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Given the mass and composition of a planet, can one determine what the radius should be?
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I'm trying to auto-generate random solar systems, and I'm basically just allocating 2% of the total system mass to planets (and moons). It provides interesting results, I always have a few gas giants, I often have many Mars-and-Mercury-massed planets.
But to calculate surface gravity, I need a radius.
This largely depends on the planet's composition... which can be quite varied (and depends quite a bit on where it formed, how it formed, etc).
But if I have a planet with 0.7 Earth mass, and 35% of that is iron/nickel (or siderophile), and 50% lithophile, and so on, can a decent estimate of radius be determined?
Do I need the breakdown on composition to be per atomic element, or can this give decent ballpark numbers if I have the mass as the ratio of lithophile/siderophile/chalcophile/volatiles?
My understanding of physics in this arena is... inadequate. I do not believe it's enough to simply look up the density of these elements on Wikipedia and calculate backwards from volume. Certainly an iron core compresses a bit such that the density is quite higher than that of an iron ingot on the surface?
planets gravity
edited 3 hours ago
kingledion
66k22217376
asked 3 hours ago
John O
1965
add a comment |Â
up vote
6
down vote
favorite
I'm trying to auto-generate random solar systems, and I'm basically just allocating 2% of the total system mass to planets (and moons). It provides interesting results, I always have a few gas giants, I often have many Mars-and-Mercury-massed planets.
But to calculate surface gravity, I need a radius.
This largely depends on the planet's composition... which can be quite varied (and depends quite a bit on where it formed, how it formed, etc).
But if I have a planet with 0.7 Earth mass, and 35% of that is iron/nickel (or siderophile), and 50% lithophile, and so on, can a decent estimate of radius be determined?
Do I need the breakdown on composition to be per atomic element, or can this give decent ballpark numbers if I have the mass as the ratio of lithophile/siderophile/chalcophile/volatiles?
My understanding of physics in this arena is... inadequate. I do not believe it's enough to simply look up the density of these elements on Wikipedia and calculate backwards from volume. Certainly an iron core compresses a bit such that the density is quite higher than that of an iron ingot on the surface?
planets gravity
edited 3 hours ago
kingledion
66k22217376
asked 3 hours ago
John O
1965
You may wish to look into hydrostatic equilibrium. Disclaimer: it is not simple if you want to do this exactly.
– Joe Bloggs
1 hour ago
@JoeBloggs Ahhh! Friends don't link friends to mobile Wikipedia! If you are on a cell phone you will be auto-redirected from the normal to mobile version...
– kingledion
12 mins ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I'm trying to auto-generate random solar systems, and I'm basically just allocating 2% of the total system mass to planets (and moons). It provides interesting results, I always have a few gas giants, I often have many Mars-and-Mercury-massed planets.
But to calculate surface gravity, I need a radius.
This largely depends on the planet's composition... which can be quite varied (and depends quite a bit on where it formed, how it formed, etc).
But if I have a planet with 0.7 Earth mass, and 35% of that is iron/nickel (or siderophile), and 50% lithophile, and so on, can a decent estimate of radius be determined?
Do I need the breakdown on composition to be per atomic element, or can this give decent ballpark numbers if I have the mass as the ratio of lithophile/siderophile/chalcophile/volatiles?
My understanding of physics in this arena is... inadequate. I do not believe it's enough to simply look up the density of these elements on Wikipedia and calculate backwards from volume. Certainly an iron core compresses a bit such that the density is quite higher than that of an iron ingot on the surface?
planets gravity
edited 3 hours ago
kingledion
66k22217376
asked 3 hours ago
John O
1965
I'm trying to auto-generate random solar systems, and I'm basically just allocating 2% of the total system mass to planets (and moons). It provides interesting results, I always have a few gas giants, I often have many Mars-and-Mercury-massed planets.
But to calculate surface gravity, I need a radius.
This largely depends on the planet's composition... which can be quite varied (and depends quite a bit on where it formed, how it formed, etc).
But if I have a planet with 0.7 Earth mass, and 35% of that is iron/nickel (or siderophile), and 50% lithophile, and so on, can a decent estimate of radius be determined?
Do I need the breakdown on composition to be per atomic element, or can this give decent ballpark numbers if I have the mass as the ratio of lithophile/siderophile/chalcophile/volatiles?
