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Would a portal to space enable propulsionless orbital launch
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I'm writing a story about a race that has space travel capabilities without rocket technology, instead is using a portal to space and was wondering whether it's possible for them to put something in orbit, and more precisely:
Is it possible to launch an object into portal A inside a vacuum tube so that when it exits the portal at the point B it would enter earth orbit?
Assumptions:
- Side A of a portal is attached to Earths surface and is perpendicular to the surface.
- Side B of a portal opens in space, on a line from the center of the earth, through side A to a point outside of the atmosphere with a (possibly) arbitrary distance. It is aligned perpendicular to the Center-A-B line.
- The velocity vector of a thing entering the portal and exiting the portal is constant.
- The side B of a portal doesn't move. It's fixed in space for the amount of time it takes for the line from earth center through side A to move more than r^2 where r is the radius of the portal. (Clarification: imagine a long pole from the earth to the B side. At portal opening, pole is at -r^2 from the center of the portal and lasts until the pole hits +r^2 from the center of the portal. So if the portal is 2m in radius, the "breakpoints" are at -4m and +4m, so at 1000km the portal would last 0.1097s)
I can't get my head around the math required to compute such an outcome, and was wondering whether you can use Earths gravity to faciliate such a launch.
They use manuevering thrusters and they do have fireworks, but their rocket technology is nowhere near as sophisticated as ours. They rely on ballistics to get the job done.
@JBH: Portal doesn't transfer gravitational forces, only matter and EM radiation (thanks for the heads up on the issue)
And yes, the idea is to make the portal further than the desired orbit and use the "falling to earth" with small manuevering thrusters to "get up to speed". I plan on using the vacuum tunnel to accelerate the rocket on the ground to reduce the fuel requirements even further, because something like even 100m/s is a whole lot of energy I don't need to bring onboard.
As per restrictions, portals are perpendicular to the surface of the sphere at the point of intersection with the ground. Thus no "falling rocketry".
@MarkOlson: The potential energy is technobabbly created from unobtainium radiation. Not something I care about, because there's a lot of ways it could work (energy used to send the matter could include that delta in potential energy) amd have not settled on the way it's done. Thanks for the heads up on this issue.
spaceships orbital-mechanics teleportation
asked Jan 30 at 15:13
Gensys LTDGensys LTD
33028
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show 1 more comment
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I'm writing a story about a race that has space travel capabilities without rocket technology, instead is using a portal to space and was wondering whether it's possible for them to put something in orbit, and more precisely:
Is it possible to launch an object into portal A inside a vacuum tube so that when it exits the portal at the point B it would enter earth orbit?
Assumptions:
- Side A of a portal is attached to Earths surface and is perpendicular to the surface.
- Side B of a portal opens in space, on a line from the center of the earth, through side A to a point outside of the atmosphere with a (possibly) arbitrary distance. It is aligned perpendicular to the Center-A-B line.
- The velocity vector of a thing entering the portal and exiting the portal is constant.
- The side B of a portal doesn't move. It's fixed in space for the amount of time it takes for the line from earth center through side A to move more than r^2 where r is the radius of the portal. (Clarification: imagine a long pole from the earth to the B side. At portal opening, pole is at -r^2 from the center of the portal and lasts until the pole hits +r^2 from the center of the portal. So if the portal is 2m in radius, the "breakpoints" are at -4m and +4m, so at 1000km the portal would last 0.1097s)
I can't get my head around the math required to compute such an outcome, and was wondering whether you can use Earths gravity to faciliate such a launch.
They use manuevering thrusters and they do have fireworks, but their rocket technology is nowhere near as sophisticated as ours. They rely on ballistics to get the job done.
@JBH: Portal doesn't transfer gravitational forces, only matter and EM radiation (thanks for the heads up on the issue)
And yes, the idea is to make the portal further than the desired orbit and use the "falling to earth" with small manuevering thrusters to "get up to speed". I plan on using the vacuum tunnel to accelerate the rocket on the ground to reduce the fuel requirements even further, because something like even 100m/s is a whole lot of energy I don't need to bring onboard.
As per restrictions, portals are perpendicular to the surface of the sphere at the point of intersection with the ground. Thus no "falling rocketry".
@MarkOlson: The potential energy is technobabbly created from unobtainium radiation. Not something I care about, because there's a lot of ways it could work (energy used to send the matter could include that delta in potential energy) amd have not settled on the way it's done. Thanks for the heads up on this issue.
spaceships orbital-mechanics teleportation
asked Jan 30 at 15:13
Gensys LTDGensys LTD
33028
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Could you clarify point 4 at all, so if the radius of the portal is 2 units of distance, then the portal stays still (relative to what?) for 2^2 units of time?
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– Agrajag
Jan 30 at 15:55
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@FaySuggers Clarified, it should now be clearer.
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– Gensys LTD
Jan 30 at 16:07
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I'm not certain I understand what the portal is still relative to, I'm guessing that you mean it's no longer orbiting, but still relative to the other portal (in which case it would, at geostationary height be moving at geostationary orbital velocity) but that doesn't fit with the pole moving past it. Fixed (still) relative to what frame of reference then?
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– Agrajag
Jan 30 at 16:22
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Note that your figure is wrong. Without a considerable boost at just the right point, the orbit will never settle into a circle. Instead, it will produce an elliptical orbit, with the high point (the apoapsis) being the portal exit point. And while I've not done the math, I suspect very strongly that any such unmodified orbit will intersect the planet.
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– WhatRoughBeast
Jan 30 at 16:34
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@FaySuggers it remains stational relative to the earth but not the earths angular momentum. so it will remain in the same position relative from earth core but will not spin.
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– Gensys LTD
Jan 30 at 16:56
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show 1 more comment
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I'm writing a story about a race that has space travel capabilities without rocket technology, instead is using a portal to space and was wondering whether it's possible for them to put something in orbit, and more precisely:
Is it possible to launch an object into portal A inside a vacuum tube so that when it exits the portal at the point B it would enter earth orbit?
Assumptions:
- Side A of a portal is attached to Earths surface and is perpendicular to the surface.
- Side B of a portal opens in space, on a line from the center of the earth, through side A to a point outside of the atmosphere with a (possibly) arbitrary distance. It is aligned perpendicular to the Center-A-B line.
- The velocity vector of a thing entering the portal and exiting the portal is constant.
- The side B of a portal doesn't move. It's fixed in space for the amount of time it takes for the line from earth center through side A to move more than r^2 where r is the radius of the portal. (Clarification: imagine a long pole from the earth to the B side. At portal opening, pole is at -r^2 from the center of the portal and lasts until the pole hits +r^2 from the center of the portal. So if the portal is 2m in radius, the "breakpoints" are at -4m and +4m, so at 1000km the portal would last 0.1097s)
I can't get my head around the math required to compute such an outcome, and was wondering whether you can use Earths gravity to faciliate such a launch.
They use manuevering thrusters and they do have fireworks, but their rocket technology is nowhere near as sophisticated as ours. They rely on ballistics to get the job done.
@JBH: Portal doesn't transfer gravitational forces, only matter and EM radiation (thanks for the heads up on the issue)
And yes, the idea is to make the portal further than the desired orbit and use the "falling to earth" with small manuevering thrusters to "get up to speed". I plan on using the vacuum tunnel to accelerate the rocket on the ground to reduce the fuel requirements even further, because something like even 100m/s is a whole lot of energy I don't need to bring onboard.
As per restrictions, portals are perpendicular to the surface of the sphere at the point of intersection with the ground. Thus no "falling rocketry".
@MarkOlson: The potential energy is technobabbly created from unobtainium radiation. Not something I care about, because there's a lot of ways it could work (energy used to send the matter could include that delta in potential energy) amd have not settled on the way it's done. Thanks for the heads up on this issue.
spaceships orbital-mechanics teleportation
asked Jan 30 at 15:13
Gensys LTDGensys LTD
33028
$endgroup$
I'm writing a story about a race that has space travel capabilities without rocket technology, instead is using a portal to space and was wondering whether it's possible for them to put something in orbit, and more precisely:
Is it possible to launch an object into portal A inside a vacuum tube so that when it exits the portal at the point B it would enter earth orbit?
Assumptions:
- Side A of a portal is attached to Earths surface and is perpendicular to the surface.
- Side B of a portal opens in space, on a line from the center of the earth, through side A to a point outside of the atmosphere with a (possibly) arbitrary distance. It is aligned perpendicular to the Center-A-B line.
- The velocity vector of a thing entering the portal and exiting the portal is constant.
- The side B of a portal doesn't move. It's fixed in space for the amount of time it takes for the line from earth center through side A to move more than r^2 where r is the radius of the portal. (Clarification: imagine a long pole from the earth to the B side. At portal opening, pole is at -r^2 from the center of the portal and lasts until the pole hits +r^2 from the center of the portal. So if the portal is 2m in radius, the "breakpoints" are at -4m and +4m, so at 1000km the portal would last 0.1097s)
I can't get my head around the math required to compute such an outcome, and was wondering whether you can use Earths gravity to faciliate such a launch.
They use manuevering thrusters and they do have fireworks, but their rocket technology is nowhere near as sophisticated as ours. They rely on ballistics to get the job done.
@JBH: Portal doesn't transfer gravitational forces, only matter and EM radiation (thanks for the heads up on the issue)
And yes, the idea is to make the portal further than the desired orbit and use the "falling to earth" with small manuevering thrusters to "get up to speed". I plan on using the vacuum tunnel to accelerate the rocket on the ground to reduce the fuel requirements even further, because something like even 100m/s is a whole lot of energy I don't need to bring onboard.
