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 13 hours ago
MuzeMuze
1,3281159
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|
show 5 more comments
$begingroup$
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 13 hours ago
MuzeMuze
1,3281159
$endgroup$
7
$begingroup$
@CBredlow that is a criticism of a very different animation which is wrong for a number of reasons not applicable to this particular gif, which seems to be broadly correct (though I haven't measured the tilt of the orbital plane to confirm that it is 60 degrees). That said, our motion through the galaxy is irrelevant to escaping the solar system.
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– asgallant
9 hours ago
2
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Where would you like to go outside the solar system?
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– Thorbjørn Ravn Andersen
8 hours ago
1
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direction. "Way" implies a 'way of means' IMO. Which would use tidel influences on a machine of planetary scale. Fuel consumption: zero.
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– Mazura
7 hours ago
2
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This is the second question in a few days about escaping the solar system. I'm wondering if there's something specific I need to be worried about...
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– Richard
6 hours ago
1
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@Richard the galaxy is on Orion's belt
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– uhoh
4 hours ago
|
show 5 more comments
$begingroup$
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 13 hours ago
MuzeMuze
1,3281159
$endgroup$
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 13 hours ago
MuzeMuze
1,3281159
asked 13 hours ago
MuzeMuze
1,3281159
asked 13 hours ago
MuzeMuze
1,3281159
asked 13 hours ago
MuzeMuze
1,3281159
asked 13 hours ago
MuzeMuze
1,3281159
1,3281159
7
$begingroup$
@CBredlow that is a criticism of a very different animation which is wrong for a number of reasons not applicable to this particular gif, which seems to be broadly correct (though I haven't measured the tilt of the orbital plane to confirm that it is 60 degrees). That said, our motion through the galaxy is irrelevant to escaping the solar system.
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– asgallant
9 hours ago
2
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Where would you like to go outside the solar system?
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– Thorbjørn Ravn Andersen
8 hours ago
1
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direction. "Way" implies a 'way of means' IMO. Which would use tidel influences on a machine of planetary scale. Fuel consumption: zero.
$endgroup$
– Mazura
7 hours ago
2
$begingroup$
This is the second question in a few days about escaping the solar system. I'm wondering if there's something specific I need to be worried about...
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– Richard
6 hours ago
1
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@Richard the galaxy is on Orion's belt
$endgroup$
– uhoh
4 hours ago
|
show 5 more comments
7
$begingroup$
@CBredlow that is a criticism of a very different animation which is wrong for a number of reasons not applicable to this particular gif, which seems to be broadly correct (though I haven't measured the tilt of the orbital plane to confirm that it is 60 degrees). That said, our motion through the galaxy is irrelevant to escaping the solar system.
$endgroup$
– asgallant
9 hours ago
2
$begingroup$
Where would you like to go outside the solar system?
$endgroup$
– Thorbjørn Ravn Andersen
8 hours ago
1
$begingroup$
direction. "Way" implies a 'way of means' IMO. Which would use tidel influences on a machine of planetary scale. Fuel consumption: zero.
$endgroup$
– Mazura
7 hours ago
2
$begingroup$
This is the second question in a few days about escaping the solar system. I'm wondering if there's something specific I need to be worried about...
$endgroup$
– Richard
6 hours ago
1
$begingroup$
@Richard the galaxy is on Orion's belt
$endgroup$
– uhoh
4 hours ago
7
7
$begingroup$
@CBredlow that is a criticism of a very different animation which is wrong for a number of reasons not applicable to this particular gif, which seems to be broadly correct (though I haven't measured the tilt of the orbital plane to confirm that it is 60 degrees). That said, our motion through the galaxy is irrelevant to escaping the solar system.
$endgroup$
– asgallant
9 hours ago
$begingroup$
@CBredlow that is a criticism of a very different animation which is wrong for a number of reasons not applicable to this particular gif, which seems to be broadly correct (though I haven't measured the tilt of the orbital plane to confirm that it is 60 degrees). That said, our motion through the galaxy is irrelevant to escaping the solar system.
$endgroup$
– asgallant
9 hours ago
2
2
$begingroup$
Where would you like to go outside the solar system?
