Do supernovas push neighboring stars outward?












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I know that a supernova can mess up the heliosphere of nearby stars, but I'm wondering if it could physically push neighboring stars off their trajectories.



It's fun to imagine all the stars surrounding a supernova being propelled outward and tumbling out of the galactic arm!



I would expect that a really close star, such as a partner in a binary pair, would get really messed up. I'm thinking more about the neighbors a few light-years away.



I realize that a supernova involves both the initial EM burst and the mass ejection which arrives later. I'm open to the effects of any of these things.










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    3












    $begingroup$


    I know that a supernova can mess up the heliosphere of nearby stars, but I'm wondering if it could physically push neighboring stars off their trajectories.



    It's fun to imagine all the stars surrounding a supernova being propelled outward and tumbling out of the galactic arm!



    I would expect that a really close star, such as a partner in a binary pair, would get really messed up. I'm thinking more about the neighbors a few light-years away.



    I realize that a supernova involves both the initial EM burst and the mass ejection which arrives later. I'm open to the effects of any of these things.










    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I know that a supernova can mess up the heliosphere of nearby stars, but I'm wondering if it could physically push neighboring stars off their trajectories.



      It's fun to imagine all the stars surrounding a supernova being propelled outward and tumbling out of the galactic arm!



      I would expect that a really close star, such as a partner in a binary pair, would get really messed up. I'm thinking more about the neighbors a few light-years away.



      I realize that a supernova involves both the initial EM burst and the mass ejection which arrives later. I'm open to the effects of any of these things.










      share|cite|improve this question









      $endgroup$




      I know that a supernova can mess up the heliosphere of nearby stars, but I'm wondering if it could physically push neighboring stars off their trajectories.



      It's fun to imagine all the stars surrounding a supernova being propelled outward and tumbling out of the galactic arm!



      I would expect that a really close star, such as a partner in a binary pair, would get really messed up. I'm thinking more about the neighbors a few light-years away.



      I realize that a supernova involves both the initial EM burst and the mass ejection which arrives later. I'm open to the effects of any of these things.







      astrophysics stars supernova






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      SlowMagicSlowMagic

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          Consider a star of mass $M$ and radius $R$ at a distance $r$ from the supernova. For a back-of-the-envelope estimate, consider how much momentum would be transferred to the star by the supernova. From that, we can estimate the star's change in velocity and decide whether or not it would be significant.



          First, for extra fun, here's a review of how a typical core-collapse supernova works [1]:




          Nuclear matter is highly incompressible. Hence once the central part of the core reaches nuclear density there is powerful resistance to further compression. That resistance is the primary source of the shock waves that turn a stellar collapse into a spectacular explosion. ... When the center of the core reaches nuclear density, it is brought to rest with a jolt. This gives rise to sound waves that propagate back through the medium of the core, rather like the vibrations in the handle of a hammer when it strikes an anvil. .. The compressibility of nuclear matter is low but not zero, and so momentum carries the collapse beyond the point of equilibrium, compressing the central core to a density even higher than that of an atomic nucleus. ... Most computer simulations suggest the highest density attained is some 50 percent greater than the equilibrium density of a nucleus. ...the sphere of nuclear matter bounces back, like a rubber ball that has been compressed.




          That "bounce" is allegedly what creates the explosion. According to [2],




          Core colapse liberates $sim 3times 10^{53}$ erg ... of gravitational binding energy of the neutron star, 99% of which is radiated in neutrinos over tens of seconds. The supernova mechanism must revive the stalled shock and convert $sim 1$% of the available energy into the energy of the explosion, which must happen within less than $sim 0.5$-$1$ s of core bounce in order to produce a typical core-collapse supernova explosion...




