1,284 Newly Validated Kepler Planets
nasa.gov
nasa.gov
http://jwst.nasa.gov/origins.html
"One of the main uses of the James Webb Space Telescope will be to study the atmospheres of exoplanets, to search for the building blocks of life elsewhere in the universe"
[1] https://en.wikipedia.org/wiki/Exoplanet [2] http://phys.org/news/2016-04-limits-uniqueness.html
https://exoplanets.jpl.nasa.gov/news/1346/
Related: "Mission Manager Q&A: Recovering The Kepler Spacecraft To Hunt For Exoplanets Again"
http://www.nasa.gov/feature/ames/kepler/mission-manager-qa-r...
If they are able to properly orient the spacecraft using solar pressure alone, couldn't this approach be purposely engineered for discharging momentum from reaction wheels in other craft, instead of expending propellant?
Source: http://www.nasa.gov/kepler/keplers-second-light-how-k2-will-...
https://www.nasa.gov/sites/default/files/thumbnails/image/fi...
These two combine to make planets closer to the sun much more likely to be seen. If there was another Sol in a perfect place and at perfect alignment and Kepler had been looking straight at it, by now we would have detected Mercury, Earth, Venus and Mars but nothing else.
Also, as someone mentioned, Kepler is only likely to find planets fairly close to their star. And if you have several large planets that are close to their star, they are more likely to interact and be knocked out into a higher orbit until only one is left to dominate the orbits close in. Smaller planets are less likely to disturb the orbits of their neighbors which means a system is more likely to have several smaller planets closer into the star, which again, makes them more likely to be detected by Kepler.
[edited for clarity]
Horrible armchair speculation: the graph ALMOST looks like there's two bell curves summed together. a larger planet forming one and a smaller planet forming one, might indicate that there's some threshold for making larger planets that has to be crossed first.
http://kepler.nasa.gov/Science/ForScientists/dataarchive/
There is also this overview:
http://exoplanetarchive.ipac.caltech.edu/docs/KeplerMission....
And bulk download site:
http://exoplanetarchive.ipac.caltech.edu/bulk_data_download/
http://www.asterank.com/galaxies/
Here's the code that I use to suck down exoplanet data, which may be useful to you:
https://github.com/typpo/asterank/blob/master/data/pipeline/...
It's so expensive in energy to do this that it's unlikely.
Even the moon, if it was made of gold bars stacked at the ready to be picked up and brought back to Earth, it wouldn't be worth it.
Even if you had technology to transport a significant force across the span of light years, why endanger that with the risk of resistance? Just go to one of the billions of habitable planets that don't have complex, radio transmitting intelligences on it.
[0] https://en.wikipedia.org/wiki/Ultra-wideband
[1] [pdf] https://www.researchgate.net/file.PostFileLoader.html?id=532...
How about we skip this dream, along with world peace, and make baby steps, like everybody using their indicators while driving?
Just to put some numbers to this, I went to my go-to quick reference for these types of problems is the Usenet Physics FAQ article on the equations for relativistic rockets:
http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/r...
The key numbers are the trip time T according to the ship's clocks, the trip time t according to clocks on Earth or Alpha Centauri, the maximum velocity v achieved by the ship, and the mass ratio M, or kg of fuel needed per kg of payload. Assuming that we accelerate for half the distance and then turn around and decelerate for the other half, and that we use an ideal rocket that converts fuel to energy at 100% efficiency and exhausts all that energy out the back with perfect collimation (so it all goes into changing the rocket's momentum and not into heat or some other waste product), the formulas for these are:
T = 2 (c/a) arccosh (ad/2c^2 + 1),
t = sqrt((d/c)^2 + 2d/a),
v = at/2 sqrt(1 + (at/2c)^2)
MR = exp(aT/c) - 1,
where a is the acceleration, d is the total distance traveled, and c is the speed of light. The equation for T, assuming that we accelerate halfway, then turn around and decelerate to arrive at the destination at rest, is
Plugging in a = 1m/s^2, d = the distance to Alpha Centauri, or about 4 x 10^16 meters, and c = 3 x 10^8 m/s, we get
T = 3.9 x 10^8 s, or about 12.7 years,
t = 4.2 x 10^8 s, or about 13.6 years,
v = 1.7 x 10^8 m/s, or about 0.57 c,
MR = 2.7
This is a pretty small mass ratio, but of course it was derived using highly idealized assumptions. More realistic assumptions would result in a much larger mass ratio. Also, a maximum speed of more than half the speed of light would create huge radiation issues requiring heavy shielding, so the payload mass would be very large.
t = 2 sqrt((d/2c)^2 + d/a)
The numerical result I gave for t is correct.
Alternatively, Bezos will probably have $100 billion to play with in the next 12 to 24 months or so ($61 billion today, most projections have Amazon climbing quite a bit higher yet, new price targets have it at $1,000 per share; whether it takes a year or six years, it'll very likely get there). He just sold off $671 million worth of shares the other day, I'd expect his sales will continue as the price climbs. So maybe he'll fund it by himself in the coming two decades.
So, how hard was this whole interstellar travel exactly?
That would be a direct invitation to other stars in a lifetime.