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Is the L2 Mars-Sun point protected from solar radiation?
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To what extent is the L2 point protected in comparison to the surface, if at all? Since the L2 point might be in the umbra of Mars, it could be shielded against the sun. It would also be helpful if someone could provide the calculation.
lagrangian-points
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Talr
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To what extent is the L2 point protected in comparison to the surface, if at all? Since the L2 point might be in the umbra of Mars, it could be shielded against the sun. It would also be helpful if someone could provide the calculation.
lagrangian-points
asked 3 hours ago
Talr
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Id assume since mars has a low magnetosphere the only reduction in radation youll recieve is eclipses of mars and the sun.
– Magic Octopus Urn
2 hours ago
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up vote
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down vote
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To what extent is the L2 point protected in comparison to the surface, if at all? Since the L2 point might be in the umbra of Mars, it could be shielded against the sun. It would also be helpful if someone could provide the calculation.
lagrangian-points
asked 3 hours ago
Talr
213
New contributor
Talr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
To what extent is the L2 point protected in comparison to the surface, if at all? Since the L2 point might be in the umbra of Mars, it could be shielded against the sun. It would also be helpful if someone could provide the calculation.
lagrangian-points
lagrangian-points
asked 3 hours ago
Talr
213
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asked 3 hours ago
Talr
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Talr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 hours ago
asked 3 hours ago
Talr
213
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Talr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 3 hours ago
Talr
213
asked 3 hours ago
Talr
213
213
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Talr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Talr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Talr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Id assume since mars has a low magnetosphere the only reduction in radation youll recieve is eclipses of mars and the sun.
– Magic Octopus Urn
2 hours ago
add a comment |Â
Id assume since mars has a low magnetosphere the only reduction in radation youll recieve is eclipses of mars and the sun.
– Magic Octopus Urn
2 hours ago
Id assume since mars has a low magnetosphere the only reduction in radation youll recieve is eclipses of mars and the sun.
– Magic Octopus Urn
2 hours ago
Id assume since mars has a low magnetosphere the only reduction in radation youll recieve is eclipses of mars and the sun.
– Magic Octopus Urn
2 hours ago
add a comment |Â
1 Answer
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The Lagrange points of Mars might be less useful than that of Earth's because Mars' orbit is more elliptical and so there are more perturbing forces.
Nonetheless we can do a simple exercise.
In this answer I show how to calculate the theoretical distances to the first two Lagrange points for a circular restricted three body problem scenario.
From geometry we can see that if a right circular cone with base radius = $r_sun$ has a radius $r$ at a distance $D$, then the vertex will be beyond that point by a distance
$$ Dfracrr_sun-r$$
Using radii for Mars, Earth and the Sun of 3396, 6378, and 696392 km, I get the following, which shows that whether or not the Lagrange points are meaningful and/or useful for an elliptical orbit like Mars, Mars' solar umbra is beyond Sun-Mars L2 by about 100 km, while for Earth's solar umbra seems to be 100 km before (short-of) Earth-Mars L2.
MARS to Sun to L1 to L2 r r/(r_sun-r) "end of umbra"
perihelion 206700 981 985 3396 0.00490 1013
aphelion 249200 1183 1187 " " 1221
Earth
perihelion 147100 1466 1476 6370 0.00926 1360
aphelion 152100 1516 1527 " " 1406
Staying within the umbra near the "L2" point would likely require less delta-v for Mars than for Earth, but either way it's an unstable point and so you need to continually do small station-keeping burns to remain protected.
The umbra protects you from line-of sight radiation like photons (light, X-rays, gamma-rays) but may be less protective for ballistic charged particles from a [CME][3] because they may be on trajectories that when projected backwards, come from points beyond the disk of the Sun for various reasons. That's an excellent *next question!
For the "normal" solar wind, it may be even more likely to expand into the shadow since wether it's ionized or not, it will have some transverse velocity distribution associated with it.
answered 34 mins ago
uhoh
29.1k1598358
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The Lagrange points of Mars might be less useful than that of Earth's because Mars' orbit is more elliptical and so there are more perturbing forces.
Nonetheless we can do a simple exercise.
In this answer I show how to calculate the theoretical distances to the first two Lagrange points for a circular restricted three body problem scenario.
From geometry we can see that if a right circular cone with base radius = $r_sun$ has a radius $r$ at a distance $D$, then the vertex will be beyond that point by a distance
$$ Dfracrr_sun-r$$
Using radii for Mars, Earth and the Sun of 3396, 6378, and 696392 km, I get the following, which shows that whether or not the Lagrange points are meaningful and/or useful for an elliptical orbit like Mars, Mars' solar umbra is beyond Sun-Mars L2 by about 100 km, while for Earth's solar umbra seems to be 100 km before (short-of) Earth-Mars L2.
MARS to Sun to L1 to L2 r r/(r_sun-r) "end of umbra"
perihelion 206700 981 985 3396 0.00490 1013
aphelion 249200 1183 1187 " " 1221
Earth
perihelion 147100 1466 1476 6370 0.00926 1360
aphelion 152100 1516 1527 " " 1406
Staying within the umbra near the "L2" point would likely require less delta-v for Mars than for Earth, but either way it's an unstable point and so you need to continually do small station-keeping burns to remain protected.
The umbra protects you from line-of sight radiation like photons (light, X-rays, gamma-rays) but may be less protective for ballistic charged particles from a [CME][3] because they may be on trajectories that when projected backwards, come from points beyond the disk of the Sun for various reasons. That's an excellent *next question!
For the "normal" solar wind, it may be even more likely to expand into the shadow since wether it's ionized or not, it will have some transverse velocity distribution associated with it.
answered 34 mins ago
uhoh
29.1k1598358
add a comment |Â
up vote
1
down vote
The Lagrange points of Mars might be less useful than that of Earth's because Mars' orbit is more elliptical and so there are more perturbing forces.
