How do I calculate the amount of sunlight a planet gets?
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I'm working on a planet and I want to know what is the maximum amount of sunlight my planet gets in lux (for Earth It's 120000 lux).
My planet orbits an M4V class red dwarf that has a mass of 0.22 M☉, radius of 0.16 R☉ and a temperature of 3000 K. The planet's semi-major axis is 2.3 AU.
Please also include the formula or program you used to calculate the answer. If you need any additional info I'll add it.
science-based planets sunlight
asked Aug 19 at 16:30
i_am_a_smart
407
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up vote
2
down vote
favorite
I'm working on a planet and I want to know what is the maximum amount of sunlight my planet gets in lux (for Earth It's 120000 lux).
My planet orbits an M4V class red dwarf that has a mass of 0.22 M☉, radius of 0.16 R☉ and a temperature of 3000 K. The planet's semi-major axis is 2.3 AU.
Please also include the formula or program you used to calculate the answer. If you need any additional info I'll add it.
science-based planets sunlight
asked Aug 19 at 16:30
i_am_a_smart
407
2
Your requirement of lux makes this complicated. Calculating radiative flux is simple math, but luminous flux depends on the interplay of the stars spectrum and human eye.
– b.Lorenz
Aug 19 at 16:56
Yeah what b.Lorenz said, I can give you the stellar luminosity very easily but the next bit is a lot more problematic.
– Ash
Aug 19 at 17:01
Also, because you have a planet orbiting a red dwarf (smaller than sun) further than the distance that mars is from, it is likely (read: certain) that the planet is too cold for liquid water to form on. So, if you want life, move it a lot closer (my estimate is about 10-20 times closer)
– JavaScriptCoder
Aug 19 at 21:17
Based on this question and its answers I think it should be moved to Physics SE.
– person27
Aug 19 at 22:58
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm working on a planet and I want to know what is the maximum amount of sunlight my planet gets in lux (for Earth It's 120000 lux).
My planet orbits an M4V class red dwarf that has a mass of 0.22 M☉, radius of 0.16 R☉ and a temperature of 3000 K. The planet's semi-major axis is 2.3 AU.
Please also include the formula or program you used to calculate the answer. If you need any additional info I'll add it.
science-based planets sunlight
asked Aug 19 at 16:30
i_am_a_smart
407
I'm working on a planet and I want to know what is the maximum amount of sunlight my planet gets in lux (for Earth It's 120000 lux).
My planet orbits an M4V class red dwarf that has a mass of 0.22 M☉, radius of 0.16 R☉ and a temperature of 3000 K. The planet's semi-major axis is 2.3 AU.
Please also include the formula or program you used to calculate the answer. If you need any additional info I'll add it.
science-based planets sunlight
science-based planets sunlight
asked Aug 19 at 16:30
i_am_a_smart
407
asked Aug 19 at 16:30
i_am_a_smart
407
asked Aug 19 at 16:30
i_am_a_smart
407
asked Aug 19 at 16:30
i_am_a_smart
407
asked Aug 19 at 16:30
i_am_a_smart
407
407
2
Your requirement of lux makes this complicated. Calculating radiative flux is simple math, but luminous flux depends on the interplay of the stars spectrum and human eye.
– b.Lorenz
Aug 19 at 16:56
Yeah what b.Lorenz said, I can give you the stellar luminosity very easily but the next bit is a lot more problematic.
– Ash
Aug 19 at 17:01
Also, because you have a planet orbiting a red dwarf (smaller than sun) further than the distance that mars is from, it is likely (read: certain) that the planet is too cold for liquid water to form on. So, if you want life, move it a lot closer (my estimate is about 10-20 times closer)
– JavaScriptCoder
Aug 19 at 21:17
Based on this question and its answers I think it should be moved to Physics SE.
– person27
Aug 19 at 22:58
add a comment |Â
2
Your requirement of lux makes this complicated. Calculating radiative flux is simple math, but luminous flux depends on the interplay of the stars spectrum and human eye.
– b.Lorenz
Aug 19 at 16:56
Yeah what b.Lorenz said, I can give you the stellar luminosity very easily but the next bit is a lot more problematic.