As per restrictions, portals are perpendicular to the surface of the sphere at the point of intersection with the ground. Thus no "falling rocketry".
@MarkOlson: The potential energy is technobabbly created from unobtainium radiation. Not something I care about, because there's a lot of ways it could work (energy used to send the matter could include that delta in potential energy) amd have not settled on the way it's done. Thanks for the heads up on this issue.
spaceships orbital-mechanics teleportation
spaceships orbital-mechanics teleportation
asked Jan 30 at 15:13
Gensys LTDGensys LTD
33028
asked Jan 30 at 15:13
Gensys LTDGensys LTD
33028
edited Jan 30 at 22:23
Gensys LTD
asked Jan 30 at 15:13
Gensys LTDGensys LTD
33028
asked Jan 30 at 15:13
Gensys LTDGensys LTD
33028
asked Jan 30 at 15:13
Gensys LTDGensys LTD
33028
33028
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Could you clarify point 4 at all, so if the radius of the portal is 2 units of distance, then the portal stays still (relative to what?) for 2^2 units of time?
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– Agrajag
Jan 30 at 15:55
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@FaySuggers Clarified, it should now be clearer.
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– Gensys LTD
Jan 30 at 16:07
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I'm not certain I understand what the portal is still relative to, I'm guessing that you mean it's no longer orbiting, but still relative to the other portal (in which case it would, at geostationary height be moving at geostationary orbital velocity) but that doesn't fit with the pole moving past it. Fixed (still) relative to what frame of reference then?
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– Agrajag
Jan 30 at 16:22
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Note that your figure is wrong. Without a considerable boost at just the right point, the orbit will never settle into a circle. Instead, it will produce an elliptical orbit, with the high point (the apoapsis) being the portal exit point. And while I've not done the math, I suspect very strongly that any such unmodified orbit will intersect the planet.
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– WhatRoughBeast
Jan 30 at 16:34
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@FaySuggers it remains stational relative to the earth but not the earths angular momentum. so it will remain in the same position relative from earth core but will not spin.
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– Gensys LTD
Jan 30 at 16:56
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show 1 more comment
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Could you clarify point 4 at all, so if the radius of the portal is 2 units of distance, then the portal stays still (relative to what?) for 2^2 units of time?
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– Agrajag
Jan 30 at 15:55
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@FaySuggers Clarified, it should now be clearer.
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– Gensys LTD
Jan 30 at 16:07
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I'm not certain I understand what the portal is still relative to, I'm guessing that you mean it's no longer orbiting, but still relative to the other portal (in which case it would, at geostationary height be moving at geostationary orbital velocity) but that doesn't fit with the pole moving past it. Fixed (still) relative to what frame of reference then?
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– Agrajag
Jan 30 at 16:22
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Note that your figure is wrong. Without a considerable boost at just the right point, the orbit will never settle into a circle. Instead, it will produce an elliptical orbit, with the high point (the apoapsis) being the portal exit point. And while I've not done the math, I suspect very strongly that any such unmodified orbit will intersect the planet.
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– WhatRoughBeast
Jan 30 at 16:34
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@FaySuggers it remains stational relative to the earth but not the earths angular momentum. so it will remain in the same position relative from earth core but will not spin.
$endgroup$
– Gensys LTD
Jan 30 at 16:56
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Could you clarify point 4 at all, so if the radius of the portal is 2 units of distance, then the portal stays still (relative to what?) for 2^2 units of time?
$endgroup$
– Agrajag
Jan 30 at 15:55
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Could you clarify point 4 at all, so if the radius of the portal is 2 units of distance, then the portal stays still (relative to what?) for 2^2 units of time?
$endgroup$
– Agrajag
Jan 30 at 15:55
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@FaySuggers Clarified, it should now be clearer.
$endgroup$
– Gensys LTD
Jan 30 at 16:07
$begingroup$
@FaySuggers Clarified, it should now be clearer.
$endgroup$
– Gensys LTD
Jan 30 at 16:07
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I'm not certain I understand what the portal is still relative to, I'm guessing that you mean it's no longer orbiting, but still relative to the other portal (in which case it would, at geostationary height be moving at geostationary orbital velocity) but that doesn't fit with the pole moving past it. Fixed (still) relative to what frame of reference then?
$endgroup$
– Agrajag
Jan 30 at 16:22
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I'm not certain I understand what the portal is still relative to, I'm guessing that you mean it's no longer orbiting, but still relative to the other portal (in which case it would, at geostationary height be moving at geostationary orbital velocity) but that doesn't fit with the pole moving past it. Fixed (still) relative to what frame of reference then?
$endgroup$
– Agrajag
Jan 30 at 16:22
$begingroup$
Note that your figure is wrong. Without a considerable boost at just the right point, the orbit will never settle into a circle. Instead, it will produce an elliptical orbit, with the high point (the apoapsis) being the portal exit point. And while I've not done the math, I suspect very strongly that any such unmodified orbit will intersect the planet.
$endgroup$
– WhatRoughBeast
Jan 30 at 16:34
$begingroup$
Note that your figure is wrong. Without a considerable boost at just the right point, the orbit will never settle into a circle. Instead, it will produce an elliptical orbit, with the high point (the apoapsis) being the portal exit point. And while I've not done the math, I suspect very strongly that any such unmodified orbit will intersect the planet.
$endgroup$
– WhatRoughBeast
Jan 30 at 16:34
$begingroup$
@FaySuggers it remains stational relative to the earth but not the earths angular momentum. so it will remain in the same position relative from earth core but will not spin.
$endgroup$
– Gensys LTD
Jan 30 at 16:56
$begingroup$
@FaySuggers it remains stational relative to the earth but not the earths angular momentum. so it will remain in the same position relative from earth core but will not spin.
$endgroup$
– Gensys LTD
Jan 30 at 16:56
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show 1 more comment
4 Answers
4
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oldest
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AtmospheriPrisonEscape's answer in this thread gives us the equations needed to calculate velocity at any particular orbital height:
Any small mass $m$ orbiting a large mass $M$ has its centrifugal force
balancing
gravitational acceleration exactly, so that
$$g = frac{v^2}{r} $$ and the gravitational acceleration $g$ is the
result of the planetary mass $M$, gravitational constant $G$ and
distance $r$ via
$$g = frac{G M}{r^2}$$
so that any velocity that fulfills the force equality is $v^2=frac{GM}{r}$, also called Keplerian or orbital velocity.
edited for MathJax formatting error
Edit upon clearer understanding of the question:
What you'd need to do is calculate the orbital velocity at the height of the portal, subtract 460m/s (the surface velocity of rotation) and accelerate your payload till it reaches that speed before entering the portal (ie for geostationary velocity: 3.08km/s - 460M/s = 2.62 km/s). Now 2.62 km/s is freaky fast to do inside the atmosphere - our current airspeed record (in a jet) is 0.97km/s in a Lockheed SR-71 Blackbird, Ok, the space shuttle's effected re-entry much faster than this but where the atmosphere is thin.
If you're looking to accelerate the payload into a stable orbit (other than at an orbital radius greater than the moon's) You're going to need to calculate the disparity between your achieved orbital velocity at whatever velocity the payload enters the portal and the needed velocity and using F=mA in newtons, gramms and metres per second - to calculate the thrust your "firework" will need to make-up the difference. - This whilst calculating for the firework itself shedding mass as it burns.
The disparity in gravitational potential energy could perhaps be explained in term of the energy the wormhole requires to run (which would be loads) - but beware, if an object enters the orbital wormhole and exits the Earthside one you'd need to dissipate the energy somehow.
answered Jan 30 at 15:44
AgrajagAgrajag
6,94911350
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Comments are not for extended discussion; this conversation has been moved to chat.
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– L.Dutch♦
Jan 31 at 5:24
add a comment |
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Most people who use portals ignore gravity, etc.
If you have a Saturn V rocket standing on its pad and open a portal right above it...
You have a massive problem with atmospheric loss through the portal. As in catastrophic, it-makes-hurricanes-look-like-spring-showers loss. Sea-level pressure at Cape Canaveral vs. zero pressure in space, literally zero. Heck, the wind through the
vortexportal might suck the rocket through without even igniting the engines. Even if the rocket survived the transition (what happens when something strikes the edge of an infinitely-thin portal entrance? It probably cuts like a hot knife through butter...), the sudden change in surface pressure on the skin of the rocket would likely burst it like an over pressurized beer can.Your portal doesn't magically suspend the effects of gravity. In fact, opening the other end of the portal so close to Earth might have serious repercussions for Earth because suddenly there's this portal at orbital height that has the same gravity as the surface of the Earth tangential to the surface of Earth. Anything from ripping atmosphere away from the planet to ripping chunks of rock off the surface could happen. But more to the point: that Saturn V must achieve 11 kps just as it would normally for the same distance — it's just doing most of that flying in "space." The gravity well is identical, the portal just shifts the location of the well.
OK, let's ignore the effects of gravity and ignore the effects of atmospheric pressure.
- The rocket falls back to Earth in a glorious fireball because it's not traveling anywhere near fast enough to maintain orbit. Now, to be fair, you could assume the rocket burns its engines long enough to come up to speed and adjust the angle of departure for the exit of the portal because the rocket is going to start falling back to Earth and that must be accommodated. Set up the angles and the length of the engine burn properly and you can survive this.