$endgroup$
– Thorbjørn Ravn Andersen
8 hours ago
$begingroup$
Where would you like to go outside the solar system?
$endgroup$
– Thorbjørn Ravn Andersen
8 hours ago
1
1
$begingroup$
direction. "Way" implies a 'way of means' IMO. Which would use tidel influences on a machine of planetary scale. Fuel consumption: zero.
$endgroup$
– Mazura
7 hours ago
$begingroup$
direction. "Way" implies a 'way of means' IMO. Which would use tidel influences on a machine of planetary scale. Fuel consumption: zero.
$endgroup$
– Mazura
7 hours ago
2
2
$begingroup$
This is the second question in a few days about escaping the solar system. I'm wondering if there's something specific I need to be worried about...
$endgroup$
– Richard
6 hours ago
$begingroup$
This is the second question in a few days about escaping the solar system. I'm wondering if there's something specific I need to be worried about...
$endgroup$
– Richard
6 hours ago
1
1
$begingroup$
@Richard the galaxy is on Orion's belt
$endgroup$
– uhoh
4 hours ago
$begingroup$
@Richard the galaxy is on Orion's belt
$endgroup$
– uhoh
4 hours ago
|
show 5 more comments
4 Answers
4
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oldest
votes
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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 and mirrors would be the most efficient propulsion method as well, however I would recommend using common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. It would work, 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 13 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
767
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AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
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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
12 hours ago
3
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@SteveLinton, the nice thing about solar sails is that both gravity and light pressure fall off as the inverse square of distance. If you've got a sail that provides acceleration at one distance from the Sun, it'll provide acceleration at all distances from the Sun.
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– Mark
7 hours ago
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@SteveLinton this is another answer. Thumbs up.
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– Muze
2 hours ago
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@Mark: Not all distances, maybe till it reaches the heliopause, I presume?
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– sampathsris
2 hours ago
1
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@sampathsris, until other stars start interfering with it. Most of the force on a solar sail comes from sunlight, not the solar wind, and that doesn't stop at the heliopause.
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– Mark
56 mins ago
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show 1 more comment
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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 13 hours ago
Steve LintonSteve Linton
8,25912246
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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.
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– Magic Octopus Urn
13 hours ago
1
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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
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– Steve Linton
13 hours ago
3
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@Muze What is the fastest way out is almost certainly not the most fuel efficient. Venus is the place we can get for the least energy and once you can reach another world you can escape by playing billiards--but that's a slow process unless everything is neatly lined up for you like with the Voyager probes.
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– Loren Pechtel
9 hours ago
1
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How about the Voyager/Grand Tour option, where your gravity assist from Jupiter is followed by ones from Saturn, Uranus, and Neptune? AFAIK, the Voyagers didn't use fuel (other than a small amount for mid-course corrections) once they had boosted from Earth orbit. Getting to Jupiter via Venus/Earth assists might make it even more fuel-efficient.
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– jamesqf
5 hours ago
2
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@Muze, If you think that the most efficient path would be a "straight line," then you are pretty far from understanding how objects move in the Solar system. Far enough that the kind of answers that you can find on a web site like this probably are going to be less than helpful. You need some book learning or some classroom learning. Good luck!
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– Solomon Slow
4 hours ago
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show 2 more comments
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That graphic is misleading. Gravity is equal in all directions, so no matter when you launch, the effort to leave the solar system remains the same.
As Steve Linton said, the most fuel-efficient way out is to get as much delta-V from gravity assists as possible.
answered 10 hours ago
HobbesHobbes
92.5k2258412
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While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's instantaneous velocity will be important as well.
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– uhoh
4 hours ago
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While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's velocity may be important as well.
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– uhoh
4 hours ago
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note: I'm leaving these duplicate comments in place temporarily since I've just reported this new instance.
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– uhoh
4 hours ago
add a comment |
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If you want to avoid gravity assists, the most fuel-efficient way out of the Solar System is to launch due East from from a launch site in the Ecuadorean Andes, sometime before local midnight on a July 4 when there's a new moon. This gives you the maximum possible benefit from the Earth's movement, leaving only about 12,000 m/s of delta-V needed in excess of Earth escape velocity.