– Ash
Aug 19 at 17:01
Also, because you have a planet orbiting a red dwarf (smaller than sun) further than the distance that mars is from, it is likely (read: certain) that the planet is too cold for liquid water to form on. So, if you want life, move it a lot closer (my estimate is about 10-20 times closer)
– JavaScriptCoder
Aug 19 at 21:17
Based on this question and its answers I think it should be moved to Physics SE.
– person27
Aug 19 at 22:58
2
2
Your requirement of lux makes this complicated. Calculating radiative flux is simple math, but luminous flux depends on the interplay of the stars spectrum and human eye.
– b.Lorenz
Aug 19 at 16:56
Your requirement of lux makes this complicated. Calculating radiative flux is simple math, but luminous flux depends on the interplay of the stars spectrum and human eye.
– b.Lorenz
Aug 19 at 16:56
Yeah what b.Lorenz said, I can give you the stellar luminosity very easily but the next bit is a lot more problematic.
– Ash
Aug 19 at 17:01
Yeah what b.Lorenz said, I can give you the stellar luminosity very easily but the next bit is a lot more problematic.
– Ash
Aug 19 at 17:01
Also, because you have a planet orbiting a red dwarf (smaller than sun) further than the distance that mars is from, it is likely (read: certain) that the planet is too cold for liquid water to form on. So, if you want life, move it a lot closer (my estimate is about 10-20 times closer)
– JavaScriptCoder
Aug 19 at 21:17
Also, because you have a planet orbiting a red dwarf (smaller than sun) further than the distance that mars is from, it is likely (read: certain) that the planet is too cold for liquid water to form on. So, if you want life, move it a lot closer (my estimate is about 10-20 times closer)
– JavaScriptCoder
Aug 19 at 21:17
Based on this question and its answers I think it should be moved to Physics SE.
– person27
Aug 19 at 22:58
Based on this question and its answers I think it should be moved to Physics SE.
– person27
Aug 19 at 22:58
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
Calculating flux
I think a slightly more helpful quantity to calculate is the flux received by the planet - the power per unit area from the star. The mean flux on Earth is the solar constant, $F_e=1.36times10^3text W m^-2$. If a planet orbits a star of luminosity $L_*$ at a distance $r_p$, the flux received is
$$F_p=fracL_*4pi r_p^2$$
The ratio of the planet's flux $F_p$ to the solar constant $F_e$ is
$$fracF_pF_e=fracL_*L_odotleft(fracr_p1text AUright)^-2$$
For an M4V dwarf, I'd expect $L_*approx0.006L_odot$. Plugging this in, we get
$$F_p=1.1times10^-3F_e=1.50text W m^-2$$
for an M4V star.
Why should we use flux instead of lux?
- It's easier to calculate. And when I say easier, I mean much easier. Lux is a lot more complicated.
- It's great for calculating things like the effective temperature of a planet.
- It's much more commonly used in this sort of scenario.
I used stellar models by Eric Mamajek to find the star's luminosity - he gives a $log L_*/L_odot=-2.2$, so $L_*approx0.006L_odot$. It's worth noting that those models list an M4V star as having $T_effapprox3200text K$ and $Rapprox0.258R_odot$ - so this is based on those figures. Using your numbers (which are for a smaller, cooler star) and the Stefan-Boltzmann law, I get $L_*approx0.00186L_odot$, leading to a flux of $F_papprox0.48text W m^-2$. I believe your values are slightly off for the given spectral type.
I've written a Python program to calculate the flux received on a planet based on a given spectral type, using those models. It should be a quick shortcut for you in the future.
Astronomical lux
It's claimed that you can convert between a star's apparent magnitude in the V-band ($m_V$) and its illuminance ($I_V$) - the value I think you're looking for. Wikipedia's formula is
$$I_V=10^(-14.18-m_V)/2.5$$
which matches values from the National Park Service. The absolute magnitude of an M4V star is - according to the same stellar models - $M_Vapprox12.80$. Apparent magnitude can be calculated from absolute magnitude by
$$m_V=M_V+5logleft(fracr_p10text parsecsright)$$
Putting this together for our case yields $m_Vapprox-16.96$, and, finally, we get $I_Vapprox12.7text lux$.
answered Aug 19 at 17:01
HDE 226868♦
60.9k12214392
Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway.