BUT! what if you take everything I just said and designed both your rocket and your story to deal with the issues. Perhaps nobody's done that before, and it would be mega-cool. The effects of gravity may mandate a minimum distance from Earth for the portal exit, meaning rockets must swim back to Earth's orbit (NOTE: You either understand this or have an innate concept of it because your diagram shows this)... Maybe you take advantage of the pressure change to lower fuel, making your rocket lighter... etc!
One more thing...
If you have the ability to open portals, why wouldn't you open the portal on the ground, below the rocket? This would let Earth's gravity pull it through. The gravity within the circumference of the portal may be zero, but the gravity around the circumference isn't. The gravity on the other side of the portal would be wonky... so much so that I'm getting a headache trying to work out exactly what it would be... but it would likely save you a ton of fuel not having to push up to the portal.
Suddenly it makes sense to "launch" rockets upside down.
answered Jan 30 at 20:23
JBHJBH
47.9k699224
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Actually, unless the portal has some magic involved in it (or some huge power source to compensate), there is an enormous potential difference across the portal and it would take the same amount of energy to move something up, say 100 miles, through the portal as going up 100 miles the normal way. (If this wasn't the case, you'd have a perpetual motion machine.)
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– Mark Olson
Jan 30 at 20:32
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@MarkOlson 😀 I'm taking the OP at his word that portals exist. Do wormholes mathematically require that same constraint? Of course, it may not be mathematically possible to open a wormhole in a gravity well, which solves the problem. (Recognizing, of course, that wormholes are theoretical, mathematical constructs that we don't actually know anything about....)
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– JBH
Jan 30 at 20:43
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@JBH wormholes are apparently valid solutions to General Relativity, so they must maintain conservation of energy. Therefore, you'd have to put in however much energy is required to move up the gravity well to the high point, plus whatever inefficiencies you have.
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– Gryphon
Jan 30 at 21:08
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@JBH Quality answer as always. For use of the word wonky +1.
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– Agrajag
Jan 30 at 21:25
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@JBH portal is encased in thick machinery that prevents edge of the portal from coming in contact with matter. Is there a way to simulate the "swimming back" somehow, or calculate it? Would love to see how I can use earths gravity to minimize the amount of fuel I need to bring to get into orbit.
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– Gensys LTD
Jan 30 at 22:25
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Yes, but there is still no free lunch
Assuming your portal is a relativity-compliant wormhole, with one end sitting on the ground and another in orbit, it would absolutely work. You cross the ground portal at walking pace, and you emerge in orbit, drifting away from the space portal at walking pace. The portal itself is pushed in the other direction, but as it is probably much, much more massive than you, it shouldn't make much of a difference. Still, keep that in mind for orbital corrections.
However, things are never that simple.
You have to balance how much mass goes both ways. Basically, each portal gains mass when an object enters, and loose mass when an object exits. If too much mass is transferred in one direction, one of the portals will loose all its mass and the wormhole will crash.
Both ends of the wormhole are created at the same place, so you have to launch the space end the hard way. This means that unless you have already lots of stuff up there (say, a conveniently redirected asteroid), you will still have to launch the same mass in orbit, but in one go. You could use another wormhole exit already up there, but it won't help you per se, as the older portal will still loose mass equivalent to the new portal.
Also, time is passing differently for both ends, as the sky one is in orbit going faster but higher in the gravity well. As we know from atomic clocks on satellites vs on the ground, there is some drift. This means that your wormhole becomes a time machine: if there has been one second in time difference between both, crossing it will take you one second in the future or in the past.
As long as both ends are far enough, that's not a problem: even light cannot make it back fast enough to arrive before it crossed the portal. In fact, if you want to use it to go to other stars, that's a serious upside! You send it on a hyperrelativistic rocket and time compression will make the Earth-decades-long travel time into a few subjective weeks. So you can start exploring the new star system only a few weeks after launch! Sure, it is decades in the future, but you don't care as it is even further away in light-years.
Problems start once time-drifted ends move closer than that.
You may create paradoxes, but it is suspected by many that this won't actually work: if a photon can make it back to its starting point, it could theoretically follow the same path again at the same time - infinite recursion! And assuming the Universe doesn't CTD with a stack overflow (after all, it hasn't until now and at least some idiot alien has to have tried at some point), then quantum fluctuations will probably crash the wormhole instead.
What happens when a wormhole crashes? Both ends turn into black holes of their respective masses. Those are probably small enough to not start eating stuff (after all, you were using them, and a too heavy one wouldn't have been practical to keep one at the surface of the planet. But light black holes pose another risk: Hawking radiation. So your black hole will turn their entire mass into energy in short order - hours, seconds or even less. The last second will look like a giant nuclear test. Probably a subterranean one, as at this point it has probably destroyed whatever kept it from falling and it is too small to care about normal matter.
The bad news is, you don't want to do loops. You could, if you are very careful, but it's probably not worth it. Better have a star network, with you at the centre of the star, getting obscenely rich by controlling the flux of everything between the branches!
The good news is, if someone else is starting their own branch, you can reinforce your own network and throw a wormhole end their way, to create a closed timelike loop. If everything goes as planned, a wormhole somewhere in their branch will crash and you just stole half their network! Locals may be unhappy about being suddenly forty years in the future, but that's what riot control drones are for.
Dr Luke Campbell, in his hard-SF Vergeworlds setting (which I -shamelessly ripped off- was inspired by for this answer) dubbed this "causality attack". If you want to look at what wormhole use would look like, definitely check it out.
answered Jan 30 at 22:35
EthEth
2,7491721
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Consider turning your portal into a Virtual Space Elevator
If I understand your description correctly, then your portal is "stationary" relative to Earth's center, but it's moving relative to Earth's surface. It's also orbiting the sun in lockstep with earth. This means that either the portal is located in an Earth-sun Lagrange point (L1 or L2) or the portal itself is using some kind of advanced anti-gravity mechanism. However, the portal on the surface in its location relative to the surface in lockstep with Earth's rotation.
So, why not place the end of the portal in Geosynchronous orbit and so your launch window becomes 24h? Unfortunately, based on Fay Suggers' answer and comments, you don't get the "Free momentum" that you earn from a real space elevator as the payload is lifted and brought back to earth. To compensate you will need an evacuated launch tube/vac-train similar to those used in a StarTram but without the floating megastructure. (The train will still be several dozen or hundreds of miles long, in a straight line, to maintain survivable G-forces! Small, circular tracks are not survivable!) Gravity Assists against earth may also help gain some extra speed.
To return back to the surface safetly, the tried-and-true method of Atmospheric re-entry will be needed. (Again, this is based on Fay Suggers' answer and comments.)
answered Jan 30 at 23:17
stuxstux
3464
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that's because the portal is not persistent. it shows up for very short spans of time, transports the ship and then disappears. Every activation sends only one ship(or a bunch of ships but very limited in size) into space.
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– Gensys LTD
Jan 31 at 1:07
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Welp, that makes for a very specific setting. I guess this is a portal-specific rule? Getting such a frame of reference would likely be very hard to do in practice. Also, sorry, I just realized that you already mentioned vacuum tunnels in your most recent edit.
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– stux
Jan 31 at 4:12
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$begingroup$
AtmospheriPrisonEscape's answer in this thread gives us the equations needed to calculate velocity at any particular orbital height:
Any small mass $m$ orbiting a large mass $M$ has its centrifugal force
balancing
gravitational acceleration exactly, so that
$$g = frac{v^2}{r} $$ and the gravitational acceleration $g$ is the
result of the planetary mass $M$, gravitational constant $G$ and
distance $r$ via
$$g = frac{G M}{r^2}$$
so that any velocity that fulfills the force equality is $v^2=frac{GM}{r}$, also called Keplerian or orbital velocity.
edited for MathJax formatting error
Edit upon clearer understanding of the question:
What you'd need to do is calculate the orbital velocity at the height of the portal, subtract 460m/s (the surface velocity of rotation) and accelerate your payload till it reaches that speed before entering the portal (ie for geostationary velocity: 3.08km/s - 460M/s = 2.62 km/s). Now 2.62 km/s is freaky fast to do inside the atmosphere - our current airspeed record (in a jet) is 0.97km/s in a Lockheed SR-71 Blackbird, Ok, the space shuttle's effected re-entry much faster than this but where the atmosphere is thin.
If you're looking to accelerate the payload into a stable orbit (other than at an orbital radius greater than the moon's) You're going to need to calculate the disparity between your achieved orbital velocity at whatever velocity the payload enters the portal and the needed velocity and using F=mA in newtons, gramms and metres per second - to calculate the thrust your "firework" will need to make-up the difference. - This whilst calculating for the firework itself shedding mass as it burns.
The disparity in gravitational potential energy could perhaps be explained in term of the energy the wormhole requires to run (which would be loads) - but beware, if an object enters the orbital wormhole and exits the Earthside one you'd need to dissipate the energy somehow.
answered Jan 30 at 15:44
AgrajagAgrajag
6,94911350
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– L.Dutch♦
Jan 31 at 5:24
add a comment |
$begingroup$
AtmospheriPrisonEscape's answer in this thread gives us the equations needed to calculate velocity at any particular orbital height:
Any small mass $m$ orbiting a large mass $M$ has its centrifugal force
balancing
gravitational acceleration exactly, so that
$$g = frac{v^2}{r} $$ and the gravitational acceleration $g$ is the
result of the planetary mass $M$, gravitational constant $G$ and
distance $r$ via
$$g = frac{G M}{r^2}$$
so that any velocity that fulfills the force equality is $v^2=frac{GM}{r}$, also called Keplerian or orbital velocity.
edited for MathJax formatting error
Edit upon clearer understanding of the question:
What you'd need to do is calculate the orbital velocity at the height of the portal, subtract 460m/s (the surface velocity of rotation) and accelerate your payload till it reaches that speed before entering the portal (ie for geostationary velocity: 3.08km/s - 460M/s = 2.62 km/s). Now 2.62 km/s is freaky fast to do inside the atmosphere - our current airspeed record (in a jet) is 0.97km/s in a Lockheed SR-71 Blackbird, Ok, the space shuttle's effected re-entry much faster than this but where the atmosphere is thin.