(Rough estimate: a Saturn V could barely put New Horizons directly into a solar escape trajectory. This is why we use gravity assists instead.)
answered 9 hours ago
MarkMark
4,0901730
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This is intriguing! I've just asked Launching east from a mountain on the equator at midnight during a new moon; ranking of each contribution?
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– uhoh
4 hours ago
add a comment |
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4 Answers
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4 Answers
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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 and mirrors would be the most efficient propulsion method as well, however I would recommend using common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. It would work, 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 13 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
767
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
12 hours ago
3
$begingroup$
@SteveLinton, the nice thing about solar sails is that both gravity and light pressure fall off as the inverse square of distance. If you've got a sail that provides acceleration at one distance from the Sun, it'll provide acceleration at all distances from the Sun.
$endgroup$
– Mark
7 hours ago
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@SteveLinton this is another answer. Thumbs up.
$endgroup$
– Muze
2 hours ago
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@Mark: Not all distances, maybe till it reaches the heliopause, I presume?
$endgroup$
– sampathsris
2 hours ago
1
$begingroup$
@sampathsris, until other stars start interfering with it. Most of the force on a solar sail comes from sunlight, not the solar wind, and that doesn't stop at the heliopause.
$endgroup$
– Mark
56 mins ago
|
show 1 more 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 and mirrors would be the most efficient propulsion method as well, however I would recommend using common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. It would work, 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 13 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
767
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
12 hours ago
3
$begingroup$
@SteveLinton, the nice thing about solar sails is that both gravity and light pressure fall off as the inverse square of distance. If you've got a sail that provides acceleration at one distance from the Sun, it'll provide acceleration at all distances from the Sun.
$endgroup$
– Mark
7 hours ago
$begingroup$
@SteveLinton this is another answer. Thumbs up.
$endgroup$
– Muze
2 hours ago
$begingroup$
@Mark: Not all distances, maybe till it reaches the heliopause, I presume?
$endgroup$
– sampathsris
2 hours ago
1
$begingroup$
@sampathsris, until other stars start interfering with it. Most of the force on a solar sail comes from sunlight, not the solar wind, and that doesn't stop at the heliopause.
$endgroup$
– Mark
56 mins ago
|
show 1 more 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 and mirrors would be the most efficient propulsion method as well, however I would recommend using common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. It would work, 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 13 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
767
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 and mirrors would be the most efficient propulsion method as well, however I would recommend using common LOXY engines to get into orbit, and then the ion engines to break orbit and head towards the sun. It would work, 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 13 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
767
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 3 hours ago
answered 13 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
767
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 13 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
767
answered 13 hours ago
AndrewMaxwellRocketsAndrewMaxwellRockets
767
767
New contributor
AndrewMaxwellRockets is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
12 hours ago
3
$begingroup$
@SteveLinton, the nice thing about solar sails is that both gravity and light pressure fall off as the inverse square of distance. If you've got a sail that provides acceleration at one distance from the Sun, it'll provide acceleration at all distances from the Sun.
$endgroup$
– Mark
7 hours ago
$begingroup$
@SteveLinton this is another answer. Thumbs up.
$endgroup$
– Muze
2 hours ago
$begingroup$
@Mark: Not all distances, maybe till it reaches the heliopause, I presume?
$endgroup$
– sampathsris
2 hours ago
1
$begingroup$
@sampathsris, until other stars start interfering with it. Most of the force on a solar sail comes from sunlight, not the solar wind, and that doesn't stop at the heliopause.
$endgroup$
– Mark
56 mins ago
|
show 1 more 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.
$endgroup$
– Steve Linton
12 hours ago
3
$begingroup$
@SteveLinton, the nice thing about solar sails is that both gravity and light pressure fall off as the inverse square of distance. If you've got a sail that provides acceleration at one distance from the Sun, it'll provide acceleration at all distances from the Sun.
$endgroup$
– Mark
7 hours ago
$begingroup$
@SteveLinton this is another answer. Thumbs up.
$endgroup$
– Muze
2 hours ago
$begingroup$
@Mark: Not all distances, maybe till it reaches the heliopause, I presume?