– Ash
Aug 19 at 17:09
@Ash I tend to use a specific grid of stellar models - I assume various ones differ by a factor of maybe two or three. An M5V dwarf is listed as being at $approx0.003L_*$; the value is sensitive to the exact choice of spectral type.
– HDE 226868♦
Aug 19 at 17:11
Yeah based on that data set it looks like the given spectral type and the specifications are incompatible.
– Ash
Aug 19 at 17:13
@Ash Ah, fair point. I've added in additional numbers based on the OP's specifications.
– HDE 226868♦
Aug 19 at 17:18
You are right. Lux is only useful to determine how bright the sky would seem to human eyes. Any other derivative effect - solar cell power, plant life, weather... depends on radiative flux.
– b.Lorenz
Aug 19 at 18:05
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Calculating flux
I think a slightly more helpful quantity to calculate is the flux received by the planet - the power per unit area from the star. The mean flux on Earth is the solar constant, $F_e=1.36times10^3text W m^-2$. If a planet orbits a star of luminosity $L_*$ at a distance $r_p$, the flux received is
$$F_p=fracL_*4pi r_p^2$$
The ratio of the planet's flux $F_p$ to the solar constant $F_e$ is
$$fracF_pF_e=fracL_*L_odotleft(fracr_p1text AUright)^-2$$
For an M4V dwarf, I'd expect $L_*approx0.006L_odot$. Plugging this in, we get
$$F_p=1.1times10^-3F_e=1.50text W m^-2$$
for an M4V star.
Why should we use flux instead of lux?
- It's easier to calculate. And when I say easier, I mean much easier. Lux is a lot more complicated.
- It's great for calculating things like the effective temperature of a planet.
- It's much more commonly used in this sort of scenario.
I used stellar models by Eric Mamajek to find the star's luminosity - he gives a $log L_*/L_odot=-2.2$, so $L_*approx0.006L_odot$. It's worth noting that those models list an M4V star as having $T_effapprox3200text K$ and $Rapprox0.258R_odot$ - so this is based on those figures. Using your numbers (which are for a smaller, cooler star) and the Stefan-Boltzmann law, I get $L_*approx0.00186L_odot$, leading to a flux of $F_papprox0.48text W m^-2$. I believe your values are slightly off for the given spectral type.
I've written a Python program to calculate the flux received on a planet based on a given spectral type, using those models. It should be a quick shortcut for you in the future.
Astronomical lux
It's claimed that you can convert between a star's apparent magnitude in the V-band ($m_V$) and its illuminance ($I_V$) - the value I think you're looking for. Wikipedia's formula is
$$I_V=10^(-14.18-m_V)/2.5$$
which matches values from the National Park Service. The absolute magnitude of an M4V star is - according to the same stellar models - $M_Vapprox12.80$. Apparent magnitude can be calculated from absolute magnitude by
$$m_V=M_V+5logleft(fracr_p10text parsecsright)$$
Putting this together for our case yields $m_Vapprox-16.96$, and, finally, we get $I_Vapprox12.7text lux$.
answered Aug 19 at 17:01
HDE 226868♦
60.9k12214392
Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway.
– Ash
Aug 19 at 17:09
@Ash I tend to use a specific grid of stellar models - I assume various ones differ by a factor of maybe two or three. An M5V dwarf is listed as being at $approx0.003L_*$; the value is sensitive to the exact choice of spectral type.
– HDE 226868♦
Aug 19 at 17:11
Yeah based on that data set it looks like the given spectral type and the specifications are incompatible.
– Ash
Aug 19 at 17:13
@Ash Ah, fair point. I've added in additional numbers based on the OP's specifications.
– HDE 226868♦
Aug 19 at 17:18
You are right. Lux is only useful to determine how bright the sky would seem to human eyes. Any other derivative effect - solar cell power, plant life, weather... depends on radiative flux.