          According to [3], one "erg" is $10^{-7}$ Joules. To give the idea the best possible chance of working, suppose that all of the $E=10^{53}text{ ergs }= 10^{46}text{ Joules}$ of energy goes into the kinetic energy of the expanding shell. The momentum $p$ is maximized by assuming that the expanding shell is massless (because $p=sqrt{(E/c)^2-(mc)^2}$), and while we're at it let's suppose that the collision of the shell with the star is perfectly elastic in order to maximize the effect on the motion of the star. Now suppose that the radius of the star is $R=7times 10^8$ meters (like the sun) and has mass $M=2times 10^{30}$ kg (like the sun), and suppose that its distance from the supernova is $r=3times 10^{16}$ meters (about 3 light-years). If the total energy in the outgoing supernova shell is $E$, then fraction intercepted by the star is the area of the star's disk ($2pi R^2$) divided by the area of the outgoing spherical shell ($4pi r^2$). So the intercepted energy $E'$ is
          $$
          E'=frac{2pi R^2}{4pi r^2}Eapprox (3times 10^{-16})E.
          $$

          Using $E=10^{46}$ Joules gives
          $$
          E'approx 3times 10^{30}text{ Joules}.
          $$

          That's a lot of energy, but is it enough? Using $capprox 3times 10^8$ m/s for the speed of light, the corresponding momentum is $p=E'/capprox 10^{22}$ kg$cdot$m/s. Optimistically assuming an elastic collision that completely reverses the direction of that part of the shell's momentum (optimistically ignoring conservation of energy), the change in the star's momentum will be twice that much. Since the star has a mass of $M=2times 10^{30}$ kg, its change in velocity (using a non-relativistic approximation, which is plenty good enough in this case) is $2p/Mapprox 10^{-8}$ meters per second, which is about 30 centimeters per year. That's probably not enough to eject the star from the galaxy. Sorry.





          References:



          [1] Page 43 in Bethe and Brown (1985), "How a Supernova Explodes," Scientific American 252: 40-48, http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/transparents/SN%20-%20Bethe%20e%20Brown.pdf



          [2] Ott $et al$ (2011), "New Aspects and Boundary Conditions of Core-Collapse Supernova Theory," http://arxiv.org/abs/1111.6282



          [3] Table 9 on page 128 in The International System of Units (SI), 8th edition, International Bureau of Weights and Measures (BIPM), http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf






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          • $begingroup$
            I guess our answers kind of complement each other. FWIW, I didn't see yours until after I submitted mine (typing & posting links can be a bit slow on the phone...)
            $endgroup$
            – PM 2Ring
            24 mins ago



















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          Probably not. Supernovae are powerful, but space is really big. ;)



          Supernova energies are often measured in foe; one foe is $10^{44}$ joules. According to Wikipedia, a big supernova can release around 100 foe as kinetic energy of ejecta, plus 1 to 5 foe for the light & other EM energy released. (The energy of the released neutrinos is higher than the EM energy, but that's only an issue if you're really close to the supernova).



          By way of comparison, the Sun's gravitational binding energy (GBE) is around $3.8 times 10^{41}$ joules. So if a sun-like star got hit by one thousandth of the light energy released by a supernova it would get seriously messed up.



          But as I said at the start, space is really big. If you spread 1 foe of energy over the surface of a sphere of 1 light-year radius, an area on that sphere equal to the Sun's cross-section would get around $1.35 times 10^{29}$ joules, which is a substantial quantity of energy, but it's around a trillionth of the Sun's GBE. So a supernova may do interesting things to the atmosphere of a star 1 light-year away, and the atmospheres of any planets in that system, but it won't disrupt the star, or cause a noticeable perturbation of its galactic orbit.



          However, supernova explosions are notoriously asymmetrical. The energy and matter they release is not spread uniformly over a nice spherical surface. So there's a chance that the damage at 1 light-year is much worse than what I stated in the previous paragraph. In particular, the supernova remnant (the neutron star or black hole produced by the collapse) may be ejected at 500 km/s or faster. If you happen to be in the path of one of those, Bad Things are likely to occur. An extreme example is Pulsar B1508+55, a neutron star heading out of the galaxy at 1100 km/s.