If you're looking to accelerate the payload into a stable orbit (other than at an orbital radius greater than the moon's) You're going to need to calculate the disparity between your achieved orbital velocity at whatever velocity the payload enters the portal and the needed velocity and using F=mA in newtons, gramms and metres per second - to calculate the thrust your "firework" will need to make-up the difference. - This whilst calculating for the firework itself shedding mass as it burns.
The disparity in gravitational potential energy could perhaps be explained in term of the energy the wormhole requires to run (which would be loads) - but beware, if an object enters the orbital wormhole and exits the Earthside one you'd need to dissipate the energy somehow.
answered Jan 30 at 15:44
AgrajagAgrajag
6,94911350
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– L.Dutch♦
Jan 31 at 5:24
add a comment |
$begingroup$
AtmospheriPrisonEscape's answer in this thread gives us the equations needed to calculate velocity at any particular orbital height:
Any small mass $m$ orbiting a large mass $M$ has its centrifugal force
balancing
gravitational acceleration exactly, so that
$$g = frac{v^2}{r} $$ and the gravitational acceleration $g$ is the
result of the planetary mass $M$, gravitational constant $G$ and
distance $r$ via
$$g = frac{G M}{r^2}$$
so that any velocity that fulfills the force equality is $v^2=frac{GM}{r}$, also called Keplerian or orbital velocity.
edited for MathJax formatting error
Edit upon clearer understanding of the question:
What you'd need to do is calculate the orbital velocity at the height of the portal, subtract 460m/s (the surface velocity of rotation) and accelerate your payload till it reaches that speed before entering the portal (ie for geostationary velocity: 3.08km/s - 460M/s = 2.62 km/s). Now 2.62 km/s is freaky fast to do inside the atmosphere - our current airspeed record (in a jet) is 0.97km/s in a Lockheed SR-71 Blackbird, Ok, the space shuttle's effected re-entry much faster than this but where the atmosphere is thin.
If you're looking to accelerate the payload into a stable orbit (other than at an orbital radius greater than the moon's) You're going to need to calculate the disparity between your achieved orbital velocity at whatever velocity the payload enters the portal and the needed velocity and using F=mA in newtons, gramms and metres per second - to calculate the thrust your "firework" will need to make-up the difference. - This whilst calculating for the firework itself shedding mass as it burns.
The disparity in gravitational potential energy could perhaps be explained in term of the energy the wormhole requires to run (which would be loads) - but beware, if an object enters the orbital wormhole and exits the Earthside one you'd need to dissipate the energy somehow.
answered Jan 30 at 15:44
AgrajagAgrajag
6,94911350
$endgroup$
AtmospheriPrisonEscape's answer in this thread gives us the equations needed to calculate velocity at any particular orbital height:
Any small mass $m$ orbiting a large mass $M$ has its centrifugal force
balancing
gravitational acceleration exactly, so that
$$g = frac{v^2}{r} $$ and the gravitational acceleration $g$ is the
result of the planetary mass $M$, gravitational constant $G$ and
distance $r$ via
$$g = frac{G M}{r^2}$$
so that any velocity that fulfills the force equality is $v^2=frac{GM}{r}$, also called Keplerian or orbital velocity.
edited for MathJax formatting error
Edit upon clearer understanding of the question:
What you'd need to do is calculate the orbital velocity at the height of the portal, subtract 460m/s (the surface velocity of rotation) and accelerate your payload till it reaches that speed before entering the portal (ie for geostationary velocity: 3.08km/s - 460M/s = 2.62 km/s). Now 2.62 km/s is freaky fast to do inside the atmosphere - our current airspeed record (in a jet) is 0.97km/s in a Lockheed SR-71 Blackbird, Ok, the space shuttle's effected re-entry much faster than this but where the atmosphere is thin.
If you're looking to accelerate the payload into a stable orbit (other than at an orbital radius greater than the moon's) You're going to need to calculate the disparity between your achieved orbital velocity at whatever velocity the payload enters the portal and the needed velocity and using F=mA in newtons, gramms and metres per second - to calculate the thrust your "firework" will need to make-up the difference. - This whilst calculating for the firework itself shedding mass as it burns.
The disparity in gravitational potential energy could perhaps be explained in term of the energy the wormhole requires to run (which would be loads) - but beware, if an object enters the orbital wormhole and exits the Earthside one you'd need to dissipate the energy somehow.
answered Jan 30 at 15:44
AgrajagAgrajag
6,94911350
edited Jan 30 at 18:53
answered Jan 30 at 15:44
AgrajagAgrajag
6,94911350
answered Jan 30 at 15:44
AgrajagAgrajag
6,94911350
answered Jan 30 at 15:44
AgrajagAgrajag
6,94911350
6,94911350
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– L.Dutch♦
Jan 31 at 5:24
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– L.Dutch♦
Jan 31 at 5:24
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– L.Dutch♦
Jan 31 at 5:24
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– L.Dutch♦
Jan 31 at 5:24
add a comment |
$begingroup$
Most people who use portals ignore gravity, etc.
If you have a Saturn V rocket standing on its pad and open a portal right above it...
You have a massive problem with atmospheric loss through the portal. As in catastrophic, it-makes-hurricanes-look-like-spring-showers loss. Sea-level pressure at Cape Canaveral vs. zero pressure in space, literally zero. Heck, the wind through the
vortexportal might suck the rocket through without even igniting the engines. Even if the rocket survived the transition (what happens when something strikes the edge of an infinitely-thin portal entrance? It probably cuts like a hot knife through butter...), the sudden change in surface pressure on the skin of the rocket would likely burst it like an over pressurized beer can.Your portal doesn't magically suspend the effects of gravity. In fact, opening the other end of the portal so close to Earth might have serious repercussions for Earth because suddenly there's this portal at orbital height that has the same gravity as the surface of the Earth tangential to the surface of Earth. Anything from ripping atmosphere away from the planet to ripping chunks of rock off the surface could happen. But more to the point: that Saturn V must achieve 11 kps just as it would normally for the same distance — it's just doing most of that flying in "space." The gravity well is identical, the portal just shifts the location of the well.
OK, let's ignore the effects of gravity and ignore the effects of atmospheric pressure.
- The rocket falls back to Earth in a glorious fireball because it's not traveling anywhere near fast enough to maintain orbit. Now, to be fair, you could assume the rocket burns its engines long enough to come up to speed and adjust the angle of departure for the exit of the portal because the rocket is going to start falling back to Earth and that must be accommodated. Set up the angles and the length of the engine burn properly and you can survive this.
BUT! what if you take everything I just said and designed both your rocket and your story to deal with the issues. Perhaps nobody's done that before, and it would be mega-cool. The effects of gravity may mandate a minimum distance from Earth for the portal exit, meaning rockets must swim back to Earth's orbit (NOTE: You either understand this or have an innate concept of it because your diagram shows this)... Maybe you take advantage of the pressure change to lower fuel, making your rocket lighter... etc!
One more thing...
If you have the ability to open portals, why wouldn't you open the portal on the ground, below the rocket? This would let Earth's gravity pull it through. The gravity within the circumference of the portal may be zero, but the gravity around the circumference isn't. The gravity on the other side of the portal would be wonky... so much so that I'm getting a headache trying to work out exactly what it would be... but it would likely save you a ton of fuel not having to push up to the portal.
Suddenly it makes sense to "launch" rockets upside down.
answered Jan 30 at 20:23
JBHJBH
47.9k699224
$endgroup$
$begingroup$
Actually, unless the portal has some magic involved in it (or some huge power source to compensate), there is an enormous potential difference across the portal and it would take the same amount of energy to move something up, say 100 miles, through the portal as going up 100 miles the normal way. (If this wasn't the case, you'd have a perpetual motion machine.)
$endgroup$
– Mark Olson
Jan 30 at 20:32
$begingroup$
@MarkOlson 😀 I'm taking the OP at his word that portals exist. Do wormholes mathematically require that same constraint? Of course, it may not be mathematically possible to open a wormhole in a gravity well, which solves the problem. (Recognizing, of course, that wormholes are theoretical, mathematical constructs that we don't actually know anything about....)
$endgroup$
– JBH
Jan 30 at 20:43
2
$begingroup$
@JBH wormholes are apparently valid solutions to General Relativity, so they must maintain conservation of energy. Therefore, you'd have to put in however much energy is required to move up the gravity well to the high point, plus whatever inefficiencies you have.
$endgroup$
– Gryphon
Jan 30 at 21:08
$begingroup$
@JBH Quality answer as always. For use of the word wonky +1.
$endgroup$
– Agrajag
Jan 30 at 21:25
$begingroup$
@JBH portal is encased in thick machinery that prevents edge of the portal from coming in contact with matter. Is there a way to simulate the "swimming back" somehow, or calculate it? Would love to see how I can use earths gravity to minimize the amount of fuel I need to bring to get into orbit.
$endgroup$
– Gensys LTD
Jan 30 at 22:25
|
show 6 more comments
$begingroup$
Most people who use portals ignore gravity, etc.
