What is the most fuel efficient way out of the Solar System?
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This Gif. shows how our solar system is traveling through space.
earthquakepredict.com
I understand with current technology we can't just fly a straight line out of the solar system but if it was possible which way out would need the least fuel?
Starting from Earth would I want to say looking at the Gif. that the best way out is to launch when Earth is traveling down and to the left.
Currently to navigate the solar system it is a dance around the planets using sling shots, gravity assist and such.
Will having more planets present on the way out always help?
fuel gravity trajectory interstellar-travel solar-system
asked 3 hours ago
MuzeMuze
1,3531158
$endgroup$
add a comment |
$begingroup$
This Gif. shows how our solar system is traveling through space.
earthquakepredict.com
I understand with current technology we can't just fly a straight line out of the solar system but if it was possible which way out would need the least fuel?
Starting from Earth would I want to say looking at the Gif. that the best way out is to launch when Earth is traveling down and to the left.
Currently to navigate the solar system it is a dance around the planets using sling shots, gravity assist and such.
Will having more planets present on the way out always help?
fuel gravity trajectory interstellar-travel solar-system
asked 3 hours ago
MuzeMuze
1,3531158
$endgroup$
$begingroup$
Unfortunately, that gif is wrong according to this: universetoday.com/107322/is-the-solar-system-really-a-vortex
$endgroup$
– CBredlow
3 hours ago
1
$begingroup$
@CBredlow How is this model wrong?
$endgroup$
– Muze
3 hours ago
add a comment |
$begingroup$
This Gif. shows how our solar system is traveling through space.
earthquakepredict.com
I understand with current technology we can't just fly a straight line out of the solar system but if it was possible which way out would need the least fuel?
Starting from Earth would I want to say looking at the Gif. that the best way out is to launch when Earth is traveling down and to the left.
Currently to navigate the solar system it is a dance around the planets using sling shots, gravity assist and such.
Will having more planets present on the way out always help?
fuel gravity trajectory interstellar-travel solar-system
asked 3 hours ago
MuzeMuze
1,3531158
$endgroup$
This Gif. shows how our solar system is traveling through space.
earthquakepredict.com
I understand with current technology we can't just fly a straight line out of the solar system but if it was possible which way out would need the least fuel?
Starting from Earth would I want to say looking at the Gif. that the best way out is to launch when Earth is traveling down and to the left.
Currently to navigate the solar system it is a dance around the planets using sling shots, gravity assist and such.
Will having more planets present on the way out always help?
fuel gravity trajectory interstellar-travel solar-system
fuel gravity trajectory interstellar-travel solar-system
asked 3 hours ago
MuzeMuze
1,3531158
asked 3 hours ago
MuzeMuze
1,3531158
asked 3 hours ago
MuzeMuze
1,3531158
asked 3 hours ago
MuzeMuze
1,3531158
asked 3 hours ago
MuzeMuze
1,3531158
1,3531158
$begingroup$
Unfortunately, that gif is wrong according to this: universetoday.com/107322/is-the-solar-system-really-a-vortex
$endgroup$
– CBredlow
3 hours ago
1
$begingroup$
@CBredlow How is this model wrong?
$endgroup$
– Muze
3 hours ago
add a comment |
$begingroup$
Unfortunately, that gif is wrong according to this: universetoday.com/107322/is-the-solar-system-really-a-vortex
$endgroup$
– CBredlow
3 hours ago
1
$begingroup$
@CBredlow How is this model wrong?
$endgroup$
– Muze
3 hours ago
$begingroup$
Unfortunately, that gif is wrong according to this: universetoday.com/107322/is-the-solar-system-really-a-vortex
$endgroup$
– CBredlow
3 hours ago
$begingroup$
Unfortunately, that gif is wrong according to this: universetoday.com/107322/is-the-solar-system-really-a-vortex
$endgroup$
– CBredlow
3 hours ago
1
1
$begingroup$
@CBredlow How is this model wrong?
$endgroup$
– Muze
3 hours ago
$begingroup$
@CBredlow How is this model wrong?