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            $begingroup$

            Consider a star of mass $M$ and radius $R$ at a distance $r$ from the supernova. For a back-of-the-envelope estimate, consider how much momentum would be transferred to the star by the supernova. From that, we can estimate the star's change in velocity and decide whether or not it would be significant.



            First, for extra fun, here's a review of how a typical core-collapse supernova works [1]:




            Nuclear matter is highly incompressible. Hence once the central part of the core reaches nuclear density there is powerful resistance to further compression. That resistance is the primary source of the shock waves that turn a stellar collapse into a spectacular explosion. ... When the center of the core reaches nuclear density, it is brought to rest with a jolt. This gives rise to sound waves that propagate back through the medium of the core, rather like the vibrations in the handle of a hammer when it strikes an anvil. .. The compressibility of nuclear matter is low but not zero, and so momentum carries the collapse beyond the point of equilibrium, compressing the central core to a density even higher than that of an atomic nucleus. ... Most computer simulations suggest the highest density attained is some 50 percent greater than the equilibrium density of a nucleus. ...the sphere of nuclear matter bounces back, like a rubber ball that has been compressed.




            That "bounce" is allegedly what creates the explosion. According to [2],




            Core colapse liberates $sim 3times 10^{53}$ erg ... of gravitational binding energy of the neutron star, 99% of which is radiated in neutrinos over tens of seconds. The supernova mechanism must revive the stalled shock and convert $sim 1$% of the available energy into the energy of the explosion, which must happen within less than $sim 0.5$-$1$ s of core bounce in order to produce a typical core-collapse supernova explosion...




            According to [3], one "erg" is $10^{-7}$ Joules. To give the idea the best possible chance of working, suppose that all of the $E=10^{53}text{ ergs }= 10^{46}text{ Joules}$ of energy goes into the kinetic energy of the expanding shell. The momentum $p$ is maximized by assuming that the expanding shell is massless (because $p=sqrt{(E/c)^2-(mc)^2}$), and while we're at it let's suppose that the collision of the shell with the star is perfectly elastic in order to maximize the effect on the motion of the star. Now suppose that the radius of the star is $R=7times 10^8$ meters (like the sun) and has mass $M=2times 10^{30}$ kg (like the sun), and suppose that its distance from the supernova is $r=3times 10^{16}$ meters (about 3 light-years). If the total energy in the outgoing supernova shell is $E$, then fraction intercepted by the star is the area of the star's disk ($2pi R^2$) divided by the area of the outgoing spherical shell ($4pi r^2$). So the intercepted energy $E'$ is
            $$
            E'=frac{2pi R^2}{4pi r^2}Eapprox (3times 10^{-16})E.
            $$

            Using $E=10^{46}$ Joules gives
            $$
            E'approx 3times 10^{30}text{ Joules}.
            $$

            That's a lot of energy, but is it enough? Using $capprox 3times 10^8$ m/s for the speed of light, the corresponding momentum is $p=E'/capprox 10^{22}$ kg$cdot$m/s. Optimistically assuming an elastic collision that completely reverses the direction of that part of the shell's momentum (optimistically ignoring conservation of energy), the change in the star's momentum will be twice that much. Since the star has a mass of $M=2times 10^{30}$ kg, its change in velocity (using a non-relativistic approximation, which is plenty good enough in this case) is $2p/Mapprox 10^{-8}$ meters per second, which is about 30 centimeters per year. That's probably not enough to eject the star from the galaxy. Sorry.





            References:



            [1] Page 43 in Bethe and Brown (1985), "How a Supernova Explodes," Scientific American 252: 40-48, http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/transparents/SN%20-%20Bethe%20e%20Brown.pdf



            [2] Ott $et al$ (2011), "New Aspects and Boundary Conditions of Core-Collapse Supernova Theory," http://arxiv.org/abs/1111.6282



            [3] Table 9 on page 128 in The International System of Units (SI), 8th edition, International Bureau of Weights and Measures (BIPM), http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I guess our answers kind of complement each other. FWIW, I didn't see yours until after I submitted mine (typing & posting links can be a bit slow on the phone...)
              $endgroup$
              – PM 2Ring
              24 mins ago
















            2












            $begingroup$

            Consider a star of mass $M$ and radius $R$ at a distance $r$ from the supernova. For a back-of-the-envelope estimate, consider how much momentum would be transferred to the star by the supernova. From that, we can estimate the star's change in velocity and decide whether or not it would be significant.



