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Proxima Centauri in 25 years

From what I understand, current theory is that most planets form from an accretion disk early in a star's own formation, dust and gas coalescing into planets on a consistent plane.

It's not impossible to have orbits at other inclinations, for instance comets which form far out beyond the normal area we think of as the solar system, the Oort cloud, come in at high inclinations, in highly elliptical orbit off the equatorial orbital plane. You see this in ring systems as well. you don't always see it with moons because moons often seem to be captured gravitationally by planets as they clear their orbit of other bodies (part of the current IAU definition of a planet) . Pluto is not on the orbital plane, which is a good sign that its origin is a bit different from the regular 8 planets that formed along the accretion disk. It's a large Trans-neptunian object with its own orbit influenced by Neptune. It's not impossible to orbit the sun in a polar orbit, either, though I don't know anything beyond a couple of probes have been shown to do that.
I say that it's already been proven that all of the planets orbit the sun in separate inclinations. Looking at the simulation of the Sol System generated in Celestia, all of the orbits did not line up with one another unless I was very far away from Sol, and this isn't even counting the Kuiper Belt Objects.
 
I say that it's already been proven that all of the planets orbit the sun in separate inclinations. Looking at the simulation of the Sol System generated in Celestia, all of the orbits did not line up with one another unless I was very far away from Sol, and this isn't even counting the Kuiper Belt Objects.
The main 8 planets all are within a few degrees inclination to equatorial plane.
 
The main 8 planets all are within a few degrees inclination to equatorial plane.
I assume you mean the equatorial plane of the Sun -- the Earth's axis is tilted by about 23 degrees relative to the plane of the Ecliptic, to which the other planets (Pluto no longer being classed as such) appear to orbit relatively closely. The Ecliptic is tilted by about 7 degrees relative to the Sun's equatorial plane.
https://en.m.wikipedia.org/wiki/Ecliptic
https://en.m.wikipedia.org/wiki/Orbital_inclination#Observations_and_theories
 
I assume you mean the equatorial plane of the Sun -- the Earth's axis is tilted by about 23 degrees relative to the plane of the Ecliptic, to which the other planets (Pluto no longer being classed as such) appear to orbit relatively closely. The Ecliptic is tilted by about 7 degrees relative to the Sun's equatorial plane.
https://en.m.wikipedia.org/wiki/Ecliptic
https://en.m.wikipedia.org/wiki/Orbital_inclination#Observations_and_theories
Yes. And this is why the planetary formation from disc accretion model seems to work. It would take a ginormous amount of delta-v change to significantly alter a planet's orbital inclination.
 
Yes. And this is why the planetary formation from disc accretion model seems to work. It would take a ginormous amount of delta-v change to significantly alter a planet's orbital inclination.
The delta-v required to change the inclination of a circular orbit is 2*v*sin(delta-i/2), where v is the orbital velocity and delta-i is the inclination change. It doesn't scale with mass.
 
alpha_centauri.png


"... and let's be honest, it's more like two and a half stars. Proxima is barely a star and barely bound to the system."
 
https://link.springer.com/article/10.1007/s10714-019-2547-9
Sad as i am to say it, it looks like while this new paper doesn't put nails in the coffin on the Woodward drive, it seems to demonstrate mathematically that Woodward's effect, while possible, requires a local mass to work, rendering the effects nearly nominal and thus useless as a space drive system, for all practicable purposes.

Only glimmer of hope is at the end 'Present cosmological measurements of the possible time variation of G are shown to occur at much lower frequencies and therefore cannot be used to rule out Woodward’s effect if G exhibits significant time-dependence at higher frequencies than observed in these cosmological measurements."

However the Fearn and Woodward device is still being tested so maybe those high frequencies (which were dependant on higher quality piezo material that I think was going to be tested under the grant) may show some positive result. But doesn't seem likely. Well, it remains an interesting phenomenon, if nothing else.
 
THANK GOD someone else recognized that. It's been bothering me for 30 years.
Either the Klingons were in our backyard, or, the probe got caught in one of the many, many cosmic phenomenon that we've seen fling ships far across the galaxy in the several Trek series. I always opted to interpret the scene as the latter, because the former seemed unlikely to me. Though now that I rethink it, I think a Bird of Prey could probably handle the two Bic razors that would get sent out to meet it from Jupiter Station (as seen as our system defense in BOBW), and when I really rethink it, I subscribe to the 'Star Trek V was Kirk's fever dream' theory, anyway. ;)
 
If we could make the 5 year trip to Alpha Centauri at light speed it's 10 years round trip but how much time passes on Earth relative to the people on the ship? Is it literally 10 years or more due to time dilation?
 
