Is there a relationship between apparent moon size and orbital distance around an earth-analogue?
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I would like to give my imaginary earth-analogue planet multiple (natural) moons.
Is the orbital period of such moons at all related to their size (or apparent size)?
In other words, am I free to have a large close-by moon, or a small far-away moon, or anything in between, without it having an effect on orbital period?
moons
asked 2 hours ago
Jack
1113
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up vote
2
down vote
favorite
I would like to give my imaginary earth-analogue planet multiple (natural) moons.
Is the orbital period of such moons at all related to their size (or apparent size)?
In other words, am I free to have a large close-by moon, or a small far-away moon, or anything in between, without it having an effect on orbital period?
moons
asked 2 hours ago
Jack
1113
See en.wikipedia.org/wiki/Orbital_speed . In most common cases, orbital period is a function of the moon's velocity, not size.
– user535733
2 hours ago
There are actually a lot of questions like this on worldbuilding SE, just use the search bar (at the top) and look for "moon size" etc.
– MParm
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I would like to give my imaginary earth-analogue planet multiple (natural) moons.
Is the orbital period of such moons at all related to their size (or apparent size)?
In other words, am I free to have a large close-by moon, or a small far-away moon, or anything in between, without it having an effect on orbital period?
moons
asked 2 hours ago
Jack
1113
I would like to give my imaginary earth-analogue planet multiple (natural) moons.
Is the orbital period of such moons at all related to their size (or apparent size)?
In other words, am I free to have a large close-by moon, or a small far-away moon, or anything in between, without it having an effect on orbital period?
moons
moons
asked 2 hours ago
Jack
1113
asked 2 hours ago
Jack
1113
asked 2 hours ago
Jack
1113
asked 2 hours ago
Jack
1113
asked 2 hours ago
Jack
1113
1113
See en.wikipedia.org/wiki/Orbital_speed . In most common cases, orbital period is a function of the moon's velocity, not size.
– user535733
2 hours ago
There are actually a lot of questions like this on worldbuilding SE, just use the search bar (at the top) and look for "moon size" etc.
– MParm
1 hour ago
add a comment |Â
See en.wikipedia.org/wiki/Orbital_speed . In most common cases, orbital period is a function of the moon's velocity, not size.
– user535733
2 hours ago
There are actually a lot of questions like this on worldbuilding SE, just use the search bar (at the top) and look for "moon size" etc.
– MParm
1 hour ago
See en.wikipedia.org/wiki/Orbital_speed . In most common cases, orbital period is a function of the moon's velocity, not size.
– user535733
2 hours ago
See en.wikipedia.org/wiki/Orbital_speed . In most common cases, orbital period is a function of the moon's velocity, not size.
– user535733
2 hours ago
There are actually a lot of questions like this on worldbuilding SE, just use the search bar (at the top) and look for "moon size" etc.
– MParm
1 hour ago
There are actually a lot of questions like this on worldbuilding SE, just use the search bar (at the top) and look for "moon size" etc.
– MParm
1 hour ago
add a comment |Â
1 Answer
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Yes and no.
There are two factors that determine an object's visible size from the planet: its physical size and its orbital distance.
Physical size is essentially independent; it can be virtually anything you like. The main thing that matters in orbital dynamics is mass, and as long as it stays low compared to the planet (I've heard 1/50th or 1/100th of the planet or less) it will have negligible effect on the orbit. Depending on the body's composition and structure, it can have varying densities, so the same mass can be more or less compact as you desire. Compare for instance Mars's two moons, which are (relatively) lightweight carbonaceous asteroids, with Earth's own moon, which is composed of heavier rock and is about twice as dense.
The other factor is orbital distance, and this plays directly into the orbital period. The further out the orbiting body is, the longer it will take to complete its orbit. You can see the equations governing this behavior here, but the gist is that the satellite's orbital period is proportionate to the one-and-a-halfth power of the orbital distance. Meanwhile the apparent diameter falls off linearly with distance.
answered 1 hour ago
Cadence
10.2k51841
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Yes and no.
There are two factors that determine an object's visible size from the planet: its physical size and its orbital distance.