            First, for extra fun, here's a review of how a typical core-collapse supernova works [1]:




            Nuclear matter is highly incompressible. Hence once the central part of the core reaches nuclear density there is powerful resistance to further compression. That resistance is the primary source of the shock waves that turn a stellar collapse into a spectacular explosion. ... When the center of the core reaches nuclear density, it is brought to rest with a jolt. This gives rise to sound waves that propagate back through the medium of the core, rather like the vibrations in the handle of a hammer when it strikes an anvil. .. The compressibility of nuclear matter is low but not zero, and so momentum carries the collapse beyond the point of equilibrium, compressing the central core to a density even higher than that of an atomic nucleus. ... Most computer simulations suggest the highest density attained is some 50 percent greater than the equilibrium density of a nucleus. ...the sphere of nuclear matter bounces back, like a rubber ball that has been compressed.




            That "bounce" is allegedly what creates the explosion. According to [2],




            Core colapse liberates $sim 3times 10^{53}$ erg ... of gravitational binding energy of the neutron star, 99% of which is radiated in neutrinos over tens of seconds. The supernova mechanism must revive the stalled shock and convert $sim 1$% of the available energy into the energy of the explosion, which must happen within less than $sim 0.5$-$1$ s of core bounce in order to produce a typical core-collapse supernova explosion...




            According to [3], one "erg" is $10^{-7}$ Joules. To give the idea the best possible chance of working, suppose that all of the $E=10^{53}text{ ergs }= 10^{46}text{ Joules}$ of energy goes into the kinetic energy of the expanding shell. The momentum $p$ is maximized by assuming that the expanding shell is massless (because $p=sqrt{(E/c)^2-(mc)^2}$), and while we're at it let's suppose that the collision of the shell with the star is perfectly elastic in order to maximize the effect on the motion of the star. Now suppose that the radius of the star is $R=7times 10^8$ meters (like the sun) and has mass $M=2times 10^{30}$ kg (like the sun), and suppose that its distance from the supernova is $r=3times 10^{16}$ meters (about 3 light-years). If the total energy in the outgoing supernova shell is $E$, then fraction intercepted by the star is the area of the star's disk ($2pi R^2$) divided by the area of the outgoing spherical shell ($4pi r^2$). So the intercepted energy $E'$ is
            $$
            E'=frac{2pi R^2}{4pi r^2}Eapprox (3times 10^{-16})E.
            $$

            Using $E=10^{46}$ Joules gives
            $$
            E'approx 3times 10^{30}text{ Joules}.
            $$

            That's a lot of energy, but is it enough? Using $capprox 3times 10^8$ m/s for the speed of light, the corresponding momentum is $p=E'/capprox 10^{22}$ kg$cdot$m/s. Optimistically assuming an elastic collision that completely reverses the direction of that part of the shell's momentum (optimistically ignoring conservation of energy), the change in the star's momentum will be twice that much. Since the star has a mass of $M=2times 10^{30}$ kg, its change in velocity (using a non-relativistic approximation, which is plenty good enough in this case) is $2p/Mapprox 10^{-8}$ meters per second, which is about 30 centimeters per year. That's probably not enough to eject the star from the galaxy. Sorry.





