[3]frac m 3M$ is a close approximation of these distances. Two additional comments: (1) The so-called Hill sphere isn't really a sphere, and (2) An orbit with a radius that is more than half the Hill sphere radius is most likely unstable.
$endgroup$
– David Hammen
Jun 20 at 13:04
1
$begingroup$
@DavidHammen: True, but the radius of the Hill "sphere" is also only approximately given by $asqrt[3]frac m 3M$; in fact, the exact expressions for the bounds of the Hill "sphere" are the roots of the same fifth-order polynomial that gives the distances to L1 and L2; see the derivation in the Wiki article. And yes, it's not actually a sphere in the geometric sense (hence the scare quotes on "sphere" in my second sentence), though I suppose you could argue that it's a "sphere" in the sense of a "sphere of influence."
$endgroup$
– Michael Seifert
Jun 20 at 13:47
4
4
$begingroup$
I love answers where you learn something new! I'd never heard of the Hill Sphere before... Also, it seems to be a common theme in orbital dynamics to forget that everything is moving. The slingshot maneuver makes no sense at all until you realise this.
$endgroup$
– Oscar Bravo
Jun 20 at 8:27
$begingroup$
I love answers where you learn something new! I'd never heard of the Hill Sphere before... Also, it seems to be a common theme in orbital dynamics to forget that everything is moving. The slingshot maneuver makes no sense at all until you realise this.
$endgroup$
– Oscar Bravo
Jun 20 at 8:27
1
1
$begingroup$
Nice answer!
+1 I think that I would say "For the Sun-Earth system..." rather than the "Earth-Moon system"; the Earth's Hill sphere is defined and exists whether or not The Moon exists, it's an artifact of the Sun-Earth system. You might also mention that the Earth-Moon Lagrange points L1 and L2 are also at about 1.5 million kilometers. It's not the same thing but it's somewhat related.$endgroup$
– uhoh
Jun 20 at 9:17
$begingroup$
Nice answer!
+1 I think that I would say "For the Sun-Earth system..." rather than the "Earth-Moon system"; the Earth's Hill sphere is defined and exists whether or not The Moon exists, it's an artifact of the Sun-Earth system. You might also mention that the Earth-Moon Lagrange points L1 and L2 are also at about 1.5 million kilometers. It's not the same thing but it's somewhat related.$endgroup$
– uhoh
Jun 20 at 9:17
2
2
$begingroup$
Michael, note that @uhoh stated "It's not the same thing but it's somewhat related." The distances between the Sun-Earth L1 and L2 points and the Earth are the real solutions to two slightly different fifth order polynomials, neither of which has a closed form expression. The expression $asqrt[3]frac m 3M$ is a close approximation of these distances. Two additional comments: (1) The so-called Hill sphere isn't really a sphere, and (2) An orbit with a radius that is more than half the Hill sphere radius is most likely unstable.
$endgroup$
– David Hammen
Jun 20 at 13:04
$begingroup$
Michael, note that @uhoh stated "It's not the same thing but it's somewhat related." The distances between the Sun-Earth L1 and L2 points and the Earth are the real solutions to two slightly different fifth order polynomials, neither of which has a closed form expression. The expression $asqrt[3]frac m 3M$ is a close approximation of these distances. Two additional comments: (1) The so-called Hill sphere isn't really a sphere, and (2) An orbit with a radius that is more than half the Hill sphere radius is most likely unstable.
$endgroup$
– David Hammen
Jun 20 at 13:04
1
1
$begingroup$
@DavidHammen: True, but the radius of the Hill "sphere" is also only approximately given by $asqrt[3]frac m 3M$; in fact, the exact expressions for the bounds of the Hill "sphere" are the roots of the same fifth-order polynomial that gives the distances to L1 and L2; see the derivation in the Wiki article. And yes, it's not actually a sphere in the geometric sense (hence the scare quotes on "sphere" in my second sentence), though I suppose you could argue that it's a "sphere" in the sense of a "sphere of influence."
$endgroup$
– Michael Seifert
Jun 20 at 13:47
$begingroup$
@DavidHammen: True, but the radius of the Hill "sphere" is also only approximately given by $asqrt[3]frac m 3M$; in fact, the exact expressions for the bounds of the Hill "sphere" are the roots of the same fifth-order polynomial that gives the distances to L1 and L2; see the derivation in the Wiki article. And yes, it's not actually a sphere in the geometric sense (hence the scare quotes on "sphere" in my second sentence), though I suppose you could argue that it's a "sphere" in the sense of a "sphere of influence."
