4 ms·

Isn't "billions" of anti protons an absurdly small number to talk about spaceflight fuel. Per http://www.answerbag.com/q_view/154104 http://www.answerbag.com/q_

by gte910h 15y ago

Isn't "billions" of anti protons an absurdly small number to talk about spaceflight fuel. Per http://www.answerbag.com/q_view/154104 http://www.answerbag.com/q_view/154104 (which looks about right) we have 2.73 X 10^26 hydrogen atoms per pound. An antiproton is basically an antihydrogen missing its positron.

DenisM 15y ago
Yeah. This might be good for scientific experiments with anti-matter though.
archgoon 15y ago
Annihilating all of it would result in about 1 tenth of a joule.
neutronicus 15y ago
One antiproton annihilation will give you about 1 GeV. Annihilate a billion of them, and you've got 10^18 eV. One eV is about 10^(-19) Joules. So, yeah. A tenth of a Joule. Underwhelming.
- spullara 15y ago
  And the larger planets might have 27 kilowatt hours worth! Hopefully something was lost in translation of this article from scientists to journalists.
- kragen 15y ago
  Here are a couple of things: • If it only persists for minutes, that's a tenth of a joule per few minutes, if you have some way to capture and confine it away from the corrupting influence of matter. That sounds a little bit more practical than a tenth of a joule, total. • The big deal with antimatter is that it has a high energy density. Even if you have a very small amount of it, it still has a high energy density. A hundred joules can accelerate a one-milligram spacecraft to 5000 meters per second, which is almost enough to stay in low earth orbit. 10 kJ can accelerate a one-microgram spacecraft to 1.6 million m/s, or 0.5% of the speed of light. And 10kJ of matter/antimatter reaction works out to 5.6 × 10⁻¹⁴ g, so it's easy to imagine containing that much in a one-microgram spacecraft. If you can gather a tenth of a joule of antimatter per hour, you only need 11 years to gather enough antimatter to launch your one-microgram probe toward Alpha Centauri.
alex_c 15y ago
It looks like there's a word missing in the quote: "We are talking about of billions of particles," Even if the missing word is another "billions", that still seems rather low.
- gte910h 15y ago
  Yeah, I hope it's a "journalist not understanding large numbers" error. It would be fantastic to have tons of anitmatter in harnessable range like that.
jessriedel 15y ago
OK, so the New Scientist article is based on THE DISCOVERY OF GEOMAGNETICALLY TRAPPED COSMIC-RAY ANTIPROTONS Astro. J. Letters 737:L29, 2011 August 20 DOI: 10.1088/2041-8205/737/2/l29 http://iopscience.iop.org/2041-8205/737/2/L29/ The key sentence from that abstract is "PAMELA data show that the magnetospheric antiproton ﬂux in the SAA exceeds the cosmic-ray antiproton ﬂux by three orders of magnitude at the present solar minimum, and exceeds the sub-cutoff antiproton ﬂux outside radiation belts by four orders of magnitude, constituting the most abundant source of antiprotons near the Earth." The cosmic ray flux is estimated in another paper on the PAMELA results: "PAMELA Results on the Cosmic-Ray Antiproton Flux from 60 MeV to 180 GeV in Kinetic Energy" Physical Review Letters 105, 121101 (2010) DOI: 10.1103/PhysRevLett.105.121101 http://prl.aps.org/abstract/PRL/v105/i12/e121101 After eyeballing Fig. 1, I'd say they measured most antiprotons to be between 1 GeV and 10 GeV, with a flux of ~2x10^-2 / (GeV m^2 s sr). (sr=steradians.) The flux peaks around 2 GeV. Rough multiplication gives an estimate of total flux: [2x10^-2 / (GeV m^2 s sr)]*[9 GeV]*[10^14 m^2]*[4 pi sr] = 2*10^14 antiprotons/second. If we ignore the kinetic energy of the antiprotons, and assume there are plenty of available protons to annihilate with them, we get about 2 GeV per antiproton. This yields [2*10^14 antiprotons/second]*[2 GeV/anitproton] = 6x10^ 4 J/s The space shuttle orbiter weighs 69,000 kg empty. The escapte velocity from earth (which doesn't change much between the surface and the edge of the atmosphere) is 11.2 km/s, so the minimum energy needed to get it out of the earth's atmosphere is [1/2]*[69,000 kg]*[11.2 km/s]^2 = 4x10^12 J OK, so it would take about 10^8 s, or 3 years, to collect enough energy from antiproton cosmic-ray rest mass to lauch a space-shuttle size object out of Earth's gravity. And you'd have to scoop up from the entire surface area above the earth. Now, the magnetospheric antiproton flux in the South Atlantic Anomaly http://en.wikipedia.org/wiki/South_Atlantic_Anomaly is said to be 3 orders of magnitude larger than the normal antiproton cosmic ray flux. By looking at the wikipedia page, it looks like the SAA covers about 10% of the surface area above the earth. If we were able to scoop up all of the antiprotons in the magnetospheric belt, it would then take about 10^6 s, or 11 days, to get enough energy to launch a space-shuttle sized object. I am highly, highly unsure about these number and the assumptions that went into them, but it at least suggests it's possible to do. Lemme know if I did something stupid.
- 15y ago

by gte910h 15y ago

DenisM 15y ago

Yeah. This might be good for scientific experiments with anti-matter though.