If we could make the 5 year trip to Alpha Centauri at light speed it's 10 years round trip but how much time passes on Earth relative to the people on the ship? Is it literally 10 years or more due to time dilation?
At light speed, no time passes for you on the ship; 10 years passes at your starting point (your initial inertial frame of reference).
 
At light speed, no time passes for you on the ship; 10 years passes at your starting point (your initial inertial frame of reference).

Oh ok so it's an exact 10 years when you return to Earth? I was under the impression that time moves faster for the people back home and when you come back many years had passed.
 
Time doesn't pass for you on the ship moving at c so your journey takes 0 seconds for you. The elapsed time measured at your starting point based on observing your journey is distance/velocity = 10 light years/c = 10 years.

If you travelled 1 million light years at c, 0 seconds would pass for you, 1 million years back home.

For a speed v less than c, the time dilation factor is 1/sqrt(1-(v/c)^2), which goes to infinity as v tends to c, as in your example.

For v less than c, one might think that as the motion is relative, each observer would observe clocks ticking more slowly for the other observer. This is true, however, it is possible to show that less time does pass on the ship, provided it starts and finishes in the same inertial frame of reference as the stay at home observer. The asymmetry arises because the ship has to accelerate and decelerate.

At 0.98c, the time dilation factor is 5, which, for the 10 light-year trip example, assuming extremely rapid acceleration and deceleration, would mean about 10.2 years would have passed on Earth and 2.04 years on the ship. The predicted effect of time dilation has been confirmed many times by the enhanced decay times of rapidly moving elementary subatomic particles such as muons.
 
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I don't know why they don't do a Voyager style probe now, and send it in that direction. It would be less costly then a manned mission somewhere wouldn't it?
But it would have several big problems, all related to the distance and travel time. It would take nearly 100,000 years to travel the distance, and that would cause big problems for powering it. For using RTG's, one would need a radionuclide with at least that long a half-life, and having a long half-life means a low decay rate, and thus much less heat. Nuclear reactors would not have that problem, but unlike RTG's, they have moving parts. Maintenance will be a big headache over such a trip -- one will need some very good AI for one's maintenance robots.

Communicating back will be a big problem, since the spacecraft will be about 2000 times farther than where Voyager 1 is now, and with the same transmission power, that translates into 4 million times less signal.

So one would need a big spaceship, something like the Project Daedalus one.
 
So at the NASA Innovative Advanced Concepts symposium this month, the team working on Dr Woodward's Mach Effect Gravity Assist (MEGA) put through a proposal to send a nuclear powered craft at .4c to Proxima Centauri. This is an actual proposal. Obviously the propulsion is a novel idea, but the craft itself is derived from Jupiter Icy Moons Orbiter (JIMO) which was a concept seriously looked at for a nuclear powered unmanned spacecraft in the early 2000's.

Since this is about the most serious near-time proposal for a starship, I thought it might be of interest. The documents are located here.

http://ssi.org/today-at-nasa-niac-2017/

a video of Doctor Fearn's presentation on the starship:
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Piezoelectric effects on internal masses resulting in pulsed gravity thrust. Sounds like MEGA Drive = Impulse Drive.:beer: or :shrug:?
 
Piezoelectric effects on internal masses resulting in pulsed gravity thrust. Sounds like MEGA Drive = Impulse Drive.:beer: or :shrug:?
It will be interesting. Professor Woodward has made a very brief succint reply to the paper I linked above, but I don't have a copy or link to it. In short he sites gravity (spacetime) would exhibit mach effects anywhere, not just close to black holes, as the Rodal paper suggests.

Fortunately for all, that's testable and it is being tested.

If we could make the 5 year trip to Alpha Centauri at light speed it's 10 years round trip but how much time passes on Earth relative to the people on the ship? Is it literally 10 years or more due to time dilation?
A 25 year trip to Proxima Centauri would never approach close enough to c for time dilation to have very noticeable effects.
 
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