If you have a Saturn V rocket standing on its pad and open a portal right above it...
You have a massive problem with atmospheric loss through the portal. As in catastrophic, it-makes-hurricanes-look-like-spring-showers loss. Sea-level pressure at Cape Canaveral vs. zero pressure in space, literally zero. Heck, the wind through the
vortexportal might suck the rocket through without even igniting the engines. Even if the rocket survived the transition (what happens when something strikes the edge of an infinitely-thin portal entrance? It probably cuts like a hot knife through butter...), the sudden change in surface pressure on the skin of the rocket would likely burst it like an over pressurized beer can.Your portal doesn't magically suspend the effects of gravity. In fact, opening the other end of the portal so close to Earth might have serious repercussions for Earth because suddenly there's this portal at orbital height that has the same gravity as the surface of the Earth tangential to the surface of Earth. Anything from ripping atmosphere away from the planet to ripping chunks of rock off the surface could happen. But more to the point: that Saturn V must achieve 11 kps just as it would normally for the same distance — it's just doing most of that flying in "space." The gravity well is identical, the portal just shifts the location of the well.
OK, let's ignore the effects of gravity and ignore the effects of atmospheric pressure.
- The rocket falls back to Earth in a glorious fireball because it's not traveling anywhere near fast enough to maintain orbit. Now, to be fair, you could assume the rocket burns its engines long enough to come up to speed and adjust the angle of departure for the exit of the portal because the rocket is going to start falling back to Earth and that must be accommodated. Set up the angles and the length of the engine burn properly and you can survive this.
BUT! what if you take everything I just said and designed both your rocket and your story to deal with the issues. Perhaps nobody's done that before, and it would be mega-cool. The effects of gravity may mandate a minimum distance from Earth for the portal exit, meaning rockets must swim back to Earth's orbit (NOTE: You either understand this or have an innate concept of it because your diagram shows this)... Maybe you take advantage of the pressure change to lower fuel, making your rocket lighter... etc!
One more thing...
If you have the ability to open portals, why wouldn't you open the portal on the ground, below the rocket? This would let Earth's gravity pull it through. The gravity within the circumference of the portal may be zero, but the gravity around the circumference isn't. The gravity on the other side of the portal would be wonky... so much so that I'm getting a headache trying to work out exactly what it would be... but it would likely save you a ton of fuel not having to push up to the portal.
Suddenly it makes sense to "launch" rockets upside down.
answered Jan 30 at 20:23
JBHJBH
47.9k699224
$endgroup$
$begingroup$
Actually, unless the portal has some magic involved in it (or some huge power source to compensate), there is an enormous potential difference across the portal and it would take the same amount of energy to move something up, say 100 miles, through the portal as going up 100 miles the normal way. (If this wasn't the case, you'd have a perpetual motion machine.)
$endgroup$
– Mark Olson
Jan 30 at 20:32
$begingroup$
@MarkOlson 😀 I'm taking the OP at his word that portals exist. Do wormholes mathematically require that same constraint? Of course, it may not be mathematically possible to open a wormhole in a gravity well, which solves the problem. (Recognizing, of course, that wormholes are theoretical, mathematical constructs that we don't actually know anything about....)
$endgroup$
– JBH
Jan 30 at 20:43
2
$begingroup$
@JBH wormholes are apparently valid solutions to General Relativity, so they must maintain conservation of energy. Therefore, you'd have to put in however much energy is required to move up the gravity well to the high point, plus whatever inefficiencies you have.
$endgroup$
– Gryphon
Jan 30 at 21:08
$begingroup$
@JBH Quality answer as always. For use of the word wonky +1.
$endgroup$
– Agrajag
Jan 30 at 21:25
$begingroup$
@JBH portal is encased in thick machinery that prevents edge of the portal from coming in contact with matter. Is there a way to simulate the "swimming back" somehow, or calculate it? Would love to see how I can use earths gravity to minimize the amount of fuel I need to bring to get into orbit.
$endgroup$
– Gensys LTD
Jan 30 at 22:25
|
show 6 more comments
$begingroup$
Most people who use portals ignore gravity, etc.
If you have a Saturn V rocket standing on its pad and open a portal right above it...
You have a massive problem with atmospheric loss through the portal. As in catastrophic, it-makes-hurricanes-look-like-spring-showers loss. Sea-level pressure at Cape Canaveral vs. zero pressure in space, literally zero. Heck, the wind through the
vortexportal might suck the rocket through without even igniting the engines. Even if the rocket survived the transition (what happens when something strikes the edge of an infinitely-thin portal entrance? It probably cuts like a hot knife through butter...), the sudden change in surface pressure on the skin of the rocket would likely burst it like an over pressurized beer can.Your portal doesn't magically suspend the effects of gravity. In fact, opening the other end of the portal so close to Earth might have serious repercussions for Earth because suddenly there's this portal at orbital height that has the same gravity as the surface of the Earth tangential to the surface of Earth. Anything from ripping atmosphere away from the planet to ripping chunks of rock off the surface could happen. But more to the point: that Saturn V must achieve 11 kps just as it would normally for the same distance — it's just doing most of that flying in "space." The gravity well is identical, the portal just shifts the location of the well.
OK, let's ignore the effects of gravity and ignore the effects of atmospheric pressure.
- The rocket falls back to Earth in a glorious fireball because it's not traveling anywhere near fast enough to maintain orbit. Now, to be fair, you could assume the rocket burns its engines long enough to come up to speed and adjust the angle of departure for the exit of the portal because the rocket is going to start falling back to Earth and that must be accommodated. Set up the angles and the length of the engine burn properly and you can survive this.
BUT! what if you take everything I just said and designed both your rocket and your story to deal with the issues. Perhaps nobody's done that before, and it would be mega-cool. The effects of gravity may mandate a minimum distance from Earth for the portal exit, meaning rockets must swim back to Earth's orbit (NOTE: You either understand this or have an innate concept of it because your diagram shows this)... Maybe you take advantage of the pressure change to lower fuel, making your rocket lighter... etc!
One more thing...
If you have the ability to open portals, why wouldn't you open the portal on the ground, below the rocket? This would let Earth's gravity pull it through. The gravity within the circumference of the portal may be zero, but the gravity around the circumference isn't. The gravity on the other side of the portal would be wonky... so much so that I'm getting a headache trying to work out exactly what it would be... but it would likely save you a ton of fuel not having to push up to the portal.
Suddenly it makes sense to "launch" rockets upside down.
answered Jan 30 at 20:23
JBHJBH
47.9k699224
$endgroup$
Most people who use portals ignore gravity, etc.
If you have a Saturn V rocket standing on its pad and open a portal right above it...
You have a massive problem with atmospheric loss through the portal. As in catastrophic, it-makes-hurricanes-look-like-spring-showers loss. Sea-level pressure at Cape Canaveral vs. zero pressure in space, literally zero. Heck, the wind through the
vortexportal might suck the rocket through without even igniting the engines. Even if the rocket survived the transition (what happens when something strikes the edge of an infinitely-thin portal entrance? It probably cuts like a hot knife through butter...), the sudden change in surface pressure on the skin of the rocket would likely burst it like an over pressurized beer can.Your portal doesn't magically suspend the effects of gravity. In fact, opening the other end of the portal so close to Earth might have serious repercussions for Earth because suddenly there's this portal at orbital height that has the same gravity as the surface of the Earth tangential to the surface of Earth. Anything from ripping atmosphere away from the planet to ripping chunks of rock off the surface could happen. But more to the point: that Saturn V must achieve 11 kps just as it would normally for the same distance — it's just doing most of that flying in "space." The gravity well is identical, the portal just shifts the location of the well.
OK, let's ignore the effects of gravity and ignore the effects of atmospheric pressure.
- The rocket falls back to Earth in a glorious fireball because it's not traveling anywhere near fast enough to maintain orbit. Now, to be fair, you could assume the rocket burns its engines long enough to come up to speed and adjust the angle of departure for the exit of the portal because the rocket is going to start falling back to Earth and that must be accommodated. Set up the angles and the length of the engine burn properly and you can survive this.
BUT! what if you take everything I just said and designed both your rocket and your story to deal with the issues. Perhaps nobody's done that before, and it would be mega-cool. The effects of gravity may mandate a minimum distance from Earth for the portal exit, meaning rockets must swim back to Earth's orbit (NOTE: You either understand this or have an innate concept of it because your diagram shows this)... Maybe you take advantage of the pressure change to lower fuel, making your rocket lighter... etc!
One more thing...
If you have the ability to open portals, why wouldn't you open the portal on the ground, below the rocket? This would let Earth's gravity pull it through. The gravity within the circumference of the portal may be zero, but the gravity around the circumference isn't. The gravity on the other side of the portal would be wonky... so much so that I'm getting a headache trying to work out exactly what it would be... but it would likely save you a ton of fuel not having to push up to the portal.
Suddenly it makes sense to "launch" rockets upside down.
answered Jan 30 at 20:23
JBHJBH
47.9k699224
edited Jan 30 at 20:28
answered Jan 30 at 20:23
JBHJBH
47.9k699224
answered Jan 30 at 20:23
JBHJBH
47.9k699224
answered Jan 30 at 20:23
JBHJBH
47.9k699224
47.9k699224
$begingroup$
Actually, unless the portal has some magic involved in it (or some huge power source to compensate), there is an enormous potential difference across the portal and it would take the same amount of energy to move something up, say 100 miles, through the portal as going up 100 miles the normal way. (If this wasn't the case, you'd have a perpetual motion machine.)