archgoon 15y ago

Annihilating all of it would result in about 1 tenth of a joule.

neutronicus 15y ago

One antiproton annihilation will give you about 1 GeV. Annihilate a billion of them, and you've got 10^18 eV. One eV is about 10^(-19) Joules. So, yeah. A tenth of a Joule. Underwhelming.

spullara 15y ago
And the larger planets might have 27 kilowatt hours worth! Hopefully something was lost in translation of this article from scientists to journalists.
kragen 15y ago
Here are a couple of things: • If it only persists for minutes, that's a tenth of a joule per few minutes, if you have some way to capture and confine it away from the corrupting influence of matter. That sounds a little bit more practical than a tenth of a joule, total. • The big deal with antimatter is that it has a high energy density. Even if you have a very small amount of it, it still has a high energy density. A hundred joules can accelerate a one-milligram spacecraft to 5000 meters per second, which is almost enough to stay in low earth orbit. 10 kJ can accelerate a one-microgram spacecraft to 1.6 million m/s, or 0.5% of the speed of light. And 10kJ of matter/antimatter reaction works out to 5.6 × 10⁻¹⁴ g, so it's easy to imagine containing that much in a one-microgram spacecraft. If you can gather a tenth of a joule of antimatter per hour, you only need 11 years to gather enough antimatter to launch your one-microgram probe toward Alpha Centauri.

alex_c 15y ago

It looks like there's a word missing in the quote: "We are talking about of billions of particles," Even if the missing word is another "billions", that still seems rather low.

gte910h 15y ago
Yeah, I hope it's a "journalist not understanding large numbers" error. It would be fantastic to have tons of anitmatter in harnessable range like that.

jessriedel 15y ago

OK, so the New Scientist article is based on THE DISCOVERY OF GEOMAGNETICALLY TRAPPED COSMIC-RAY ANTIPROTONS Astro. J. Letters 737:L29, 2011 August 20 DOI: 10.1088/2041-8205/737/2/l29 http://iopscience.iop.org/2041-8205/737/2/L29/ The key sentence from that abstract is "PAMELA data show that the magnetospheric antiproton ﬂux in the SAA exceeds the cosmic-ray antiproton ﬂux by three orders of magnitude at the present solar minimum, and exceeds the sub-cutoff antiproton ﬂux outside radiation belts by four orders of magnitude, constituting the most abundant source of antiprotons near the Earth." The cosmic ray flux is estimated in another paper on the PAMELA results: "PAMELA Results on the Cosmic-Ray Antiproton Flux from 60 MeV to 180 GeV in Kinetic Energy" Physical Review Letters 105, 121101 (2010) DOI: 10.1103/PhysRevLett.105.121101 http://prl.aps.org/abstract/PRL/v105/i12/e121101 After eyeballing Fig. 1, I'd say they measured most antiprotons to be between 1 GeV and 10 GeV, with a flux of ~2x10^-2 / (GeV m^2 s sr). (sr=steradians.) The flux peaks around 2 GeV. Rough multiplication gives an estimate of total flux: [2x10^-2 / (GeV m^2 s sr)]*[9 GeV]*[10^14 m^2]*[4 pi sr] = 2*10^14 antiprotons/second. If we ignore the kinetic energy of the antiprotons, and assume there are plenty of available protons to annihilate with them, we get about 2 GeV per antiproton. This yields [2*10^14 antiprotons/second]*[2 GeV/anitproton] = 6x10^ 4 J/s The space shuttle orbiter weighs 69,000 kg empty. The escapte velocity from earth (which doesn't change much between the surface and the edge of the atmosphere) is 11.2 km/s, so the minimum energy needed to get it out of the earth's atmosphere is [1/2]*[69,000 kg]*[11.2 km/s]^2 = 4x10^12 J OK, so it would take about 10^8 s, or 3 years, to collect enough energy from antiproton cosmic-ray rest mass to lauch a space-shuttle size object out of Earth's gravity. And you'd have to scoop up from the entire surface area above the earth. Now, the magnetospheric antiproton flux in the South Atlantic Anomaly http://en.wikipedia.org/wiki/South_Atlantic_Anomaly is said to be 3 orders of magnitude larger than the normal antiproton cosmic ray flux. By looking at the wikipedia page, it looks like the SAA covers about 10% of the surface area above the earth. If we were able to scoop up all of the antiprotons in the magnetospheric belt, it would then take about 10^6 s, or 11 days, to get enough energy to launch a space-shuttle sized object. I am highly, highly unsure about these number and the assumptions that went into them, but it at least suggests it's possible to do. Lemme know if I did something stupid.

15y ago