– b.Lorenz
Aug 19 at 18:05
 |Â
show 2 more comments
up vote
8
down vote
accepted
Calculating flux
I think a slightly more helpful quantity to calculate is the flux received by the planet - the power per unit area from the star. The mean flux on Earth is the solar constant, $F_e=1.36times10^3text W m^-2$. If a planet orbits a star of luminosity $L_*$ at a distance $r_p$, the flux received is
$$F_p=fracL_*4pi r_p^2$$
The ratio of the planet's flux $F_p$ to the solar constant $F_e$ is
$$fracF_pF_e=fracL_*L_odotleft(fracr_p1text AUright)^-2$$
For an M4V dwarf, I'd expect $L_*approx0.006L_odot$. Plugging this in, we get
$$F_p=1.1times10^-3F_e=1.50text W m^-2$$
for an M4V star.
Why should we use flux instead of lux?
- It's easier to calculate. And when I say easier, I mean much easier. Lux is a lot more complicated.
- It's great for calculating things like the effective temperature of a planet.
- It's much more commonly used in this sort of scenario.
I used stellar models by Eric Mamajek to find the star's luminosity - he gives a $log L_*/L_odot=-2.2$, so $L_*approx0.006L_odot$. It's worth noting that those models list an M4V star as having $T_effapprox3200text K$ and $Rapprox0.258R_odot$ - so this is based on those figures. Using your numbers (which are for a smaller, cooler star) and the Stefan-Boltzmann law, I get $L_*approx0.00186L_odot$, leading to a flux of $F_papprox0.48text W m^-2$. I believe your values are slightly off for the given spectral type.
I've written a Python program to calculate the flux received on a planet based on a given spectral type, using those models. It should be a quick shortcut for you in the future.
Astronomical lux
It's claimed that you can convert between a star's apparent magnitude in the V-band ($m_V$) and its illuminance ($I_V$) - the value I think you're looking for. Wikipedia's formula is
$$I_V=10^(-14.18-m_V)/2.5$$
which matches values from the National Park Service. The absolute magnitude of an M4V star is - according to the same stellar models - $M_Vapprox12.80$. Apparent magnitude can be calculated from absolute magnitude by
$$m_V=M_V+5logleft(fracr_p10text parsecsright)$$
Putting this together for our case yields $m_Vapprox-16.96$, and, finally, we get $I_Vapprox12.7text lux$.
answered Aug 19 at 17:01
HDE 226868♦
60.9k12214392
Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway.
– Ash
Aug 19 at 17:09
@Ash I tend to use a specific grid of stellar models - I assume various ones differ by a factor of maybe two or three. An M5V dwarf is listed as being at $approx0.003L_*$; the value is sensitive to the exact choice of spectral type.
– HDE 226868♦
Aug 19 at 17:11
Yeah based on that data set it looks like the given spectral type and the specifications are incompatible.
– Ash
Aug 19 at 17:13
@Ash Ah, fair point. I've added in additional numbers based on the OP's specifications.
– HDE 226868♦
Aug 19 at 17:18
You are right. Lux is only useful to determine how bright the sky would seem to human eyes. Any other derivative effect - solar cell power, plant life, weather... depends on radiative flux.
– b.Lorenz
Aug 19 at 18:05
 |Â
show 2 more comments
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Calculating flux
I think a slightly more helpful quantity to calculate is the flux received by the planet - the power per unit area from the star. The mean flux on Earth is the solar constant, $F_e=1.36times10^3text W m^-2$. If a planet orbits a star of luminosity $L_*$ at a distance $r_p$, the flux received is
$$F_p=fracL_*4pi r_p^2$$
The ratio of the planet's flux $F_p$ to the solar constant $F_e$ is
$$fracF_pF_e=fracL_*L_odotleft(fracr_p1text AUright)^-2$$
For an M4V dwarf, I'd expect $L_*approx0.006L_odot$. Plugging this in, we get
$$F_p=1.1times10^-3F_e=1.50text W m^-2$$
for an M4V star.
Why should we use flux instead of lux?