$endgroup$
– Muze
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
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I'm not sure the GIF is especially relevant, even if correct. The most fuel efficient way to leave the solar system at present, is to launch into a trajectory that (like that used for Gallileo) may well involve one or several gravity assists from Earth or Venus, but which eventually gets you to Jupiter. You use the gravity of Jupiter to drop you as close to the Sun as your systems can survive (it may alternatively be possible to get there just using Earth and Venus assists, as the Parker Solar Probe did). Once there you burn all your remaining fuel as close to the Sun as you can and then you coast. You might gain a little from encounters with Jupiter and/or Saturn on the way out, but you are moving so fast it doesn't really matter.
The Thousand AU study reckoned you could achieve exit from the inner solar system at a velocity of perhaps 10-15 AU/year using this technique with a large current launcher.
answered 3 hours ago
Steve LintonSteve Linton
8,19912245
$endgroup$
$begingroup$
+1 that would depend on how close Venus is to Earth. If the scale of the gif is correct and you used a straight line out what would be the fastest way out?
$endgroup$
– Muze
3 hours ago
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@Muze check out the parker solar probe's trajectory, except at the closest point of approach instead of doing science you'd do a burn. I'd love to see the actual numbers though, given a particular craft and a particular perigee.
$endgroup$
– Magic Octopus Urn
3 hours ago
$begingroup$
If for some reason you don't want to do any gravity well maneuvers at all then you want to launch tangentially to Earth's orbit to get maximum benefit from Earth's orbital velocity. You'd want to launch at midnight so that Earth's rotational velocity added in as well. But this is still much worse than going via a close solar encounter
$endgroup$
– Steve Linton
2 hours ago
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@SteveLinton I will read on this in depth thanks again.
$endgroup$
– Muze
2 hours ago
add a comment |
$begingroup$
Launching from the west side of Earth and taking a slingshot course by the sun for a turn of about 90*, then doing another sling past Jupiter(burning all your fuel) would, I think, be the most efficient way to do this. Ion engines would be the most efficient propulsion method as well, however I would recommend using a common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. That is, if you were to launch your rocket today, along the orbital path of the planets...
However, another way to do it(again with the help of the sun) would be to go into LEO, and then set your periapsis to the 'night side' of Earth. Then, at periapsis, have your engines at full blast until the other side of your orbit goes to the opposite side of Sol, and then coast till you reach the bottom, and, again, go full blast. by the time you are done accelerating, your apoapsis should be well out of the solar system.
There is no way to go "straight" out of the solar system without being accelerated and decelerated. Gravitational forces will always be acting upon you, and until you get to the point where the nearest extrasolar body's gravity is stronger than the sun's, you won't be "out" of the solar system.
answered 3 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
517
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Light pressure at Earth's distance from the Sun is about $6mu Pa$. An alumnium sail 50 nm thick (much thinner and it would be transparent) has areal density about $1.3 times 10^{-4} kg/m^2$ so with no payload would accelerate at $4times 10^{-2} m/s^2$ (abotut 4 milli gravities). By the time it reached the orbit of Mars (ignoring the Sun's gravity) it would be moving at something in the ballpark of $100 km/s$. So yes, a very thin solar sail with no allowance for rigging or payload could escape the solar system.
$endgroup$
– Steve Linton
2 hours ago
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
I'm not sure the GIF is especially relevant, even if correct. The most fuel efficient way to leave the solar system at present, is to launch into a trajectory that (like that used for Gallileo) may well involve one or several gravity assists from Earth or Venus, but which eventually gets you to Jupiter. You use the gravity of Jupiter to drop you as close to the Sun as your systems can survive (it may alternatively be possible to get there just using Earth and Venus assists, as the Parker Solar Probe did). Once there you burn all your remaining fuel as close to the Sun as you can and then you coast. You might gain a little from encounters with Jupiter and/or Saturn on the way out, but you are moving so fast it doesn't really matter.