$endgroup$
– Mark Olson
Jan 30 at 20:32
$begingroup$
@MarkOlson 😀 I'm taking the OP at his word that portals exist. Do wormholes mathematically require that same constraint? Of course, it may not be mathematically possible to open a wormhole in a gravity well, which solves the problem. (Recognizing, of course, that wormholes are theoretical, mathematical constructs that we don't actually know anything about....)
$endgroup$
– JBH
Jan 30 at 20:43
2
$begingroup$
@JBH wormholes are apparently valid solutions to General Relativity, so they must maintain conservation of energy. Therefore, you'd have to put in however much energy is required to move up the gravity well to the high point, plus whatever inefficiencies you have.
$endgroup$
– Gryphon
Jan 30 at 21:08
$begingroup$
@JBH Quality answer as always. For use of the word wonky +1.
$endgroup$
– Agrajag
Jan 30 at 21:25
$begingroup$
@JBH portal is encased in thick machinery that prevents edge of the portal from coming in contact with matter. Is there a way to simulate the "swimming back" somehow, or calculate it? Would love to see how I can use earths gravity to minimize the amount of fuel I need to bring to get into orbit.
$endgroup$
– Gensys LTD
Jan 30 at 22:25
|
show 6 more comments
$begingroup$
Actually, unless the portal has some magic involved in it (or some huge power source to compensate), there is an enormous potential difference across the portal and it would take the same amount of energy to move something up, say 100 miles, through the portal as going up 100 miles the normal way. (If this wasn't the case, you'd have a perpetual motion machine.)
$endgroup$
– Mark Olson
Jan 30 at 20:32
$begingroup$
@MarkOlson 😀 I'm taking the OP at his word that portals exist. Do wormholes mathematically require that same constraint? Of course, it may not be mathematically possible to open a wormhole in a gravity well, which solves the problem. (Recognizing, of course, that wormholes are theoretical, mathematical constructs that we don't actually know anything about....)
$endgroup$
– JBH
Jan 30 at 20:43
2
$begingroup$
@JBH wormholes are apparently valid solutions to General Relativity, so they must maintain conservation of energy. Therefore, you'd have to put in however much energy is required to move up the gravity well to the high point, plus whatever inefficiencies you have.
$endgroup$
– Gryphon
Jan 30 at 21:08
$begingroup$
@JBH Quality answer as always. For use of the word wonky +1.
$endgroup$
– Agrajag
Jan 30 at 21:25
$begingroup$
@JBH portal is encased in thick machinery that prevents edge of the portal from coming in contact with matter. Is there a way to simulate the "swimming back" somehow, or calculate it? Would love to see how I can use earths gravity to minimize the amount of fuel I need to bring to get into orbit.
$endgroup$
– Gensys LTD
Jan 30 at 22:25
$begingroup$
Actually, unless the portal has some magic involved in it (or some huge power source to compensate), there is an enormous potential difference across the portal and it would take the same amount of energy to move something up, say 100 miles, through the portal as going up 100 miles the normal way. (If this wasn't the case, you'd have a perpetual motion machine.)
$endgroup$
– Mark Olson
Jan 30 at 20:32
$begingroup$
Actually, unless the portal has some magic involved in it (or some huge power source to compensate), there is an enormous potential difference across the portal and it would take the same amount of energy to move something up, say 100 miles, through the portal as going up 100 miles the normal way. (If this wasn't the case, you'd have a perpetual motion machine.)
$endgroup$
– Mark Olson
Jan 30 at 20:32
$begingroup$
@MarkOlson 😀 I'm taking the OP at his word that portals exist. Do wormholes mathematically require that same constraint? Of course, it may not be mathematically possible to open a wormhole in a gravity well, which solves the problem. (Recognizing, of course, that wormholes are theoretical, mathematical constructs that we don't actually know anything about....)
$endgroup$
– JBH
Jan 30 at 20:43
$begingroup$
@MarkOlson 😀 I'm taking the OP at his word that portals exist. Do wormholes mathematically require that same constraint? Of course, it may not be mathematically possible to open a wormhole in a gravity well, which solves the problem. (Recognizing, of course, that wormholes are theoretical, mathematical constructs that we don't actually know anything about....)
$endgroup$
– JBH
Jan 30 at 20:43
2
2
$begingroup$
@JBH wormholes are apparently valid solutions to General Relativity, so they must maintain conservation of energy. Therefore, you'd have to put in however much energy is required to move up the gravity well to the high point, plus whatever inefficiencies you have.
$endgroup$
– Gryphon
Jan 30 at 21:08
$begingroup$
@JBH wormholes are apparently valid solutions to General Relativity, so they must maintain conservation of energy. Therefore, you'd have to put in however much energy is required to move up the gravity well to the high point, plus whatever inefficiencies you have.
$endgroup$
– Gryphon
Jan 30 at 21:08
$begingroup$
@JBH Quality answer as always. For use of the word wonky +1.
$endgroup$
– Agrajag
Jan 30 at 21:25
$begingroup$
@JBH Quality answer as always. For use of the word wonky +1.
$endgroup$
– Agrajag
Jan 30 at 21:25
$begingroup$
@JBH portal is encased in thick machinery that prevents edge of the portal from coming in contact with matter. Is there a way to simulate the "swimming back" somehow, or calculate it? Would love to see how I can use earths gravity to minimize the amount of fuel I need to bring to get into orbit.
$endgroup$
– Gensys LTD
Jan 30 at 22:25
$begingroup$
@JBH portal is encased in thick machinery that prevents edge of the portal from coming in contact with matter. Is there a way to simulate the "swimming back" somehow, or calculate it? Would love to see how I can use earths gravity to minimize the amount of fuel I need to bring to get into orbit.
$endgroup$
– Gensys LTD
Jan 30 at 22:25
|
show 6 more comments
$begingroup$
Yes, but there is still no free lunch
Assuming your portal is a relativity-compliant wormhole, with one end sitting on the ground and another in orbit, it would absolutely work. You cross the ground portal at walking pace, and you emerge in orbit, drifting away from the space portal at walking pace. The portal itself is pushed in the other direction, but as it is probably much, much more massive than you, it shouldn't make much of a difference. Still, keep that in mind for orbital corrections.
However, things are never that simple.
You have to balance how much mass goes both ways. Basically, each portal gains mass when an object enters, and loose mass when an object exits. If too much mass is transferred in one direction, one of the portals will loose all its mass and the wormhole will crash.
Both ends of the wormhole are created at the same place, so you have to launch the space end the hard way. This means that unless you have already lots of stuff up there (say, a conveniently redirected asteroid), you will still have to launch the same mass in orbit, but in one go. You could use another wormhole exit already up there, but it won't help you per se, as the older portal will still loose mass equivalent to the new portal.
Also, time is passing differently for both ends, as the sky one is in orbit going faster but higher in the gravity well. As we know from atomic clocks on satellites vs on the ground, there is some drift. This means that your wormhole becomes a time machine: if there has been one second in time difference between both, crossing it will take you one second in the future or in the past.
As long as both ends are far enough, that's not a problem: even light cannot make it back fast enough to arrive before it crossed the portal. In fact, if you want to use it to go to other stars, that's a serious upside! You send it on a hyperrelativistic rocket and time compression will make the Earth-decades-long travel time into a few subjective weeks. So you can start exploring the new star system only a few weeks after launch! Sure, it is decades in the future, but you don't care as it is even further away in light-years.
Problems start once time-drifted ends move closer than that.
You may create paradoxes, but it is suspected by many that this won't actually work: if a photon can make it back to its starting point, it could theoretically follow the same path again at the same time - infinite recursion! And assuming the Universe doesn't CTD with a stack overflow (after all, it hasn't until now and at least some idiot alien has to have tried at some point), then quantum fluctuations will probably crash the wormhole instead.
What happens when a wormhole crashes? Both ends turn into black holes of their respective masses. Those are probably small enough to not start eating stuff (after all, you were using them, and a too heavy one wouldn't have been practical to keep one at the surface of the planet. But light black holes pose another risk: Hawking radiation. So your black hole will turn their entire mass into energy in short order - hours, seconds or even less. The last second will look like a giant nuclear test. Probably a subterranean one, as at this point it has probably destroyed whatever kept it from falling and it is too small to care about normal matter.
The bad news is, you don't want to do loops. You could, if you are very careful, but it's probably not worth it. Better have a star network, with you at the centre of the star, getting obscenely rich by controlling the flux of everything between the branches!
The good news is, if someone else is starting their own branch, you can reinforce your own network and throw a wormhole end their way, to create a closed timelike loop. If everything goes as planned, a wormhole somewhere in their branch will crash and you just stole half their network! Locals may be unhappy about being suddenly forty years in the future, but that's what riot control drones are for.
Dr Luke Campbell, in his hard-SF Vergeworlds setting (which I -shamelessly ripped off- was inspired by for this answer) dubbed this "causality attack". If you want to look at what wormhole use would look like, definitely check it out.
answered Jan 30 at 22:35
EthEth
2,7491721
$endgroup$
add a comment |
$begingroup$
Yes, but there is still no free lunch
Assuming your portal is a relativity-compliant wormhole, with one end sitting on the ground and another in orbit, it would absolutely work. You cross the ground portal at walking pace, and you emerge in orbit, drifting away from the space portal at walking pace. The portal itself is pushed in the other direction, but as it is probably much, much more massive than you, it shouldn't make much of a difference. Still, keep that in mind for orbital corrections.
However, things are never that simple.