$endgroup$
– sampathsris
2 hours ago
1
$begingroup$
@sampathsris, until other stars start interfering with it. Most of the force on a solar sail comes from sunlight, not the solar wind, and that doesn't stop at the heliopause.
$endgroup$
– Mark
56 mins 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.
$endgroup$
– Steve Linton
12 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.
$endgroup$
– Steve Linton
12 hours ago
3
3
$begingroup$
@SteveLinton, the nice thing about solar sails is that both gravity and light pressure fall off as the inverse square of distance. If you've got a sail that provides acceleration at one distance from the Sun, it'll provide acceleration at all distances from the Sun.
$endgroup$
– Mark
7 hours ago
$begingroup$
@SteveLinton, the nice thing about solar sails is that both gravity and light pressure fall off as the inverse square of distance. If you've got a sail that provides acceleration at one distance from the Sun, it'll provide acceleration at all distances from the Sun.
$endgroup$
– Mark
7 hours ago
$begingroup$
@SteveLinton this is another answer. Thumbs up.
$endgroup$
– Muze
2 hours ago
$begingroup$
@SteveLinton this is another answer. Thumbs up.
$endgroup$
– Muze
2 hours ago
$begingroup$
@Mark: Not all distances, maybe till it reaches the heliopause, I presume?
$endgroup$
– sampathsris
2 hours ago
$begingroup$
@Mark: Not all distances, maybe till it reaches the heliopause, I presume?
$endgroup$
– sampathsris
2 hours ago
1
1
$begingroup$
@sampathsris, until other stars start interfering with it. Most of the force on a solar sail comes from sunlight, not the solar wind, and that doesn't stop at the heliopause.
$endgroup$
– Mark
56 mins ago
$begingroup$
@sampathsris, until other stars start interfering with it. Most of the force on a solar sail comes from sunlight, not the solar wind, and that doesn't stop at the heliopause.
$endgroup$
– Mark
56 mins ago
|
show 1 more 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 13 hours ago
Steve LintonSteve Linton
8,25912246
$endgroup$
1
$begingroup$
@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
13 hours ago
1
$begingroup$
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
13 hours ago
3
$begingroup$
@Muze What is the fastest way out is almost certainly not the most fuel efficient. Venus is the place we can get for the least energy and once you can reach another world you can escape by playing billiards--but that's a slow process unless everything is neatly lined up for you like with the Voyager probes.
$endgroup$
– Loren Pechtel
9 hours ago
1
$begingroup$
How about the Voyager/Grand Tour option, where your gravity assist from Jupiter is followed by ones from Saturn, Uranus, and Neptune? AFAIK, the Voyagers didn't use fuel (other than a small amount for mid-course corrections) once they had boosted from Earth orbit. Getting to Jupiter via Venus/Earth assists might make it even more fuel-efficient.
$endgroup$
– jamesqf
5 hours ago
2
$begingroup$
@Muze, If you think that the most efficient path would be a "straight line," then you are pretty far from understanding how objects move in the Solar system. Far enough that the kind of answers that you can find on a web site like this probably are going to be less than helpful. You need some book learning or some classroom learning. Good luck!
$endgroup$
– Solomon Slow
4 hours ago
|
show 2 more comments
$begingroup$
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 13 hours ago
Steve LintonSteve Linton
8,25912246
$endgroup$
1
$begingroup$
@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
13 hours ago
1
$begingroup$
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
13 hours ago
3
$begingroup$
@Muze What is the fastest way out is almost certainly not the most fuel efficient. Venus is the place we can get for the least energy and once you can reach another world you can escape by playing billiards--but that's a slow process unless everything is neatly lined up for you like with the Voyager probes.
$endgroup$
– Loren Pechtel
9 hours ago
1
$begingroup$
How about the Voyager/Grand Tour option, where your gravity assist from Jupiter is followed by ones from Saturn, Uranus, and Neptune? AFAIK, the Voyagers didn't use fuel (other than a small amount for mid-course corrections) once they had boosted from Earth orbit. Getting to Jupiter via Venus/Earth assists might make it even more fuel-efficient.