The Thousand AU study reckoned you could achieve exit from the inner solar system at a velocity of perhaps 10-15 AU/year using this technique with a large current launcher.
answered 3 hours ago
Steve LintonSteve Linton
8,19912245
$endgroup$
$begingroup$
+1 that would depend on how close Venus is to Earth. If the scale of the gif is correct and you used a straight line out what would be the fastest way out?
$endgroup$
– Muze
3 hours ago
$begingroup$
@Muze check out the parker solar probe's trajectory, except at the closest point of approach instead of doing science you'd do a burn. I'd love to see the actual numbers though, given a particular craft and a particular perigee.
$endgroup$
– Magic Octopus Urn
3 hours ago
$begingroup$
If for some reason you don't want to do any gravity well maneuvers at all then you want to launch tangentially to Earth's orbit to get maximum benefit from Earth's orbital velocity. You'd want to launch at midnight so that Earth's rotational velocity added in as well. But this is still much worse than going via a close solar encounter
$endgroup$
– Steve Linton
2 hours ago
$begingroup$
@SteveLinton I will read on this in depth thanks again.
$endgroup$
– Muze
2 hours ago
add a comment |
$begingroup$
I'm not sure the GIF is especially relevant, even if correct. The most fuel efficient way to leave the solar system at present, is to launch into a trajectory that (like that used for Gallileo) may well involve one or several gravity assists from Earth or Venus, but which eventually gets you to Jupiter. You use the gravity of Jupiter to drop you as close to the Sun as your systems can survive (it may alternatively be possible to get there just using Earth and Venus assists, as the Parker Solar Probe did). Once there you burn all your remaining fuel as close to the Sun as you can and then you coast. You might gain a little from encounters with Jupiter and/or Saturn on the way out, but you are moving so fast it doesn't really matter.
The Thousand AU study reckoned you could achieve exit from the inner solar system at a velocity of perhaps 10-15 AU/year using this technique with a large current launcher.
answered 3 hours ago
Steve LintonSteve Linton
8,19912245
$endgroup$
$begingroup$
+1 that would depend on how close Venus is to Earth. If the scale of the gif is correct and you used a straight line out what would be the fastest way out?
$endgroup$
– Muze
3 hours ago
$begingroup$
@Muze check out the parker solar probe's trajectory, except at the closest point of approach instead of doing science you'd do a burn. I'd love to see the actual numbers though, given a particular craft and a particular perigee.
$endgroup$
– Magic Octopus Urn
3 hours ago
$begingroup$
If for some reason you don't want to do any gravity well maneuvers at all then you want to launch tangentially to Earth's orbit to get maximum benefit from Earth's orbital velocity. You'd want to launch at midnight so that Earth's rotational velocity added in as well. But this is still much worse than going via a close solar encounter
$endgroup$
– Steve Linton
2 hours ago
$begingroup$
@SteveLinton I will read on this in depth thanks again.
$endgroup$
– Muze
2 hours ago
add a comment |
$begingroup$
I'm not sure the GIF is especially relevant, even if correct. The most fuel efficient way to leave the solar system at present, is to launch into a trajectory that (like that used for Gallileo) may well involve one or several gravity assists from Earth or Venus, but which eventually gets you to Jupiter. You use the gravity of Jupiter to drop you as close to the Sun as your systems can survive (it may alternatively be possible to get there just using Earth and Venus assists, as the Parker Solar Probe did). Once there you burn all your remaining fuel as close to the Sun as you can and then you coast. You might gain a little from encounters with Jupiter and/or Saturn on the way out, but you are moving so fast it doesn't really matter.
The Thousand AU study reckoned you could achieve exit from the inner solar system at a velocity of perhaps 10-15 AU/year using this technique with a large current launcher.
answered 3 hours ago
Steve LintonSteve Linton
8,19912245
$endgroup$
I'm not sure the GIF is especially relevant, even if correct. The most fuel efficient way to leave the solar system at present, is to launch into a trajectory that (like that used for Gallileo) may well involve one or several gravity assists from Earth or Venus, but which eventually gets you to Jupiter. You use the gravity of Jupiter to drop you as close to the Sun as your systems can survive (it may alternatively be possible to get there just using Earth and Venus assists, as the Parker Solar Probe did). Once there you burn all your remaining fuel as close to the Sun as you can and then you coast. You might gain a little from encounters with Jupiter and/or Saturn on the way out, but you are moving so fast it doesn't really matter.