Physical size is essentially independent; it can be virtually anything you like. The main thing that matters in orbital dynamics is mass, and as long as it stays low compared to the planet (I've heard 1/50th or 1/100th of the planet or less) it will have negligible effect on the orbit. Depending on the body's composition and structure, it can have varying densities, so the same mass can be more or less compact as you desire. Compare for instance Mars's two moons, which are (relatively) lightweight carbonaceous asteroids, with Earth's own moon, which is composed of heavier rock and is about twice as dense.
The other factor is orbital distance, and this plays directly into the orbital period. The further out the orbiting body is, the longer it will take to complete its orbit. You can see the equations governing this behavior here, but the gist is that the satellite's orbital period is proportionate to the one-and-a-halfth power of the orbital distance. Meanwhile the apparent diameter falls off linearly with distance.
answered 1 hour ago
Cadence
10.2k51841
add a comment |Â
up vote
3
down vote
Yes and no.
There are two factors that determine an object's visible size from the planet: its physical size and its orbital distance.
Physical size is essentially independent; it can be virtually anything you like. The main thing that matters in orbital dynamics is mass, and as long as it stays low compared to the planet (I've heard 1/50th or 1/100th of the planet or less) it will have negligible effect on the orbit. Depending on the body's composition and structure, it can have varying densities, so the same mass can be more or less compact as you desire. Compare for instance Mars's two moons, which are (relatively) lightweight carbonaceous asteroids, with Earth's own moon, which is composed of heavier rock and is about twice as dense.
The other factor is orbital distance, and this plays directly into the orbital period. The further out the orbiting body is, the longer it will take to complete its orbit. You can see the equations governing this behavior here, but the gist is that the satellite's orbital period is proportionate to the one-and-a-halfth power of the orbital distance. Meanwhile the apparent diameter falls off linearly with distance.
answered 1 hour ago
Cadence
10.2k51841
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Yes and no.
There are two factors that determine an object's visible size from the planet: its physical size and its orbital distance.
Physical size is essentially independent; it can be virtually anything you like. The main thing that matters in orbital dynamics is mass, and as long as it stays low compared to the planet (I've heard 1/50th or 1/100th of the planet or less) it will have negligible effect on the orbit. Depending on the body's composition and structure, it can have varying densities, so the same mass can be more or less compact as you desire. Compare for instance Mars's two moons, which are (relatively) lightweight carbonaceous asteroids, with Earth's own moon, which is composed of heavier rock and is about twice as dense.
The other factor is orbital distance, and this plays directly into the orbital period. The further out the orbiting body is, the longer it will take to complete its orbit. You can see the equations governing this behavior here, but the gist is that the satellite's orbital period is proportionate to the one-and-a-halfth power of the orbital distance. Meanwhile the apparent diameter falls off linearly with distance.
answered 1 hour ago
Cadence
10.2k51841
Yes and no.
There are two factors that determine an object's visible size from the planet: its physical size and its orbital distance.
Physical size is essentially independent; it can be virtually anything you like. The main thing that matters in orbital dynamics is mass, and as long as it stays low compared to the planet (I've heard 1/50th or 1/100th of the planet or less) it will have negligible effect on the orbit. Depending on the body's composition and structure, it can have varying densities, so the same mass can be more or less compact as you desire. Compare for instance Mars's two moons, which are (relatively) lightweight carbonaceous asteroids, with Earth's own moon, which is composed of heavier rock and is about twice as dense.
The other factor is orbital distance, and this plays directly into the orbital period. The further out the orbiting body is, the longer it will take to complete its orbit. You can see the equations governing this behavior here, but the gist is that the satellite's orbital period is proportionate to the one-and-a-halfth power of the orbital distance. Meanwhile the apparent diameter falls off linearly with distance.
answered 1 hour ago
Cadence
10.2k51841
answered 1 hour ago
Cadence
10.2k51841
answered 1 hour ago
Cadence
10.2k51841
answered 1 hour ago
Cadence
10.2k51841
10.2k51841
add a comment |Â
add a comment |Â
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See en.wikipedia.org/wiki/Orbital_speed . In most common cases, orbital period is a function of the moon's velocity, not size.
– user535733
2 hours ago
There are actually a lot of questions like this on worldbuilding SE, just use the search bar (at the top) and look for "moon size" etc.
– MParm
1 hour ago