$endgroup$
– jamesqf
5 hours ago
2
$begingroup$
@Muze, If you think that the most efficient path would be a "straight line," then you are pretty far from understanding how objects move in the Solar system. Far enough that the kind of answers that you can find on a web site like this probably are going to be less than helpful. You need some book learning or some classroom learning. Good luck!
$endgroup$
– Solomon Slow
4 hours ago
|
show 2 more comments
$begingroup$
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 13 hours ago
Steve LintonSteve Linton
8,25912246
$endgroup$
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 13 hours ago
Steve LintonSteve Linton
8,25912246
answered 13 hours ago
Steve LintonSteve Linton
8,25912246
answered 13 hours ago
Steve LintonSteve Linton
8,25912246
answered 13 hours ago
Steve LintonSteve Linton
8,25912246
8,25912246
1
$begingroup$
@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
13 hours ago
1
$begingroup$
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
13 hours ago
3
$begingroup$
@Muze What is the fastest way out is almost certainly not the most fuel efficient. Venus is the place we can get for the least energy and once you can reach another world you can escape by playing billiards--but that's a slow process unless everything is neatly lined up for you like with the Voyager probes.
$endgroup$
– Loren Pechtel
9 hours ago
1
$begingroup$
How about the Voyager/Grand Tour option, where your gravity assist from Jupiter is followed by ones from Saturn, Uranus, and Neptune? AFAIK, the Voyagers didn't use fuel (other than a small amount for mid-course corrections) once they had boosted from Earth orbit. Getting to Jupiter via Venus/Earth assists might make it even more fuel-efficient.
$endgroup$
– jamesqf
5 hours ago
2
$begingroup$
@Muze, If you think that the most efficient path would be a "straight line," then you are pretty far from understanding how objects move in the Solar system. Far enough that the kind of answers that you can find on a web site like this probably are going to be less than helpful. You need some book learning or some classroom learning. Good luck!
$endgroup$
– Solomon Slow
4 hours ago
|
show 2 more comments
1
$begingroup$
@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
13 hours ago
1
$begingroup$
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
13 hours ago
3
$begingroup$
@Muze What is the fastest way out is almost certainly not the most fuel efficient. Venus is the place we can get for the least energy and once you can reach another world you can escape by playing billiards--but that's a slow process unless everything is neatly lined up for you like with the Voyager probes.
$endgroup$
– Loren Pechtel
9 hours ago
1
$begingroup$
How about the Voyager/Grand Tour option, where your gravity assist from Jupiter is followed by ones from Saturn, Uranus, and Neptune? AFAIK, the Voyagers didn't use fuel (other than a small amount for mid-course corrections) once they had boosted from Earth orbit. Getting to Jupiter via Venus/Earth assists might make it even more fuel-efficient.
$endgroup$
– jamesqf
5 hours ago
2
$begingroup$
@Muze, If you think that the most efficient path would be a "straight line," then you are pretty far from understanding how objects move in the Solar system. Far enough that the kind of answers that you can find on a web site like this probably are going to be less than helpful. You need some book learning or some classroom learning. Good luck!
$endgroup$
– Solomon Slow
4 hours ago
1
1
$begingroup$
@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
13 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
13 hours ago
1
1
$begingroup$
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
13 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
13 hours ago
3
3
$begingroup$
@Muze What is the fastest way out is almost certainly not the most fuel efficient. Venus is the place we can get for the least energy and once you can reach another world you can escape by playing billiards--but that's a slow process unless everything is neatly lined up for you like with the Voyager probes.
$endgroup$
– Loren Pechtel
9 hours ago
$begingroup$
@Muze What is the fastest way out is almost certainly not the most fuel efficient. Venus is the place we can get for the least energy and once you can reach another world you can escape by playing billiards--but that's a slow process unless everything is neatly lined up for you like with the Voyager probes.
$endgroup$
– Loren Pechtel
9 hours ago
1
1
$begingroup$
How about the Voyager/Grand Tour option, where your gravity assist from Jupiter is followed by ones from Saturn, Uranus, and Neptune? AFAIK, the Voyagers didn't use fuel (other than a small amount for mid-course corrections) once they had boosted from Earth orbit. Getting to Jupiter via Venus/Earth assists might make it even more fuel-efficient.