            References:



            [1] Page 43 in Bethe and Brown (1985), "How a Supernova Explodes," Scientific American 252: 40-48, http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/transparents/SN%20-%20Bethe%20e%20Brown.pdf



            [2] Ott $et al$ (2011), "New Aspects and Boundary Conditions of Core-Collapse Supernova Theory," http://arxiv.org/abs/1111.6282



            [3] Table 9 on page 128 in The International System of Units (SI), 8th edition, International Bureau of Weights and Measures (BIPM), http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I guess our answers kind of complement each other. FWIW, I didn't see yours until after I submitted mine (typing & posting links can be a bit slow on the phone...)
              $endgroup$
              – PM 2Ring
              24 mins ago














            2












            2








            2





            $begingroup$

            Consider a star of mass $M$ and radius $R$ at a distance $r$ from the supernova. For a back-of-the-envelope estimate, consider how much momentum would be transferred to the star by the supernova. From that, we can estimate the star's change in velocity and decide whether or not it would be significant.



            First, for extra fun, here's a review of how a typical core-collapse supernova works [1]:




            Nuclear matter is highly incompressible. Hence once the central part of the core reaches nuclear density there is powerful resistance to further compression. That resistance is the primary source of the shock waves that turn a stellar collapse into a spectacular explosion. ... When the center of the core reaches nuclear density, it is brought to rest with a jolt. This gives rise to sound waves that propagate back through the medium of the core, rather like the vibrations in the handle of a hammer when it strikes an anvil. .. The compressibility of nuclear matter is low but not zero, and so momentum carries the collapse beyond the point of equilibrium, compressing the central core to a density even higher than that of an atomic nucleus. ... Most computer simulations suggest the highest density attained is some 50 percent greater than the equilibrium density of a nucleus. ...the sphere of nuclear matter bounces back, like a rubber ball that has been compressed.




            That "bounce" is allegedly what creates the explosion. According to [2],




            Core colapse liberates $sim 3times 10^{53}$ erg ... of gravitational binding energy of the neutron star, 99% of which is radiated in neutrinos over tens of seconds. The supernova mechanism must revive the stalled shock and convert $sim 1$% of the available energy into the energy of the explosion, which must happen within less than $sim 0.5$-$1$ s of core bounce in order to produce a typical core-collapse supernova explosion...




            According to [3], one "erg" is $10^{-7}$ Joules. To give the idea the best possible chance of working, suppose that all of the $E=10^{53}text{ ergs }= 10^{46}text{ Joules}$ of energy goes into the kinetic energy of the expanding shell. The momentum $p$ is maximized by assuming that the expanding shell is massless (because $p=sqrt{(E/c)^2-(mc)^2}$), and while we're at it let's suppose that the collision of the shell with the star is perfectly elastic in order to maximize the effect on the motion of the star. Now suppose that the radius of the star is $R=7times 10^8$ meters (like the sun) and has mass $M=2times 10^{30}$ kg (like the sun), and suppose that its distance from the supernova is $r=3times 10^{16}$ meters (about 3 light-years). If the total energy in the outgoing supernova shell is $E$, then fraction intercepted by the star is the area of the star's disk ($2pi R^2$) divided by the area of the outgoing spherical shell ($4pi r^2$). So the intercepted energy $E'$ is
            $$
            E'=frac{2pi R^2}{4pi r^2}Eapprox (3times 10^{-16})E.
            $$

            Using $E=10^{46}$ Joules gives
            $$
            E'approx 3times 10^{30}text{ Joules}.
            $$

            That's a lot of energy, but is it enough? Using $capprox 3times 10^8$ m/s for the speed of light, the corresponding momentum is $p=E'/capprox 10^{22}$ kg$cdot$m/s. Optimistically assuming an elastic collision that completely reverses the direction of that part of the shell's momentum (optimistically ignoring conservation of energy), the change in the star's momentum will be twice that much. Since the star has a mass of $M=2times 10^{30}$ kg, its change in velocity (using a non-relativistic approximation, which is plenty good enough in this case) is $2p/Mapprox 10^{-8}$ meters per second, which is about 30 centimeters per year. That's probably not enough to eject the star from the galaxy. Sorry.