You have to balance how much mass goes both ways. Basically, each portal gains mass when an object enters, and loose mass when an object exits. If too much mass is transferred in one direction, one of the portals will loose all its mass and the wormhole will crash.
Both ends of the wormhole are created at the same place, so you have to launch the space end the hard way. This means that unless you have already lots of stuff up there (say, a conveniently redirected asteroid), you will still have to launch the same mass in orbit, but in one go. You could use another wormhole exit already up there, but it won't help you per se, as the older portal will still loose mass equivalent to the new portal.
Also, time is passing differently for both ends, as the sky one is in orbit going faster but higher in the gravity well. As we know from atomic clocks on satellites vs on the ground, there is some drift. This means that your wormhole becomes a time machine: if there has been one second in time difference between both, crossing it will take you one second in the future or in the past.
As long as both ends are far enough, that's not a problem: even light cannot make it back fast enough to arrive before it crossed the portal. In fact, if you want to use it to go to other stars, that's a serious upside! You send it on a hyperrelativistic rocket and time compression will make the Earth-decades-long travel time into a few subjective weeks. So you can start exploring the new star system only a few weeks after launch! Sure, it is decades in the future, but you don't care as it is even further away in light-years.
Problems start once time-drifted ends move closer than that.
You may create paradoxes, but it is suspected by many that this won't actually work: if a photon can make it back to its starting point, it could theoretically follow the same path again at the same time - infinite recursion! And assuming the Universe doesn't CTD with a stack overflow (after all, it hasn't until now and at least some idiot alien has to have tried at some point), then quantum fluctuations will probably crash the wormhole instead.
What happens when a wormhole crashes? Both ends turn into black holes of their respective masses. Those are probably small enough to not start eating stuff (after all, you were using them, and a too heavy one wouldn't have been practical to keep one at the surface of the planet. But light black holes pose another risk: Hawking radiation. So your black hole will turn their entire mass into energy in short order - hours, seconds or even less. The last second will look like a giant nuclear test. Probably a subterranean one, as at this point it has probably destroyed whatever kept it from falling and it is too small to care about normal matter.
The bad news is, you don't want to do loops. You could, if you are very careful, but it's probably not worth it. Better have a star network, with you at the centre of the star, getting obscenely rich by controlling the flux of everything between the branches!
The good news is, if someone else is starting their own branch, you can reinforce your own network and throw a wormhole end their way, to create a closed timelike loop. If everything goes as planned, a wormhole somewhere in their branch will crash and you just stole half their network! Locals may be unhappy about being suddenly forty years in the future, but that's what riot control drones are for.
Dr Luke Campbell, in his hard-SF Vergeworlds setting (which I -shamelessly ripped off- was inspired by for this answer) dubbed this "causality attack". If you want to look at what wormhole use would look like, definitely check it out.
answered Jan 30 at 22:35
EthEth
2,7491721
$endgroup$
add a comment |
$begingroup$
Yes, but there is still no free lunch
Assuming your portal is a relativity-compliant wormhole, with one end sitting on the ground and another in orbit, it would absolutely work. You cross the ground portal at walking pace, and you emerge in orbit, drifting away from the space portal at walking pace. The portal itself is pushed in the other direction, but as it is probably much, much more massive than you, it shouldn't make much of a difference. Still, keep that in mind for orbital corrections.
However, things are never that simple.
You have to balance how much mass goes both ways. Basically, each portal gains mass when an object enters, and loose mass when an object exits. If too much mass is transferred in one direction, one of the portals will loose all its mass and the wormhole will crash.
Both ends of the wormhole are created at the same place, so you have to launch the space end the hard way. This means that unless you have already lots of stuff up there (say, a conveniently redirected asteroid), you will still have to launch the same mass in orbit, but in one go. You could use another wormhole exit already up there, but it won't help you per se, as the older portal will still loose mass equivalent to the new portal.
Also, time is passing differently for both ends, as the sky one is in orbit going faster but higher in the gravity well. As we know from atomic clocks on satellites vs on the ground, there is some drift. This means that your wormhole becomes a time machine: if there has been one second in time difference between both, crossing it will take you one second in the future or in the past.
As long as both ends are far enough, that's not a problem: even light cannot make it back fast enough to arrive before it crossed the portal. In fact, if you want to use it to go to other stars, that's a serious upside! You send it on a hyperrelativistic rocket and time compression will make the Earth-decades-long travel time into a few subjective weeks. So you can start exploring the new star system only a few weeks after launch! Sure, it is decades in the future, but you don't care as it is even further away in light-years.
Problems start once time-drifted ends move closer than that.
You may create paradoxes, but it is suspected by many that this won't actually work: if a photon can make it back to its starting point, it could theoretically follow the same path again at the same time - infinite recursion! And assuming the Universe doesn't CTD with a stack overflow (after all, it hasn't until now and at least some idiot alien has to have tried at some point), then quantum fluctuations will probably crash the wormhole instead.
What happens when a wormhole crashes? Both ends turn into black holes of their respective masses. Those are probably small enough to not start eating stuff (after all, you were using them, and a too heavy one wouldn't have been practical to keep one at the surface of the planet. But light black holes pose another risk: Hawking radiation. So your black hole will turn their entire mass into energy in short order - hours, seconds or even less. The last second will look like a giant nuclear test. Probably a subterranean one, as at this point it has probably destroyed whatever kept it from falling and it is too small to care about normal matter.
The bad news is, you don't want to do loops. You could, if you are very careful, but it's probably not worth it. Better have a star network, with you at the centre of the star, getting obscenely rich by controlling the flux of everything between the branches!
The good news is, if someone else is starting their own branch, you can reinforce your own network and throw a wormhole end their way, to create a closed timelike loop. If everything goes as planned, a wormhole somewhere in their branch will crash and you just stole half their network! Locals may be unhappy about being suddenly forty years in the future, but that's what riot control drones are for.
Dr Luke Campbell, in his hard-SF Vergeworlds setting (which I -shamelessly ripped off- was inspired by for this answer) dubbed this "causality attack". If you want to look at what wormhole use would look like, definitely check it out.
answered Jan 30 at 22:35
EthEth
2,7491721
$endgroup$
Yes, but there is still no free lunch
Assuming your portal is a relativity-compliant wormhole, with one end sitting on the ground and another in orbit, it would absolutely work. You cross the ground portal at walking pace, and you emerge in orbit, drifting away from the space portal at walking pace. The portal itself is pushed in the other direction, but as it is probably much, much more massive than you, it shouldn't make much of a difference. Still, keep that in mind for orbital corrections.
However, things are never that simple.
You have to balance how much mass goes both ways. Basically, each portal gains mass when an object enters, and loose mass when an object exits. If too much mass is transferred in one direction, one of the portals will loose all its mass and the wormhole will crash.
Both ends of the wormhole are created at the same place, so you have to launch the space end the hard way. This means that unless you have already lots of stuff up there (say, a conveniently redirected asteroid), you will still have to launch the same mass in orbit, but in one go. You could use another wormhole exit already up there, but it won't help you per se, as the older portal will still loose mass equivalent to the new portal.
Also, time is passing differently for both ends, as the sky one is in orbit going faster but higher in the gravity well. As we know from atomic clocks on satellites vs on the ground, there is some drift. This means that your wormhole becomes a time machine: if there has been one second in time difference between both, crossing it will take you one second in the future or in the past.
As long as both ends are far enough, that's not a problem: even light cannot make it back fast enough to arrive before it crossed the portal. In fact, if you want to use it to go to other stars, that's a serious upside! You send it on a hyperrelativistic rocket and time compression will make the Earth-decades-long travel time into a few subjective weeks. So you can start exploring the new star system only a few weeks after launch! Sure, it is decades in the future, but you don't care as it is even further away in light-years.
Problems start once time-drifted ends move closer than that.
You may create paradoxes, but it is suspected by many that this won't actually work: if a photon can make it back to its starting point, it could theoretically follow the same path again at the same time - infinite recursion! And assuming the Universe doesn't CTD with a stack overflow (after all, it hasn't until now and at least some idiot alien has to have tried at some point), then quantum fluctuations will probably crash the wormhole instead.
What happens when a wormhole crashes? Both ends turn into black holes of their respective masses. Those are probably small enough to not start eating stuff (after all, you were using them, and a too heavy one wouldn't have been practical to keep one at the surface of the planet. But light black holes pose another risk: Hawking radiation. So your black hole will turn their entire mass into energy in short order - hours, seconds or even less. The last second will look like a giant nuclear test. Probably a subterranean one, as at this point it has probably destroyed whatever kept it from falling and it is too small to care about normal matter.
The bad news is, you don't want to do loops. You could, if you are very careful, but it's probably not worth it. Better have a star network, with you at the centre of the star, getting obscenely rich by controlling the flux of everything between the branches!
The good news is, if someone else is starting their own branch, you can reinforce your own network and throw a wormhole end their way, to create a closed timelike loop. If everything goes as planned, a wormhole somewhere in their branch will crash and you just stole half their network! Locals may be unhappy about being suddenly forty years in the future, but that's what riot control drones are for.
Dr Luke Campbell, in his hard-SF Vergeworlds setting (which I -shamelessly ripped off- was inspired by for this answer) dubbed this "causality attack". If you want to look at what wormhole use would look like, definitely check it out.
answered Jan 30 at 22:35
EthEth
2,7491721
answered Jan 30 at 22:35
EthEth
2,7491721
answered Jan 30 at 22:35
EthEth
2,7491721
answered Jan 30 at 22:35
EthEth
2,7491721
2,7491721
add a comment |
add a comment |
$begingroup$
Consider turning your portal into a Virtual Space Elevator
If I understand your description correctly, then your portal is "stationary" relative to Earth's center, but it's moving relative to Earth's surface. It's also orbiting the sun in lockstep with earth. This means that either the portal is located in an Earth-sun Lagrange point (L1 or L2) or the portal itself is using some kind of advanced anti-gravity mechanism. However, the portal on the surface in its location relative to the surface in lockstep with Earth's rotation.