$endgroup$
– jamesqf
5 hours ago
$begingroup$
How about the Voyager/Grand Tour option, where your gravity assist from Jupiter is followed by ones from Saturn, Uranus, and Neptune? AFAIK, the Voyagers didn't use fuel (other than a small amount for mid-course corrections) once they had boosted from Earth orbit. Getting to Jupiter via Venus/Earth assists might make it even more fuel-efficient.
$endgroup$
– jamesqf
5 hours ago
2
2
$begingroup$
@Muze, If you think that the most efficient path would be a "straight line," then you are pretty far from understanding how objects move in the Solar system. Far enough that the kind of answers that you can find on a web site like this probably are going to be less than helpful. You need some book learning or some classroom learning. Good luck!
$endgroup$
– Solomon Slow
4 hours ago
$begingroup$
@Muze, If you think that the most efficient path would be a "straight line," then you are pretty far from understanding how objects move in the Solar system. Far enough that the kind of answers that you can find on a web site like this probably are going to be less than helpful. You need some book learning or some classroom learning. Good luck!
$endgroup$
– Solomon Slow
4 hours ago
|
show 2 more comments
$begingroup$
That graphic is misleading. Gravity is equal in all directions, so no matter when you launch, the effort to leave the solar system remains the same.
As Steve Linton said, the most fuel-efficient way out is to get as much delta-V from gravity assists as possible.
answered 10 hours ago
HobbesHobbes
92.5k2258412
$endgroup$
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's instantaneous velocity will be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's velocity may be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
note: I'm leaving these duplicate comments in place temporarily since I've just reported this new instance.
$endgroup$
– uhoh
4 hours ago
add a comment |
$begingroup$
That graphic is misleading. Gravity is equal in all directions, so no matter when you launch, the effort to leave the solar system remains the same.
As Steve Linton said, the most fuel-efficient way out is to get as much delta-V from gravity assists as possible.
answered 10 hours ago
HobbesHobbes
92.5k2258412
$endgroup$
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's instantaneous velocity will be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's velocity may be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
note: I'm leaving these duplicate comments in place temporarily since I've just reported this new instance.
$endgroup$
– uhoh
4 hours ago
add a comment |
$begingroup$
That graphic is misleading. Gravity is equal in all directions, so no matter when you launch, the effort to leave the solar system remains the same.
As Steve Linton said, the most fuel-efficient way out is to get as much delta-V from gravity assists as possible.
answered 10 hours ago
HobbesHobbes
92.5k2258412
$endgroup$
That graphic is misleading. Gravity is equal in all directions, so no matter when you launch, the effort to leave the solar system remains the same.
As Steve Linton said, the most fuel-efficient way out is to get as much delta-V from gravity assists as possible.
answered 10 hours ago
HobbesHobbes
92.5k2258412
answered 10 hours ago
HobbesHobbes
92.5k2258412
answered 10 hours ago
HobbesHobbes
92.5k2258412
answered 10 hours ago
HobbesHobbes
92.5k2258412
92.5k2258412
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's instantaneous velocity will be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's velocity may be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
note: I'm leaving these duplicate comments in place temporarily since I've just reported this new instance.
$endgroup$
– uhoh
4 hours ago
add a comment |
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's instantaneous velocity will be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's velocity may be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
note: I'm leaving these duplicate comments in place temporarily since I've just reported this new instance.
$endgroup$
– uhoh
4 hours ago
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's instantaneous velocity will be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's instantaneous velocity will be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's velocity may be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
While "(g)ravity is equal in all directions..." for a single body, this answer suggests that a) it isn't when you consider the Sun and the Moon, and b) Earth's velocity may be important as well.
$endgroup$
– uhoh
4 hours ago
$begingroup$
note: I'm leaving these duplicate comments in place temporarily since I've just reported this new instance.
$endgroup$
– uhoh
4 hours ago
$begingroup$
note: I'm leaving these duplicate comments in place temporarily since I've just reported this new instance.
$endgroup$
– uhoh
4 hours ago
add a comment |
$begingroup$
If you want to avoid gravity assists, the most fuel-efficient way out of the Solar System is to launch due East from from a launch site in the Ecuadorean Andes, sometime before local midnight on a July 4 when there's a new moon. This gives you the maximum possible benefit from the Earth's movement, leaving only about 12,000 m/s of delta-V needed in excess of Earth escape velocity.