            References:



            [1] Page 43 in Bethe and Brown (1985), "How a Supernova Explodes," Scientific American 252: 40-48, http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/transparents/SN%20-%20Bethe%20e%20Brown.pdf



            [2] Ott $et al$ (2011), "New Aspects and Boundary Conditions of Core-Collapse Supernova Theory," http://arxiv.org/abs/1111.6282



            [3] Table 9 on page 128 in The International System of Units (SI), 8th edition, International Bureau of Weights and Measures (BIPM), http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf






            share|cite|improve this answer











            $endgroup$



            Consider a star of mass $M$ and radius $R$ at a distance $r$ from the supernova. For a back-of-the-envelope estimate, consider how much momentum would be transferred to the star by the supernova. From that, we can estimate the star's change in velocity and decide whether or not it would be significant.



            First, for extra fun, here's a review of how a typical core-collapse supernova works [1]:




            Nuclear matter is highly incompressible. Hence once the central part of the core reaches nuclear density there is powerful resistance to further compression. That resistance is the primary source of the shock waves that turn a stellar collapse into a spectacular explosion. ... When the center of the core reaches nuclear density, it is brought to rest with a jolt. This gives rise to sound waves that propagate back through the medium of the core, rather like the vibrations in the handle of a hammer when it strikes an anvil. .. The compressibility of nuclear matter is low but not zero, and so momentum carries the collapse beyond the point of equilibrium, compressing the central core to a density even higher than that of an atomic nucleus. ... Most computer simulations suggest the highest density attained is some 50 percent greater than the equilibrium density of a nucleus. ...the sphere of nuclear matter bounces back, like a rubber ball that has been compressed.




            That "bounce" is allegedly what creates the explosion. According to [2],




            Core colapse liberates $sim 3times 10^{53}$ erg ... of gravitational binding energy of the neutron star, 99% of which is radiated in neutrinos over tens of seconds. The supernova mechanism must revive the stalled shock and convert $sim 1$% of the available energy into the energy of the explosion, which must happen within less than $sim 0.5$-$1$ s of core bounce in order to produce a typical core-collapse supernova explosion...




            According to [3], one "erg" is $10^{-7}$ Joules. To give the idea the best possible chance of working, suppose that all of the $E=10^{53}text{ ergs }= 10^{46}text{ Joules}$ of energy goes into the kinetic energy of the expanding shell. The momentum $p$ is maximized by assuming that the expanding shell is massless (because $p=sqrt{(E/c)^2-(mc)^2}$), and while we're at it let's suppose that the collision of the shell with the star is perfectly elastic in order to maximize the effect on the motion of the star. Now suppose that the radius of the star is $R=7times 10^8$ meters (like the sun) and has mass $M=2times 10^{30}$ kg (like the sun), and suppose that its distance from the supernova is $r=3times 10^{16}$ meters (about 3 light-years). If the total energy in the outgoing supernova shell is $E$, then fraction intercepted by the star is the area of the star's disk ($2pi R^2$) divided by the area of the outgoing spherical shell ($4pi r^2$). So the intercepted energy $E'$ is
            $$
            E'=frac{2pi R^2}{4pi r^2}Eapprox (3times 10^{-16})E.
            $$

            Using $E=10^{46}$ Joules gives
            $$
            E'approx 3times 10^{30}text{ Joules}.
            $$

            That's a lot of energy, but is it enough? Using $capprox 3times 10^8$ m/s for the speed of light, the corresponding momentum is $p=E'/capprox 10^{22}$ kg$cdot$m/s. Optimistically assuming an elastic collision that completely reverses the direction of that part of the shell's momentum (optimistically ignoring conservation of energy), the change in the star's momentum will be twice that much. Since the star has a mass of $M=2times 10^{30}$ kg, its change in velocity (using a non-relativistic approximation, which is plenty good enough in this case) is $2p/Mapprox 10^{-8}$ meters per second, which is about 30 centimeters per year. That's probably not enough to eject the star from the galaxy. Sorry.