- It's easier to calculate. And when I say easier, I mean much easier. Lux is a lot more complicated.
- It's great for calculating things like the effective temperature of a planet.
- It's much more commonly used in this sort of scenario.
I used stellar models by Eric Mamajek to find the star's luminosity - he gives a $log L_*/L_odot=-2.2$, so $L_*approx0.006L_odot$. It's worth noting that those models list an M4V star as having $T_effapprox3200text K$ and $Rapprox0.258R_odot$ - so this is based on those figures. Using your numbers (which are for a smaller, cooler star) and the Stefan-Boltzmann law, I get $L_*approx0.00186L_odot$, leading to a flux of $F_papprox0.48text W m^-2$. I believe your values are slightly off for the given spectral type.
I've written a Python program to calculate the flux received on a planet based on a given spectral type, using those models. It should be a quick shortcut for you in the future.
Astronomical lux
It's claimed that you can convert between a star's apparent magnitude in the V-band ($m_V$) and its illuminance ($I_V$) - the value I think you're looking for. Wikipedia's formula is
$$I_V=10^(-14.18-m_V)/2.5$$
which matches values from the National Park Service. The absolute magnitude of an M4V star is - according to the same stellar models - $M_Vapprox12.80$. Apparent magnitude can be calculated from absolute magnitude by
$$m_V=M_V+5logleft(fracr_p10text parsecsright)$$
Putting this together for our case yields $m_Vapprox-16.96$, and, finally, we get $I_Vapprox12.7text lux$.
answered Aug 19 at 17:01
HDE 226868♦
60.9k12214392
Calculating flux
I think a slightly more helpful quantity to calculate is the flux received by the planet - the power per unit area from the star. The mean flux on Earth is the solar constant, $F_e=1.36times10^3text W m^-2$. If a planet orbits a star of luminosity $L_*$ at a distance $r_p$, the flux received is
$$F_p=fracL_*4pi r_p^2$$
The ratio of the planet's flux $F_p$ to the solar constant $F_e$ is
$$fracF_pF_e=fracL_*L_odotleft(fracr_p1text AUright)^-2$$
For an M4V dwarf, I'd expect $L_*approx0.006L_odot$. Plugging this in, we get
$$F_p=1.1times10^-3F_e=1.50text W m^-2$$
for an M4V star.
Why should we use flux instead of lux?
- It's easier to calculate. And when I say easier, I mean much easier. Lux is a lot more complicated.
- It's great for calculating things like the effective temperature of a planet.
- It's much more commonly used in this sort of scenario.
I used stellar models by Eric Mamajek to find the star's luminosity - he gives a $log L_*/L_odot=-2.2$, so $L_*approx0.006L_odot$. It's worth noting that those models list an M4V star as having $T_effapprox3200text K$ and $Rapprox0.258R_odot$ - so this is based on those figures. Using your numbers (which are for a smaller, cooler star) and the Stefan-Boltzmann law, I get $L_*approx0.00186L_odot$, leading to a flux of $F_papprox0.48text W m^-2$. I believe your values are slightly off for the given spectral type.
I've written a Python program to calculate the flux received on a planet based on a given spectral type, using those models. It should be a quick shortcut for you in the future.
Astronomical lux
It's claimed that you can convert between a star's apparent magnitude in the V-band ($m_V$) and its illuminance ($I_V$) - the value I think you're looking for. Wikipedia's formula is
$$I_V=10^(-14.18-m_V)/2.5$$
which matches values from the National Park Service. The absolute magnitude of an M4V star is - according to the same stellar models - $M_Vapprox12.80$. Apparent magnitude can be calculated from absolute magnitude by
$$m_V=M_V+5logleft(fracr_p10text parsecsright)$$
Putting this together for our case yields $m_Vapprox-16.96$, and, finally, we get $I_Vapprox12.7text lux$.
answered Aug 19 at 17:01
HDE 226868♦
60.9k12214392
edited Aug 19 at 21:09
answered Aug 19 at 17:01
HDE 226868♦
60.9k12214392
answered Aug 19 at 17:01
HDE 226868♦
60.9k12214392
answered Aug 19 at 17:01
HDE 226868♦
60.9k12214392
60.9k12214392
Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway.