(Rough estimate: a Saturn V could barely put New Horizons directly into a solar escape trajectory. This is why we use gravity assists instead.)
answered 9 hours ago
MarkMark
4,0901730
$endgroup$
$begingroup$
This is intriguing! I've just asked Launching east from a mountain on the equator at midnight during a new moon; ranking of each contribution?
$endgroup$
– uhoh
4 hours ago
add a comment |
$begingroup$
If you want to avoid gravity assists, the most fuel-efficient way out of the Solar System is to launch due East from from a launch site in the Ecuadorean Andes, sometime before local midnight on a July 4 when there's a new moon. This gives you the maximum possible benefit from the Earth's movement, leaving only about 12,000 m/s of delta-V needed in excess of Earth escape velocity.
(Rough estimate: a Saturn V could barely put New Horizons directly into a solar escape trajectory. This is why we use gravity assists instead.)
answered 9 hours ago
MarkMark
4,0901730
$endgroup$
$begingroup$
This is intriguing! I've just asked Launching east from a mountain on the equator at midnight during a new moon; ranking of each contribution?
$endgroup$
– uhoh
4 hours ago
add a comment |
$begingroup$
If you want to avoid gravity assists, the most fuel-efficient way out of the Solar System is to launch due East from from a launch site in the Ecuadorean Andes, sometime before local midnight on a July 4 when there's a new moon. This gives you the maximum possible benefit from the Earth's movement, leaving only about 12,000 m/s of delta-V needed in excess of Earth escape velocity.
(Rough estimate: a Saturn V could barely put New Horizons directly into a solar escape trajectory. This is why we use gravity assists instead.)
answered 9 hours ago
MarkMark
4,0901730
$endgroup$
If you want to avoid gravity assists, the most fuel-efficient way out of the Solar System is to launch due East from from a launch site in the Ecuadorean Andes, sometime before local midnight on a July 4 when there's a new moon. This gives you the maximum possible benefit from the Earth's movement, leaving only about 12,000 m/s of delta-V needed in excess of Earth escape velocity.
(Rough estimate: a Saturn V could barely put New Horizons directly into a solar escape trajectory. This is why we use gravity assists instead.)
answered 9 hours ago
MarkMark
4,0901730
answered 9 hours ago
MarkMark
4,0901730
answered 9 hours ago
MarkMark
4,0901730
answered 9 hours ago
MarkMark
4,0901730
4,0901730
$begingroup$
This is intriguing! I've just asked Launching east from a mountain on the equator at midnight during a new moon; ranking of each contribution?
$endgroup$
– uhoh
4 hours ago
add a comment |
$begingroup$
This is intriguing! I've just asked Launching east from a mountain on the equator at midnight during a new moon; ranking of each contribution?
$endgroup$
– uhoh
4 hours ago
$begingroup$
This is intriguing! I've just asked Launching east from a mountain on the equator at midnight during a new moon; ranking of each contribution?
$endgroup$
– uhoh
4 hours ago
$begingroup$
This is intriguing! I've just asked Launching east from a mountain on the equator at midnight during a new moon; ranking of each contribution?
$endgroup$
– uhoh
4 hours ago
add a comment |
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7
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@CBredlow that is a criticism of a very different animation which is wrong for a number of reasons not applicable to this particular gif, which seems to be broadly correct (though I haven't measured the tilt of the orbital plane to confirm that it is 60 degrees). That said, our motion through the galaxy is irrelevant to escaping the solar system.
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– asgallant
9 hours ago
2
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Where would you like to go outside the solar system?
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– Thorbjørn Ravn Andersen
8 hours ago
1
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direction. "Way" implies a 'way of means' IMO. Which would use tidel influences on a machine of planetary scale. Fuel consumption: zero.
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– Mazura
7 hours ago
2
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This is the second question in a few days about escaping the solar system. I'm wondering if there's something specific I need to be worried about...
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– Richard
6 hours ago
1
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@Richard the galaxy is on Orion's belt
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– uhoh
4 hours ago