So, why not place the end of the portal in Geosynchronous orbit and so your launch window becomes 24h? Unfortunately, based on Fay Suggers' answer and comments, you don't get the "Free momentum" that you earn from a real space elevator as the payload is lifted and brought back to earth. To compensate you will need an evacuated launch tube/vac-train similar to those used in a StarTram but without the floating megastructure. (The train will still be several dozen or hundreds of miles long, in a straight line, to maintain survivable G-forces! Small, circular tracks are not survivable!) Gravity Assists against earth may also help gain some extra speed.
To return back to the surface safetly, the tried-and-true method of Atmospheric re-entry will be needed. (Again, this is based on Fay Suggers' answer and comments.)
answered Jan 30 at 23:17
stuxstux
3464
$endgroup$
$begingroup$
that's because the portal is not persistent. it shows up for very short spans of time, transports the ship and then disappears. Every activation sends only one ship(or a bunch of ships but very limited in size) into space.
$endgroup$
– Gensys LTD
Jan 31 at 1:07
$begingroup$
Welp, that makes for a very specific setting. I guess this is a portal-specific rule? Getting such a frame of reference would likely be very hard to do in practice. Also, sorry, I just realized that you already mentioned vacuum tunnels in your most recent edit.
$endgroup$
– stux
Jan 31 at 4:12
add a comment |
$begingroup$
Consider turning your portal into a Virtual Space Elevator
If I understand your description correctly, then your portal is "stationary" relative to Earth's center, but it's moving relative to Earth's surface. It's also orbiting the sun in lockstep with earth. This means that either the portal is located in an Earth-sun Lagrange point (L1 or L2) or the portal itself is using some kind of advanced anti-gravity mechanism. However, the portal on the surface in its location relative to the surface in lockstep with Earth's rotation.
So, why not place the end of the portal in Geosynchronous orbit and so your launch window becomes 24h? Unfortunately, based on Fay Suggers' answer and comments, you don't get the "Free momentum" that you earn from a real space elevator as the payload is lifted and brought back to earth. To compensate you will need an evacuated launch tube/vac-train similar to those used in a StarTram but without the floating megastructure. (The train will still be several dozen or hundreds of miles long, in a straight line, to maintain survivable G-forces! Small, circular tracks are not survivable!) Gravity Assists against earth may also help gain some extra speed.
To return back to the surface safetly, the tried-and-true method of Atmospheric re-entry will be needed. (Again, this is based on Fay Suggers' answer and comments.)
answered Jan 30 at 23:17
stuxstux
3464
$endgroup$
$begingroup$
that's because the portal is not persistent. it shows up for very short spans of time, transports the ship and then disappears. Every activation sends only one ship(or a bunch of ships but very limited in size) into space.
$endgroup$
– Gensys LTD
Jan 31 at 1:07
$begingroup$
Welp, that makes for a very specific setting. I guess this is a portal-specific rule? Getting such a frame of reference would likely be very hard to do in practice. Also, sorry, I just realized that you already mentioned vacuum tunnels in your most recent edit.
$endgroup$
– stux
Jan 31 at 4:12
add a comment |
$begingroup$
Consider turning your portal into a Virtual Space Elevator
If I understand your description correctly, then your portal is "stationary" relative to Earth's center, but it's moving relative to Earth's surface. It's also orbiting the sun in lockstep with earth. This means that either the portal is located in an Earth-sun Lagrange point (L1 or L2) or the portal itself is using some kind of advanced anti-gravity mechanism. However, the portal on the surface in its location relative to the surface in lockstep with Earth's rotation.
So, why not place the end of the portal in Geosynchronous orbit and so your launch window becomes 24h? Unfortunately, based on Fay Suggers' answer and comments, you don't get the "Free momentum" that you earn from a real space elevator as the payload is lifted and brought back to earth. To compensate you will need an evacuated launch tube/vac-train similar to those used in a StarTram but without the floating megastructure. (The train will still be several dozen or hundreds of miles long, in a straight line, to maintain survivable G-forces! Small, circular tracks are not survivable!) Gravity Assists against earth may also help gain some extra speed.
To return back to the surface safetly, the tried-and-true method of Atmospheric re-entry will be needed. (Again, this is based on Fay Suggers' answer and comments.)
answered Jan 30 at 23:17
stuxstux
3464
$endgroup$
Consider turning your portal into a Virtual Space Elevator
If I understand your description correctly, then your portal is "stationary" relative to Earth's center, but it's moving relative to Earth's surface. It's also orbiting the sun in lockstep with earth. This means that either the portal is located in an Earth-sun Lagrange point (L1 or L2) or the portal itself is using some kind of advanced anti-gravity mechanism. However, the portal on the surface in its location relative to the surface in lockstep with Earth's rotation.
So, why not place the end of the portal in Geosynchronous orbit and so your launch window becomes 24h? Unfortunately, based on Fay Suggers' answer and comments, you don't get the "Free momentum" that you earn from a real space elevator as the payload is lifted and brought back to earth. To compensate you will need an evacuated launch tube/vac-train similar to those used in a StarTram but without the floating megastructure. (The train will still be several dozen or hundreds of miles long, in a straight line, to maintain survivable G-forces! Small, circular tracks are not survivable!) Gravity Assists against earth may also help gain some extra speed.
To return back to the surface safetly, the tried-and-true method of Atmospheric re-entry will be needed. (Again, this is based on Fay Suggers' answer and comments.)
answered Jan 30 at 23:17
stuxstux
3464
edited Jan 30 at 23:23
answered Jan 30 at 23:17
stuxstux
3464
answered Jan 30 at 23:17
stuxstux
3464
answered Jan 30 at 23:17
stuxstux
3464
3464
$begingroup$
that's because the portal is not persistent. it shows up for very short spans of time, transports the ship and then disappears. Every activation sends only one ship(or a bunch of ships but very limited in size) into space.
$endgroup$
– Gensys LTD
Jan 31 at 1:07
$begingroup$
Welp, that makes for a very specific setting. I guess this is a portal-specific rule? Getting such a frame of reference would likely be very hard to do in practice. Also, sorry, I just realized that you already mentioned vacuum tunnels in your most recent edit.
$endgroup$
– stux
Jan 31 at 4:12
add a comment |
$begingroup$
that's because the portal is not persistent. it shows up for very short spans of time, transports the ship and then disappears. Every activation sends only one ship(or a bunch of ships but very limited in size) into space.
$endgroup$
– Gensys LTD
Jan 31 at 1:07
$begingroup$
Welp, that makes for a very specific setting. I guess this is a portal-specific rule? Getting such a frame of reference would likely be very hard to do in practice. Also, sorry, I just realized that you already mentioned vacuum tunnels in your most recent edit.
$endgroup$
– stux
Jan 31 at 4:12
$begingroup$
that's because the portal is not persistent. it shows up for very short spans of time, transports the ship and then disappears. Every activation sends only one ship(or a bunch of ships but very limited in size) into space.
$endgroup$
– Gensys LTD
Jan 31 at 1:07
$begingroup$
that's because the portal is not persistent. it shows up for very short spans of time, transports the ship and then disappears. Every activation sends only one ship(or a bunch of ships but very limited in size) into space.
$endgroup$
– Gensys LTD
Jan 31 at 1:07
$begingroup$
Welp, that makes for a very specific setting. I guess this is a portal-specific rule? Getting such a frame of reference would likely be very hard to do in practice. Also, sorry, I just realized that you already mentioned vacuum tunnels in your most recent edit.
$endgroup$
– stux
Jan 31 at 4:12
$begingroup$
Welp, that makes for a very specific setting. I guess this is a portal-specific rule? Getting such a frame of reference would likely be very hard to do in practice. Also, sorry, I just realized that you already mentioned vacuum tunnels in your most recent edit.
$endgroup$
– stux
Jan 31 at 4:12
add a comment |
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$begingroup$
Could you clarify point 4 at all, so if the radius of the portal is 2 units of distance, then the portal stays still (relative to what?) for 2^2 units of time?
$endgroup$
– Agrajag
Jan 30 at 15:55
$begingroup$
@FaySuggers Clarified, it should now be clearer.
$endgroup$
– Gensys LTD
Jan 30 at 16:07
$begingroup$
I'm not certain I understand what the portal is still relative to, I'm guessing that you mean it's no longer orbiting, but still relative to the other portal (in which case it would, at geostationary height be moving at geostationary orbital velocity) but that doesn't fit with the pole moving past it. Fixed (still) relative to what frame of reference then?
$endgroup$
– Agrajag
Jan 30 at 16:22
$begingroup$
Note that your figure is wrong. Without a considerable boost at just the right point, the orbit will never settle into a circle. Instead, it will produce an elliptical orbit, with the high point (the apoapsis) being the portal exit point. And while I've not done the math, I suspect very strongly that any such unmodified orbit will intersect the planet.
$endgroup$
– WhatRoughBeast
Jan 30 at 16:34
$begingroup$
@FaySuggers it remains stational relative to the earth but not the earths angular momentum. so it will remain in the same position relative from earth core but will not spin.
$endgroup$
– Gensys LTD
Jan 30 at 16:56