$endgroup$
– Michael Seifert
Jun 20 at 13:47
add a comment |
$begingroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered Jun 19 at 21:00
Maury MarkowitzMaury Markowitz
5,1041 gold badge6 silver badges29 bronze badges
$endgroup$
3
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
Jun 19 at 21:01
2
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
Jun 19 at 21:02
7
$begingroup$
It's always reassuring to have two concurring opinions ;-).
$endgroup$
– Peter A. Schneider
Jun 20 at 15:41
add a comment |
$begingroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered Jun 19 at 21:00
Maury MarkowitzMaury Markowitz
5,1041 gold badge6 silver badges29 bronze badges
$endgroup$
3
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
Jun 19 at 21:01
2
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
Jun 19 at 21:02
7
$begingroup$
It's always reassuring to have two concurring opinions ;-).
$endgroup$
– Peter A. Schneider
Jun 20 at 15:41
add a comment |
$begingroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered Jun 19 at 21:00
Maury MarkowitzMaury Markowitz
5,1041 gold badge6 silver badges29 bronze badges
$endgroup$
What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
answered Jun 19 at 21:00
Maury MarkowitzMaury Markowitz
5,1041 gold badge6 silver badges29 bronze badges
answered Jun 19 at 21:00
Maury MarkowitzMaury Markowitz
5,1041 gold badge6 silver badges29 bronze badges
answered Jun 19 at 21:00
Maury MarkowitzMaury Markowitz
5,1041 gold badge6 silver badges29 bronze badges
answered Jun 19 at 21:00
Maury MarkowitzMaury Markowitz
5,1041 gold badge6 silver badges29 bronze badges
5,1041 gold badge6 silver badges29 bronze badges
3
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
Jun 19 at 21:01
2
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
Jun 19 at 21:02
7
$begingroup$
It's always reassuring to have two concurring opinions ;-).
$endgroup$
– Peter A. Schneider
Jun 20 at 15:41
add a comment |
3
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
Jun 19 at 21:01
2
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
Jun 19 at 21:02
7
$begingroup$
It's always reassuring to have two concurring opinions ;-).
$endgroup$
– Peter A. Schneider
Jun 20 at 15:41
3
3
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
Jun 19 at 21:01
$begingroup$
Sigh, posted while I was typing.
$endgroup$
– Maury Markowitz
Jun 19 at 21:01
2
2
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
Jun 19 at 21:02
$begingroup$
Heh, that's happened to me more times than I can count...
$endgroup$
– Michael Seifert
Jun 19 at 21:02
7
7
$begingroup$
It's always reassuring to have two concurring opinions ;-).
$endgroup$
– Peter A. Schneider
Jun 20 at 15:41
$begingroup$
It's always reassuring to have two concurring opinions ;-).
$endgroup$
– Peter A. Schneider
Jun 20 at 15:41
add a comment |
$begingroup$
There are several different ways of characterizing whether something is "really" in orbit around the sun or Earth. One is asking which gravitational force is larger. As you found, by that standard, the moon primarily orbits around the sun, with its orbit being next to Earth's. Another criteria would be looking at which body's gravitational potential energy dominates: if you wanted to fly a rocket from the moon to outside the solar system, which body's gravity would contribute more to the energy you would need? For that question, the moon is even more clearly in orbit around the sun rather than Earth. But you can also ask at what point would Earth's gravity be so weak that perturbations from the sun would take the moon out of a stable orbit, which gives you the Hill sphere, of which the moon is well within. There are other pairs of bodies, such as Pluto and Charon, for which these different definitions also give different answers.
answered Jun 21 at 15:57
AcccumulationAcccumulation
4,0837 silver badges16 bronze badges
$endgroup$
add a comment |
$begingroup$
There are several different ways of characterizing whether something is "really" in orbit around the sun or Earth. One is asking which gravitational force is larger. As you found, by that standard, the moon primarily orbits around the sun, with its orbit being next to Earth's. Another criteria would be looking at which body's gravitational potential energy dominates: if you wanted to fly a rocket from the moon to outside the solar system, which body's gravity would contribute more to the energy you would need? For that question, the moon is even more clearly in orbit around the sun rather than Earth. But you can also ask at what point would Earth's gravity be so weak that perturbations from the sun would take the moon out of a stable orbit, which gives you the Hill sphere, of which the moon is well within. There are other pairs of bodies, such as Pluto and Charon, for which these different definitions also give different answers.