– Ash
Aug 19 at 17:09
@Ash I tend to use a specific grid of stellar models - I assume various ones differ by a factor of maybe two or three. An M5V dwarf is listed as being at $approx0.003L_*$; the value is sensitive to the exact choice of spectral type.
– HDE 226868♦
Aug 19 at 17:11
Yeah based on that data set it looks like the given spectral type and the specifications are incompatible.
– Ash
Aug 19 at 17:13
@Ash Ah, fair point. I've added in additional numbers based on the OP's specifications.
– HDE 226868♦
Aug 19 at 17:18
You are right. Lux is only useful to determine how bright the sky would seem to human eyes. Any other derivative effect - solar cell power, plant life, weather... depends on radiative flux.
– b.Lorenz
Aug 19 at 18:05
 |Â
show 2 more comments
Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway.
– Ash
Aug 19 at 17:09
@Ash I tend to use a specific grid of stellar models - I assume various ones differ by a factor of maybe two or three. An M5V dwarf is listed as being at $approx0.003L_*$; the value is sensitive to the exact choice of spectral type.
– HDE 226868♦
Aug 19 at 17:11
Yeah based on that data set it looks like the given spectral type and the specifications are incompatible.
– Ash
Aug 19 at 17:13
@Ash Ah, fair point. I've added in additional numbers based on the OP's specifications.
– HDE 226868♦
Aug 19 at 17:18
You are right. Lux is only useful to determine how bright the sky would seem to human eyes. Any other derivative effect - solar cell power, plant life, weather... depends on radiative flux.
– b.Lorenz
Aug 19 at 18:05
Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway.
– Ash
Aug 19 at 17:09
Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway.
– Ash
Aug 19 at 17:09
@Ash I tend to use a specific grid of stellar models - I assume various ones differ by a factor of maybe two or three. An M5V dwarf is listed as being at $approx0.003L_*$; the value is sensitive to the exact choice of spectral type.
– HDE 226868♦
Aug 19 at 17:11
@Ash I tend to use a specific grid of stellar models - I assume various ones differ by a factor of maybe two or three. An M5V dwarf is listed as being at $approx0.003L_*$; the value is sensitive to the exact choice of spectral type.
– HDE 226868♦
Aug 19 at 17:11
Yeah based on that data set it looks like the given spectral type and the specifications are incompatible.
– Ash
Aug 19 at 17:13
Yeah based on that data set it looks like the given spectral type and the specifications are incompatible.
– Ash
Aug 19 at 17:13
@Ash Ah, fair point. I've added in additional numbers based on the OP's specifications.
– HDE 226868♦
Aug 19 at 17:18
@Ash Ah, fair point. I've added in additional numbers based on the OP's specifications.
– HDE 226868♦
Aug 19 at 17:18
You are right. Lux is only useful to determine how bright the sky would seem to human eyes. Any other derivative effect - solar cell power, plant life, weather... depends on radiative flux.
– b.Lorenz
Aug 19 at 18:05
You are right. Lux is only useful to determine how bright the sky would seem to human eyes. Any other derivative effect - solar cell power, plant life, weather... depends on radiative flux.
– b.Lorenz
Aug 19 at 18:05
 |Â
show 2 more comments
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2
Your requirement of lux makes this complicated. Calculating radiative flux is simple math, but luminous flux depends on the interplay of the stars spectrum and human eye.
– b.Lorenz
Aug 19 at 16:56
Yeah what b.Lorenz said, I can give you the stellar luminosity very easily but the next bit is a lot more problematic.
– Ash
Aug 19 at 17:01
Also, because you have a planet orbiting a red dwarf (smaller than sun) further than the distance that mars is from, it is likely (read: certain) that the planet is too cold for liquid water to form on. So, if you want life, move it a lot closer (my estimate is about 10-20 times closer)
– JavaScriptCoder
Aug 19 at 21:17
Based on this question and its answers I think it should be moved to Physics SE.
– person27
Aug 19 at 22:58