The Thousand AU study reckoned you could achieve exit from the inner solar system at a velocity of perhaps 10-15 AU/year using this technique with a large current launcher.
answered 3 hours ago
Steve LintonSteve Linton
8,19912245
answered 3 hours ago
Steve LintonSteve Linton
8,19912245
answered 3 hours ago
Steve LintonSteve Linton
8,19912245
answered 3 hours ago
Steve LintonSteve Linton
8,19912245
8,19912245
$begingroup$
+1 that would depend on how close Venus is to Earth. If the scale of the gif is correct and you used a straight line out what would be the fastest way out?
$endgroup$
– Muze
3 hours ago
$begingroup$
@Muze check out the parker solar probe's trajectory, except at the closest point of approach instead of doing science you'd do a burn. I'd love to see the actual numbers though, given a particular craft and a particular perigee.
$endgroup$
– Magic Octopus Urn
3 hours ago
$begingroup$
If for some reason you don't want to do any gravity well maneuvers at all then you want to launch tangentially to Earth's orbit to get maximum benefit from Earth's orbital velocity. You'd want to launch at midnight so that Earth's rotational velocity added in as well. But this is still much worse than going via a close solar encounter
$endgroup$
– Steve Linton
2 hours ago
$begingroup$
@SteveLinton I will read on this in depth thanks again.
$endgroup$
– Muze
2 hours ago
add a comment |
$begingroup$
+1 that would depend on how close Venus is to Earth. If the scale of the gif is correct and you used a straight line out what would be the fastest way out?
$endgroup$
– Muze
3 hours ago
$begingroup$
@Muze check out the parker solar probe's trajectory, except at the closest point of approach instead of doing science you'd do a burn. I'd love to see the actual numbers though, given a particular craft and a particular perigee.
$endgroup$
– Magic Octopus Urn
3 hours ago
$begingroup$
If for some reason you don't want to do any gravity well maneuvers at all then you want to launch tangentially to Earth's orbit to get maximum benefit from Earth's orbital velocity. You'd want to launch at midnight so that Earth's rotational velocity added in as well. But this is still much worse than going via a close solar encounter
$endgroup$
– Steve Linton
2 hours ago
$begingroup$
@SteveLinton I will read on this in depth thanks again.
$endgroup$
– Muze
2 hours ago
$begingroup$
+1 that would depend on how close Venus is to Earth. If the scale of the gif is correct and you used a straight line out what would be the fastest way out?
$endgroup$
– Muze
3 hours ago
$begingroup$
+1 that would depend on how close Venus is to Earth. If the scale of the gif is correct and you used a straight line out what would be the fastest way out?
$endgroup$
– Muze
3 hours ago
$begingroup$
@Muze check out the parker solar probe's trajectory, except at the closest point of approach instead of doing science you'd do a burn. I'd love to see the actual numbers though, given a particular craft and a particular perigee.
$endgroup$
– Magic Octopus Urn
3 hours ago
$begingroup$
@Muze check out the parker solar probe's trajectory, except at the closest point of approach instead of doing science you'd do a burn. I'd love to see the actual numbers though, given a particular craft and a particular perigee.
$endgroup$
– Magic Octopus Urn
3 hours ago
$begingroup$
If for some reason you don't want to do any gravity well maneuvers at all then you want to launch tangentially to Earth's orbit to get maximum benefit from Earth's orbital velocity. You'd want to launch at midnight so that Earth's rotational velocity added in as well. But this is still much worse than going via a close solar encounter
$endgroup$
– Steve Linton
2 hours ago
$begingroup$
If for some reason you don't want to do any gravity well maneuvers at all then you want to launch tangentially to Earth's orbit to get maximum benefit from Earth's orbital velocity. You'd want to launch at midnight so that Earth's rotational velocity added in as well. But this is still much worse than going via a close solar encounter
$endgroup$
– Steve Linton
2 hours ago
$begingroup$
@SteveLinton I will read on this in depth thanks again.