My understanding of physics in this arena is... inadequate. I do not believe it's enough to simply look up the density of these elements on Wikipedia and calculate backwards from volume. Certainly an iron core compresses a bit such that the density is quite higher than that of an iron ingot on the surface?
planets gravity
planets gravity
edited 3 hours ago
kingledion
66k22217376
asked 3 hours ago
John O
1965
edited 3 hours ago
kingledion
66k22217376
asked 3 hours ago
John O
1965
edited 3 hours ago
kingledion
66k22217376
edited 3 hours ago
kingledion
66k22217376
edited 3 hours ago
kingledion
66k22217376
66k22217376
asked 3 hours ago
John O
1965
asked 3 hours ago
John O
1965
asked 3 hours ago
John O
1965
1965
You may wish to look into hydrostatic equilibrium. Disclaimer: it is not simple if you want to do this exactly.
– Joe Bloggs
1 hour ago
@JoeBloggs Ahhh! Friends don't link friends to mobile Wikipedia! If you are on a cell phone you will be auto-redirected from the normal to mobile version...
– kingledion
12 mins ago
add a comment |Â
You may wish to look into hydrostatic equilibrium. Disclaimer: it is not simple if you want to do this exactly.
– Joe Bloggs
1 hour ago
@JoeBloggs Ahhh! Friends don't link friends to mobile Wikipedia! If you are on a cell phone you will be auto-redirected from the normal to mobile version...
– kingledion
12 mins ago
You may wish to look into hydrostatic equilibrium. Disclaimer: it is not simple if you want to do this exactly.
– Joe Bloggs
1 hour ago
You may wish to look into hydrostatic equilibrium. Disclaimer: it is not simple if you want to do this exactly.
– Joe Bloggs
1 hour ago
@JoeBloggs Ahhh! Friends don't link friends to mobile Wikipedia! If you are on a cell phone you will be auto-redirected from the normal to mobile version...
– kingledion
12 mins ago
@JoeBloggs Ahhh! Friends don't link friends to mobile Wikipedia! If you are on a cell phone you will be auto-redirected from the normal to mobile version...
– kingledion
12 mins ago
add a comment |Â
1 Answer
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Mass, density and radius are related
Let $m$ be the mass of a planet.
For a given a density $rho$, the relationship between mass and volume is
$$beginalign
V &= frac43pi r^3\
m &= rho V = frac43pirho r^3
endalign$$
The bottom equation gives you your relations. Of the two variables you are interested in, mass ($m$) and radius ($r$), the solutions in terms of one another are:
$$beginalign
m&=frac43pirho r^3\
r&=sqrt[3]frac3m4pirho
endalign$$
Be careful with units! Always convert mass to kg, radius to m, and density to kg/m$^3$ to be safe.
What should density be?
Since you need density in both these equations, what are some reasonable densities for a planet?
Object/Planet Density (kg/m^3)
Earth's Inner Core 12800
Earth's Outer Core 9900
Earth 5510
Mercury 5430
Venus 5240
Mars 3930
Vesta (densest asteroid) 3420
Luna 3340
Ceres (largest asteroid) 2080
Ganymede 1940
Titan 1880
Neptune 1640
Jupiter 1330
Uranus 1270
A small chunk of water ice 934
Saturn 690
A lot goes into planetary density, and that could totally be its own question. Mass drives density; the bigger a planet the more gravity will compress it. Mercury has more iron (relatively) than Earth, but is less dense because Earth's gravity compresses its core more. But these values are some guidelines for solving the above equations.
answered 3 hours ago
kingledion
66k22217376
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Mass, density and radius are related
Let $m$ be the mass of a planet.
For a given a density $rho$, the relationship between mass and volume is
$$beginalign
V &= frac43pi r^3\
m &= rho V = frac43pirho r^3
endalign$$
The bottom equation gives you your relations. Of the two variables you are interested in, mass ($m$) and radius ($r$), the solutions in terms of one another are:
$$beginalign
m&=frac43pirho r^3\
r&=sqrt[3]frac3m4pirho
endalign$$
Be careful with units! Always convert mass to kg, radius to m, and density to kg/m$^3$ to be safe.
What should density be?
Since you need density in both these equations, what are some reasonable densities for a planet?
Object/Planet Density (kg/m^3)
Earth's Inner Core 12800
Earth's Outer Core 9900
Earth 5510
Mercury 5430
Venus 5240
Mars 3930
Vesta (densest asteroid) 3420
Luna 3340
Ceres (largest asteroid) 2080
Ganymede 1940
Titan 1880
Neptune 1640
Jupiter 1330
Uranus 1270
A small chunk of water ice 934
Saturn 690
A lot goes into planetary density, and that could totally be its own question. Mass drives density; the bigger a planet the more gravity will compress it. Mercury has more iron (relatively) than Earth, but is less dense because Earth's gravity compresses its core more. But these values are some guidelines for solving the above equations.
answered 3 hours ago
kingledion
66k22217376
add a comment |Â
up vote
6
down vote
Mass, density and radius are related
Let $m$ be the mass of a planet.