            References:



            [1] Page 43 in Bethe and Brown (1985), "How a Supernova Explodes," Scientific American 252: 40-48, http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/transparents/SN%20-%20Bethe%20e%20Brown.pdf



            [2] Ott $et al$ (2011), "New Aspects and Boundary Conditions of Core-Collapse Supernova Theory," http://arxiv.org/abs/1111.6282



            [3] Table 9 on page 128 in The International System of Units (SI), 8th edition, International Bureau of Weights and Measures (BIPM), http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf







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            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 1 hour ago









            Dan YandDan Yand

            8,71811335




            8,71811335












            • $begingroup$
              I guess our answers kind of complement each other. FWIW, I didn't see yours until after I submitted mine (typing & posting links can be a bit slow on the phone...)
              $endgroup$
              – PM 2Ring
              24 mins ago


















            • $begingroup$
              I guess our answers kind of complement each other. FWIW, I didn't see yours until after I submitted mine (typing & posting links can be a bit slow on the phone...)
              $endgroup$
              – PM 2Ring
              24 mins ago
















            $begingroup$
            I guess our answers kind of complement each other. FWIW, I didn't see yours until after I submitted mine (typing & posting links can be a bit slow on the phone...)
            $endgroup$
            – PM 2Ring
            24 mins ago




            $begingroup$
            I guess our answers kind of complement each other. FWIW, I didn't see yours until after I submitted mine (typing & posting links can be a bit slow on the phone...)
            $endgroup$
            – PM 2Ring
            24 mins ago











            0












            $begingroup$

            Probably not. Supernovae are powerful, but space is really big. ;)



            Supernova energies are often measured in foe; one foe is $10^{44}$ joules. According to Wikipedia, a big supernova can release around 100 foe as kinetic energy of ejecta, plus 1 to 5 foe for the light & other EM energy released. (The energy of the released neutrinos is higher than the EM energy, but that's only an issue if you're really close to the supernova).



            By way of comparison, the Sun's gravitational binding energy (GBE) is around $3.8 times 10^{41}$ joules. So if a sun-like star got hit by one thousandth of the light energy released by a supernova it would get seriously messed up.



            But as I said at the start, space is really big. If you spread 1 foe of energy over the surface of a sphere of 1 light-year radius, an area on that sphere equal to the Sun's cross-section would get around $1.35 times 10^{29}$ joules, which is a substantial quantity of energy, but it's around a trillionth of the Sun's GBE. So a supernova may do interesting things to the atmosphere of a star 1 light-year away, and the atmospheres of any planets in that system, but it won't disrupt the star, or cause a noticeable perturbation of its galactic orbit.



            However, supernova explosions are notoriously asymmetrical. The energy and matter they release is not spread uniformly over a nice spherical surface. So there's a chance that the damage at 1 light-year is much worse than what I stated in the previous paragraph. In particular, the supernova remnant (the neutron star or black hole produced by the collapse) may be ejected at 500 km/s or faster. If you happen to be in the path of one of those, Bad Things are likely to occur. An extreme example is Pulsar B1508+55, a neutron star heading out of the galaxy at 1100 km/s.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Probably not. Supernovae are powerful, but space is really big. ;)



              Supernova energies are often measured in foe; one foe is $10^{44}$ joules. According to Wikipedia, a big supernova can release around 100 foe as kinetic energy of ejecta, plus 1 to 5 foe for the light & other EM energy released. (The energy of the released neutrinos is higher than the EM energy, but that's only an issue if you're really close to the supernova).



              By way of comparison, the Sun's gravitational binding energy (GBE) is around $3.8 times 10^{41}$ joules. So if a sun-like star got hit by one thousandth of the light energy released by a supernova it would get seriously messed up.



              But as I said at the start, space is really big. If you spread 1 foe of energy over the surface of a sphere of 1 light-year radius, an area on that sphere equal to the Sun's cross-section would get around $1.35 times 10^{29}$ joules, which is a substantial quantity of energy, but it's around a trillionth of the Sun's GBE. So a supernova may do interesting things to the atmosphere of a star 1 light-year away, and the atmospheres of any planets in that system, but it won't disrupt the star, or cause a noticeable perturbation of its galactic orbit.