$endgroup$
– Muze
2 hours ago
$begingroup$
@SteveLinton I will read on this in depth thanks again.
$endgroup$
– Muze
2 hours ago
add a comment |
$begingroup$
Launching from the west side of Earth and taking a slingshot course by the sun for a turn of about 90*, then doing another sling past Jupiter(burning all your fuel) would, I think, be the most efficient way to do this. Ion engines would be the most efficient propulsion method as well, however I would recommend using a common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. That is, if you were to launch your rocket today, along the orbital path of the planets...
However, another way to do it(again with the help of the sun) would be to go into LEO, and then set your periapsis to the 'night side' of Earth. Then, at periapsis, have your engines at full blast until the other side of your orbit goes to the opposite side of Sol, and then coast till you reach the bottom, and, again, go full blast. by the time you are done accelerating, your apoapsis should be well out of the solar system.
There is no way to go "straight" out of the solar system without being accelerated and decelerated. Gravitational forces will always be acting upon you, and until you get to the point where the nearest extrasolar body's gravity is stronger than the sun's, you won't be "out" of the solar system.
answered 3 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
517
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Light pressure at Earth's distance from the Sun is about $6mu Pa$. An alumnium sail 50 nm thick (much thinner and it would be transparent) has areal density about $1.3 times 10^{-4} kg/m^2$ so with no payload would accelerate at $4times 10^{-2} m/s^2$ (abotut 4 milli gravities). By the time it reached the orbit of Mars (ignoring the Sun's gravity) it would be moving at something in the ballpark of $100 km/s$. So yes, a very thin solar sail with no allowance for rigging or payload could escape the solar system.
$endgroup$
– Steve Linton
2 hours ago
add a comment |
$begingroup$
Launching from the west side of Earth and taking a slingshot course by the sun for a turn of about 90*, then doing another sling past Jupiter(burning all your fuel) would, I think, be the most efficient way to do this. Ion engines would be the most efficient propulsion method as well, however I would recommend using a common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. That is, if you were to launch your rocket today, along the orbital path of the planets...
However, another way to do it(again with the help of the sun) would be to go into LEO, and then set your periapsis to the 'night side' of Earth. Then, at periapsis, have your engines at full blast until the other side of your orbit goes to the opposite side of Sol, and then coast till you reach the bottom, and, again, go full blast. by the time you are done accelerating, your apoapsis should be well out of the solar system.
There is no way to go "straight" out of the solar system without being accelerated and decelerated. Gravitational forces will always be acting upon you, and until you get to the point where the nearest extrasolar body's gravity is stronger than the sun's, you won't be "out" of the solar system.
answered 3 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
517
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Light pressure at Earth's distance from the Sun is about $6mu Pa$. An alumnium sail 50 nm thick (much thinner and it would be transparent) has areal density about $1.3 times 10^{-4} kg/m^2$ so with no payload would accelerate at $4times 10^{-2} m/s^2$ (abotut 4 milli gravities). By the time it reached the orbit of Mars (ignoring the Sun's gravity) it would be moving at something in the ballpark of $100 km/s$. So yes, a very thin solar sail with no allowance for rigging or payload could escape the solar system.
$endgroup$
– Steve Linton
2 hours ago
add a comment |
$begingroup$
Launching from the west side of Earth and taking a slingshot course by the sun for a turn of about 90*, then doing another sling past Jupiter(burning all your fuel) would, I think, be the most efficient way to do this. Ion engines would be the most efficient propulsion method as well, however I would recommend using a common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. That is, if you were to launch your rocket today, along the orbital path of the planets...