For a given a density $rho$, the relationship between mass and volume is
$$beginalign
V &= frac43pi r^3\
m &= rho V = frac43pirho r^3
endalign$$
The bottom equation gives you your relations. Of the two variables you are interested in, mass ($m$) and radius ($r$), the solutions in terms of one another are:
$$beginalign
m&=frac43pirho r^3\
r&=sqrt[3]frac3m4pirho
endalign$$
Be careful with units! Always convert mass to kg, radius to m, and density to kg/m$^3$ to be safe.
What should density be?
Since you need density in both these equations, what are some reasonable densities for a planet?
Object/Planet Density (kg/m^3)
Earth's Inner Core 12800
Earth's Outer Core 9900
Earth 5510
Mercury 5430
Venus 5240
Mars 3930
Vesta (densest asteroid) 3420
Luna 3340
Ceres (largest asteroid) 2080
Ganymede 1940
Titan 1880
Neptune 1640
Jupiter 1330
Uranus 1270
A small chunk of water ice 934
Saturn 690
A lot goes into planetary density, and that could totally be its own question. Mass drives density; the bigger a planet the more gravity will compress it. Mercury has more iron (relatively) than Earth, but is less dense because Earth's gravity compresses its core more. But these values are some guidelines for solving the above equations.
answered 3 hours ago
kingledion
66k22217376
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Mass, density and radius are related
Let $m$ be the mass of a planet.
For a given a density $rho$, the relationship between mass and volume is
$$beginalign
V &= frac43pi r^3\
m &= rho V = frac43pirho r^3
endalign$$
The bottom equation gives you your relations. Of the two variables you are interested in, mass ($m$) and radius ($r$), the solutions in terms of one another are:
$$beginalign
m&=frac43pirho r^3\
r&=sqrt[3]frac3m4pirho
endalign$$
Be careful with units! Always convert mass to kg, radius to m, and density to kg/m$^3$ to be safe.
What should density be?
Since you need density in both these equations, what are some reasonable densities for a planet?
Object/Planet Density (kg/m^3)
Earth's Inner Core 12800
Earth's Outer Core 9900
Earth 5510
Mercury 5430
Venus 5240
Mars 3930
Vesta (densest asteroid) 3420
Luna 3340
Ceres (largest asteroid) 2080
Ganymede 1940
Titan 1880
Neptune 1640
Jupiter 1330
Uranus 1270
A small chunk of water ice 934
Saturn 690
A lot goes into planetary density, and that could totally be its own question. Mass drives density; the bigger a planet the more gravity will compress it. Mercury has more iron (relatively) than Earth, but is less dense because Earth's gravity compresses its core more. But these values are some guidelines for solving the above equations.
answered 3 hours ago
kingledion
66k22217376
Mass, density and radius are related
Let $m$ be the mass of a planet.
For a given a density $rho$, the relationship between mass and volume is
$$beginalign
V &= frac43pi r^3\
m &= rho V = frac43pirho r^3
endalign$$
The bottom equation gives you your relations. Of the two variables you are interested in, mass ($m$) and radius ($r$), the solutions in terms of one another are:
$$beginalign
m&=frac43pirho r^3\
r&=sqrt[3]frac3m4pirho
endalign$$
Be careful with units! Always convert mass to kg, radius to m, and density to kg/m$^3$ to be safe.
What should density be?
Since you need density in both these equations, what are some reasonable densities for a planet?
Object/Planet Density (kg/m^3)
Earth's Inner Core 12800
Earth's Outer Core 9900
Earth 5510
Mercury 5430
Venus 5240
Mars 3930
Vesta (densest asteroid) 3420
Luna 3340
Ceres (largest asteroid) 2080
Ganymede 1940
Titan 1880
Neptune 1640
Jupiter 1330
Uranus 1270
A small chunk of water ice 934
Saturn 690
A lot goes into planetary density, and that could totally be its own question. Mass drives density; the bigger a planet the more gravity will compress it. Mercury has more iron (relatively) than Earth, but is less dense because Earth's gravity compresses its core more. But these values are some guidelines for solving the above equations.
answered 3 hours ago
kingledion
66k22217376
edited 2 hours ago
answered 3 hours ago
kingledion
66k22217376
answered 3 hours ago
kingledion
66k22217376
answered 3 hours ago
kingledion
66k22217376
66k22217376
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You may wish to look into hydrostatic equilibrium. Disclaimer: it is not simple if you want to do this exactly.
– Joe Bloggs
1 hour ago
@JoeBloggs Ahhh! Friends don't link friends to mobile Wikipedia! If you are on a cell phone you will be auto-redirected from the normal to mobile version...
– kingledion
12 mins ago