Nonetheless we can do a simple exercise.
In this answer I show how to calculate the theoretical distances to the first two Lagrange points for a circular restricted three body problem scenario.
From geometry we can see that if a right circular cone with base radius = $r_sun$ has a radius $r$ at a distance $D$, then the vertex will be beyond that point by a distance
$$ Dfracrr_sun-r$$
Using radii for Mars, Earth and the Sun of 3396, 6378, and 696392 km, I get the following, which shows that whether or not the Lagrange points are meaningful and/or useful for an elliptical orbit like Mars, Mars' solar umbra is beyond Sun-Mars L2 by about 100 km, while for Earth's solar umbra seems to be 100 km before (short-of) Earth-Mars L2.
MARS to Sun to L1 to L2 r r/(r_sun-r) "end of umbra"
perihelion 206700 981 985 3396 0.00490 1013
aphelion 249200 1183 1187 " " 1221
Earth
perihelion 147100 1466 1476 6370 0.00926 1360
aphelion 152100 1516 1527 " " 1406
Staying within the umbra near the "L2" point would likely require less delta-v for Mars than for Earth, but either way it's an unstable point and so you need to continually do small station-keeping burns to remain protected.
The umbra protects you from line-of sight radiation like photons (light, X-rays, gamma-rays) but may be less protective for ballistic charged particles from a [CME][3] because they may be on trajectories that when projected backwards, come from points beyond the disk of the Sun for various reasons. That's an excellent *next question!
For the "normal" solar wind, it may be even more likely to expand into the shadow since wether it's ionized or not, it will have some transverse velocity distribution associated with it.
answered 34 mins ago
uhoh
29.1k1598358
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The Lagrange points of Mars might be less useful than that of Earth's because Mars' orbit is more elliptical and so there are more perturbing forces.
Nonetheless we can do a simple exercise.
In this answer I show how to calculate the theoretical distances to the first two Lagrange points for a circular restricted three body problem scenario.
From geometry we can see that if a right circular cone with base radius = $r_sun$ has a radius $r$ at a distance $D$, then the vertex will be beyond that point by a distance
$$ Dfracrr_sun-r$$
Using radii for Mars, Earth and the Sun of 3396, 6378, and 696392 km, I get the following, which shows that whether or not the Lagrange points are meaningful and/or useful for an elliptical orbit like Mars, Mars' solar umbra is beyond Sun-Mars L2 by about 100 km, while for Earth's solar umbra seems to be 100 km before (short-of) Earth-Mars L2.
MARS to Sun to L1 to L2 r r/(r_sun-r) "end of umbra"
perihelion 206700 981 985 3396 0.00490 1013
aphelion 249200 1183 1187 " " 1221
Earth
perihelion 147100 1466 1476 6370 0.00926 1360
aphelion 152100 1516 1527 " " 1406
Staying within the umbra near the "L2" point would likely require less delta-v for Mars than for Earth, but either way it's an unstable point and so you need to continually do small station-keeping burns to remain protected.
The umbra protects you from line-of sight radiation like photons (light, X-rays, gamma-rays) but may be less protective for ballistic charged particles from a [CME][3] because they may be on trajectories that when projected backwards, come from points beyond the disk of the Sun for various reasons. That's an excellent *next question!
For the "normal" solar wind, it may be even more likely to expand into the shadow since wether it's ionized or not, it will have some transverse velocity distribution associated with it.
answered 34 mins ago
uhoh
29.1k1598358
The Lagrange points of Mars might be less useful than that of Earth's because Mars' orbit is more elliptical and so there are more perturbing forces.
Nonetheless we can do a simple exercise.
In this answer I show how to calculate the theoretical distances to the first two Lagrange points for a circular restricted three body problem scenario.
From geometry we can see that if a right circular cone with base radius = $r_sun$ has a radius $r$ at a distance $D$, then the vertex will be beyond that point by a distance
$$ Dfracrr_sun-r$$
Using radii for Mars, Earth and the Sun of 3396, 6378, and 696392 km, I get the following, which shows that whether or not the Lagrange points are meaningful and/or useful for an elliptical orbit like Mars, Mars' solar umbra is beyond Sun-Mars L2 by about 100 km, while for Earth's solar umbra seems to be 100 km before (short-of) Earth-Mars L2.
MARS to Sun to L1 to L2 r r/(r_sun-r) "end of umbra"
perihelion 206700 981 985 3396 0.00490 1013
aphelion 249200 1183 1187 " " 1221
Earth
perihelion 147100 1466 1476 6370 0.00926 1360
aphelion 152100 1516 1527 " " 1406
Staying within the umbra near the "L2" point would likely require less delta-v for Mars than for Earth, but either way it's an unstable point and so you need to continually do small station-keeping burns to remain protected.
The umbra protects you from line-of sight radiation like photons (light, X-rays, gamma-rays) but may be less protective for ballistic charged particles from a [CME][3] because they may be on trajectories that when projected backwards, come from points beyond the disk of the Sun for various reasons. That's an excellent *next question!
For the "normal" solar wind, it may be even more likely to expand into the shadow since wether it's ionized or not, it will have some transverse velocity distribution associated with it.
answered 34 mins ago
uhoh
29.1k1598358
edited 27 mins ago
answered 34 mins ago
uhoh
29.1k1598358
answered 34 mins ago
uhoh
29.1k1598358
answered 34 mins ago
uhoh
29.1k1598358
29.1k1598358
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Id assume since mars has a low magnetosphere the only reduction in radation youll recieve is eclipses of mars and the sun.
– Magic Octopus Urn
2 hours ago