However, another way to do it(again with the help of the sun) would be to go into LEO, and then set your periapsis to the 'night side' of Earth. Then, at periapsis, have your engines at full blast until the other side of your orbit goes to the opposite side of Sol, and then coast till you reach the bottom, and, again, go full blast. by the time you are done accelerating, your apoapsis should be well out of the solar system.
There is no way to go "straight" out of the solar system without being accelerated and decelerated. Gravitational forces will always be acting upon you, and until you get to the point where the nearest extrasolar body's gravity is stronger than the sun's, you won't be "out" of the solar system.
answered 3 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
517
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Launching from the west side of Earth and taking a slingshot course by the sun for a turn of about 90*, then doing another sling past Jupiter(burning all your fuel) would, I think, be the most efficient way to do this. Ion engines would be the most efficient propulsion method as well, however I would recommend using a common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. That is, if you were to launch your rocket today, along the orbital path of the planets...
However, another way to do it(again with the help of the sun) would be to go into LEO, and then set your periapsis to the 'night side' of Earth. Then, at periapsis, have your engines at full blast until the other side of your orbit goes to the opposite side of Sol, and then coast till you reach the bottom, and, again, go full blast. by the time you are done accelerating, your apoapsis should be well out of the solar system.
There is no way to go "straight" out of the solar system without being accelerated and decelerated. Gravitational forces will always be acting upon you, and until you get to the point where the nearest extrasolar body's gravity is stronger than the sun's, you won't be "out" of the solar system.
answered 3 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
517
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
answered 3 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
517
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
517
answered 3 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
517
517
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Light pressure at Earth's distance from the Sun is about $6mu Pa$. An alumnium sail 50 nm thick (much thinner and it would be transparent) has areal density about $1.3 times 10^{-4} kg/m^2$ so with no payload would accelerate at $4times 10^{-2} m/s^2$ (abotut 4 milli gravities). By the time it reached the orbit of Mars (ignoring the Sun's gravity) it would be moving at something in the ballpark of $100 km/s$. So yes, a very thin solar sail with no allowance for rigging or payload could escape the solar system.
$endgroup$
– Steve Linton
2 hours ago
add a comment |
2
$begingroup$
Light pressure at Earth's distance from the Sun is about $6mu Pa$. An alumnium sail 50 nm thick (much thinner and it would be transparent) has areal density about $1.3 times 10^{-4} kg/m^2$ so with no payload would accelerate at $4times 10^{-2} m/s^2$ (abotut 4 milli gravities). By the time it reached the orbit of Mars (ignoring the Sun's gravity) it would be moving at something in the ballpark of $100 km/s$. So yes, a very thin solar sail with no allowance for rigging or payload could escape the solar system.
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– Steve Linton
2 hours ago
2
2
$begingroup$
Light pressure at Earth's distance from the Sun is about $6mu Pa$. An alumnium sail 50 nm thick (much thinner and it would be transparent) has areal density about $1.3 times 10^{-4} kg/m^2$ so with no payload would accelerate at $4times 10^{-2} m/s^2$ (abotut 4 milli gravities). By the time it reached the orbit of Mars (ignoring the Sun's gravity) it would be moving at something in the ballpark of $100 km/s$. So yes, a very thin solar sail with no allowance for rigging or payload could escape the solar system.
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– Steve Linton
2 hours ago
$begingroup$
Light pressure at Earth's distance from the Sun is about $6mu Pa$. An alumnium sail 50 nm thick (much thinner and it would be transparent) has areal density about $1.3 times 10^{-4} kg/m^2$ so with no payload would accelerate at $4times 10^{-2} m/s^2$ (abotut 4 milli gravities). By the time it reached the orbit of Mars (ignoring the Sun's gravity) it would be moving at something in the ballpark of $100 km/s$. So yes, a very thin solar sail with no allowance for rigging or payload could escape the solar system.
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– Steve Linton
2 hours ago
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Unfortunately, that gif is wrong according to this: universetoday.com/107322/is-the-solar-system-really-a-vortex
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– CBredlow
3 hours ago
1
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@CBredlow How is this model wrong?
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– Muze
3 hours ago