              However, supernova explosions are notoriously asymmetrical. The energy and matter they release is not spread uniformly over a nice spherical surface. So there's a chance that the damage at 1 light-year is much worse than what I stated in the previous paragraph. In particular, the supernova remnant (the neutron star or black hole produced by the collapse) may be ejected at 500 km/s or faster. If you happen to be in the path of one of those, Bad Things are likely to occur. An extreme example is Pulsar B1508+55, a neutron star heading out of the galaxy at 1100 km/s.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Probably not. Supernovae are powerful, but space is really big. ;)



                Supernova energies are often measured in foe; one foe is $10^{44}$ joules. According to Wikipedia, a big supernova can release around 100 foe as kinetic energy of ejecta, plus 1 to 5 foe for the light & other EM energy released. (The energy of the released neutrinos is higher than the EM energy, but that's only an issue if you're really close to the supernova).



                By way of comparison, the Sun's gravitational binding energy (GBE) is around $3.8 times 10^{41}$ joules. So if a sun-like star got hit by one thousandth of the light energy released by a supernova it would get seriously messed up.



                But as I said at the start, space is really big. If you spread 1 foe of energy over the surface of a sphere of 1 light-year radius, an area on that sphere equal to the Sun's cross-section would get around $1.35 times 10^{29}$ joules, which is a substantial quantity of energy, but it's around a trillionth of the Sun's GBE. So a supernova may do interesting things to the atmosphere of a star 1 light-year away, and the atmospheres of any planets in that system, but it won't disrupt the star, or cause a noticeable perturbation of its galactic orbit.



                However, supernova explosions are notoriously asymmetrical. The energy and matter they release is not spread uniformly over a nice spherical surface. So there's a chance that the damage at 1 light-year is much worse than what I stated in the previous paragraph. In particular, the supernova remnant (the neutron star or black hole produced by the collapse) may be ejected at 500 km/s or faster. If you happen to be in the path of one of those, Bad Things are likely to occur. An extreme example is Pulsar B1508+55, a neutron star heading out of the galaxy at 1100 km/s.






                share|cite|improve this answer









                $endgroup$



                Probably not. Supernovae are powerful, but space is really big. ;)



                Supernova energies are often measured in foe; one foe is $10^{44}$ joules. According to Wikipedia, a big supernova can release around 100 foe as kinetic energy of ejecta, plus 1 to 5 foe for the light & other EM energy released. (The energy of the released neutrinos is higher than the EM energy, but that's only an issue if you're really close to the supernova).



                By way of comparison, the Sun's gravitational binding energy (GBE) is around $3.8 times 10^{41}$ joules. So if a sun-like star got hit by one thousandth of the light energy released by a supernova it would get seriously messed up.



                But as I said at the start, space is really big. If you spread 1 foe of energy over the surface of a sphere of 1 light-year radius, an area on that sphere equal to the Sun's cross-section would get around $1.35 times 10^{29}$ joules, which is a substantial quantity of energy, but it's around a trillionth of the Sun's GBE. So a supernova may do interesting things to the atmosphere of a star 1 light-year away, and the atmospheres of any planets in that system, but it won't disrupt the star, or cause a noticeable perturbation of its galactic orbit.



                However, supernova explosions are notoriously asymmetrical. The energy and matter they release is not spread uniformly over a nice spherical surface. So there's a chance that the damage at 1 light-year is much worse than what I stated in the previous paragraph. In particular, the supernova remnant (the neutron star or black hole produced by the collapse) may be ejected at 500 km/s or faster. If you happen to be in the path of one of those, Bad Things are likely to occur. An extreme example is Pulsar B1508+55, a neutron star heading out of the galaxy at 1100 km/s.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 32 mins ago









                PM 2RingPM 2Ring

                2,3822716




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