answered Jun 21 at 15:57
AcccumulationAcccumulation
4,0837 silver badges16 bronze badges
$endgroup$
add a comment |
$begingroup$
There are several different ways of characterizing whether something is "really" in orbit around the sun or Earth. One is asking which gravitational force is larger. As you found, by that standard, the moon primarily orbits around the sun, with its orbit being next to Earth's. Another criteria would be looking at which body's gravitational potential energy dominates: if you wanted to fly a rocket from the moon to outside the solar system, which body's gravity would contribute more to the energy you would need? For that question, the moon is even more clearly in orbit around the sun rather than Earth. But you can also ask at what point would Earth's gravity be so weak that perturbations from the sun would take the moon out of a stable orbit, which gives you the Hill sphere, of which the moon is well within. There are other pairs of bodies, such as Pluto and Charon, for which these different definitions also give different answers.
answered Jun 21 at 15:57
AcccumulationAcccumulation
4,0837 silver badges16 bronze badges
$endgroup$
There are several different ways of characterizing whether something is "really" in orbit around the sun or Earth. One is asking which gravitational force is larger. As you found, by that standard, the moon primarily orbits around the sun, with its orbit being next to Earth's. Another criteria would be looking at which body's gravitational potential energy dominates: if you wanted to fly a rocket from the moon to outside the solar system, which body's gravity would contribute more to the energy you would need? For that question, the moon is even more clearly in orbit around the sun rather than Earth. But you can also ask at what point would Earth's gravity be so weak that perturbations from the sun would take the moon out of a stable orbit, which gives you the Hill sphere, of which the moon is well within. There are other pairs of bodies, such as Pluto and Charon, for which these different definitions also give different answers.
answered Jun 21 at 15:57
AcccumulationAcccumulation
4,0837 silver badges16 bronze badges
answered Jun 21 at 15:57
AcccumulationAcccumulation
4,0837 silver badges16 bronze badges
answered Jun 21 at 15:57
AcccumulationAcccumulation
4,0837 silver badges16 bronze badges
answered Jun 21 at 15:57
AcccumulationAcccumulation
4,0837 silver badges16 bronze badges
4,0837 silver badges16 bronze badges
add a comment |
add a comment |
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OK, I'm not a physicist or a mathematician, but isn't the question flawed? Although we think of moons orbiting planets and planets orbiting stars, don't they actually orbit around their common centres of gravity. Therefore, don't moons already orbit the star (because they are all orbiting each other)?
$endgroup$
– Gareth
Jun 20 at 8:20
5
$begingroup$
@Gareth It's not that simple. The Moon doesn't orbit the Sun, but the Earth-Moon system does. As long as the Earth and the Moon are gravitationally bound, neither is individually orbiting the Sun. We tend to say that the Earth orbits the Sun, but that can be taken two ways - 1) As massive as the Moon is, it's still just a tiny fraction of the mass of Earth, so it can usually be ignored, just like all the other junk orbiting Earth, 2) We say "Earth", but really mean the "Earth-Moon system" (including all the other tiny bits of stuff gravitationally bound to the Earth).
$endgroup$
– Luaan
Jun 20 at 8:56
3
$begingroup$
@Gareth If you look at trajectories of Earth and the Moon relative to the Sun, they're very similar - the distance between the Earth and the Moon is tiny compared to their distance to the Sun. Both wobble, with the Earth's wobble being much smaller than the Moon's, but both of those wobble's amplitudes are too small to have much impact relative to the distance to the Sun. So in this sense, both the Earth and the Moon orbit the Sun, with some perturbations. However, all this is OT, since the OP obviously cares about when the moon stops orbiting the planet, not starts orbiting the star :P
$endgroup$
– Luaan
Jun 20 at 9:17
$begingroup$
@Luann Fair enough. And perhaps this needs a question of it's own, but if the moon were to become a trojan or greek would it be considered to be orbiting the star or the planet? Or neither?
$endgroup$
– Gareth
Jun 20 at 9:29
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That all seems like a matter definitions and choices what is a moon and planet etc. If you don't care about technicalities and astronomical vocabulary exactness (like is Pluto a planet) really you just have to look at the constraints necessary to get definitive answers. Like say given a velocity, masses, then it is determined if it is a stable orbit, hyperbole (escape velocity) or parabolic trajectory. Or basically different eccentricity cases, which there are three possibilities for any body in question.
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– marshal craft
Jun 20 at 12:56