4 ms·

Lol, it's pretty ambitious to try tracking planes with gravitational effects. But now that gravitation wave detection works, I think there are plans (or at leas

Lol, it's pretty ambitious to try tracking planes with gravitational effects. But now that gravitation wave detection works, I think there are plans (or at least ideas) to build such devices in earth or solar orbit for triangulation of distant objects like stars and black holes. Detection arms could be much longer and disruptions much lower than on earth, so results would me much more accurate.

jcims 9y ago
I did some noodling a while back to see if LIGO could detect the mass energy conversion of a fission 'event' in North Korea. If memory serves it was definitely within the range in terms of impact to the gravitational field, but correlating a infinitessimally short pop of noise would be much more difficult than the death waltz of binary black holes. However, a larger array of more sensitive detectors that are immune from the tremendous amount of terrestrial noise might make short work of it.
- StormChaser_5 9y ago
  Why would an event in NK cause a change in gravity? Wouldn't the energy liberated in a nuclear event would cause the same amount of gravity to be felt elsewhere as the mass it came from?
  - techdragon 9y ago
    Two theories off hand without research to double check. 1 - nuclear explosion involves the conversion of a tiny amount of mass into rapidly expanding cloud of energy conveyed by various particles. The significance being the drastic density difference before and after it’s blown to bits. This difference is equivalent to a sudden moment of regular mass since energy and mass are equivalent and energy has gravitational attraction. This is roughly the same as attempting to catch the gravitational waves theorised to be produced by a supernova. We can’t catch these yet due to insufficient sensitivity, but despite the nuclear explosion being ridiculously tiny compared to a supernova, we are also ridiculously close compared to the supernova, so the proximity to the event may “win” and make it strong enough for the detectors to see it. 2 - the equivalence of nuclear explosions to earthquakes in terms of energy displacement into the earths crust, and in particular underground tests of them, has been established. (This is obviously dependent on the size of the explosion and if it’s an underground test or not, since underground tests dump basically all their energy into the surrounding geology they make it a “more effective earthquake”). Earthquakes can speed up or slow down the earths rotation, which has tidal effects which involve the moon, so between the seismic and seismically induced tidal changes, we get to affecting the moons orbit and consequently we get a tiny gravitational wave which once again is detectable only because of how much closer we are to the source compared to the things we built LIGO to detect. Of course I might be wrong on both of these. They were just fun to think about on my ride home from work. :-)
  - DoctorOetker 9y ago
    Initially I thought the same for a couple of seconds because of mass energy equivalence, but then I realized: assume the fission event to occur in empty space, assume the liberated energy is isotropically radiated, and assume all the energy is thus expanding spherically, then the mass/energy that was initially in the mother isotope is now after some time distributed on a spherical surface. As long as the observer is outside of the shell of escaping energy, to the observer the gravitational field looks like all the mass is concentrated at the center of the expanding shell of energy, exactly where the mass/energy originally was, and hence gravitationally undetectable. After a while the expanding surface has past the observer, and the observer is inside a spherically symmetric shell of mass/energy. Even if the observer is not at the center there will no gravitational field at all according to the observer. (consult any physicist or physics textbook, or do the exercise for the gravitational field in a 'hollow earth'). So to the observer this looks like a bit of disappearing mass at the location of fission event. It seems extremely difficult to me to actually measure such an event gravitationally though..
danbruc 9y ago
You are probably thinking of the planned Laser Interferometer Space Antenna (LISA) [1]. Launch is scheduled for 2034 but the technical demonstration mission LISA Pathfinder [2] was already done between 2015 and 2017 and exceeded its targets with regard to sensitivity. Other proposed space-based gravitational wave detectors are TianQin [3] and DECIGO [4]. [1] https://en.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna https://en.wikipedia.org/wiki/Laser_Interferometer_Space_Ant... [2] https://en.wikipedia.org/wiki/LISA_Pathfinder https://en.wikipedia.org/wiki/LISA_Pathfinder [3] https://en.wikipedia.org/wiki/TianQin https://en.wikipedia.org/wiki/TianQin [4] https://en.wikipedia.org/wiki/Deci-hertz_Interferometer_Gravitational_wave_Observatory https://en.wikipedia.org/wiki/Deci-hertz_Interferometer_Grav...

jcims 9y ago

I did some noodling a while back to see if LIGO could detect the mass energy conversion of a fission 'event' in North Korea. If memory serves it was definitely within the range in terms of impact to the gravitational field, but correlating a infinitessimally short pop of noise would be much more difficult than the death waltz of binary black holes. However, a larger array of more sensitive detectors that are immune from the tremendous amount of terrestrial noise might make short work of it.

StormChaser_5 9y ago
Why would an event in NK cause a change in gravity? Wouldn't the energy liberated in a nuclear event would cause the same amount of gravity to be felt elsewhere as the mass it came from?
- techdragon 9y ago
  Two theories off hand without research to double check. 1 - nuclear explosion involves the conversion of a tiny amount of mass into rapidly expanding cloud of energy conveyed by various particles. The significance being the drastic density difference before and after it’s blown to bits. This difference is equivalent to a sudden moment of regular mass since energy and mass are equivalent and energy has gravitational attraction. This is roughly the same as attempting to catch the gravitational waves theorised to be produced by a supernova. We can’t catch these yet due to insufficient sensitivity, but despite the nuclear explosion being ridiculously tiny compared to a supernova, we are also ridiculously close compared to the supernova, so the proximity to the event may “win” and make it strong enough for the detectors to see it. 2 - the equivalence of nuclear explosions to earthquakes in terms of energy displacement into the earths crust, and in particular underground tests of them, has been established. (This is obviously dependent on the size of the explosion and if it’s an underground test or not, since underground tests dump basically all their energy into the surrounding geology they make it a “more effective earthquake”). Earthquakes can speed up or slow down the earths rotation, which has tidal effects which involve the moon, so between the seismic and seismically induced tidal changes, we get to affecting the moons orbit and consequently we get a tiny gravitational wave which once again is detectable only because of how much closer we are to the source compared to the things we built LIGO to detect. Of course I might be wrong on both of these. They were just fun to think about on my ride home from work. :-)
- DoctorOetker 9y ago
  Initially I thought the same for a couple of seconds because of mass energy equivalence, but then I realized: assume the fission event to occur in empty space, assume the liberated energy is isotropically radiated, and assume all the energy is thus expanding spherically, then the mass/energy that was initially in the mother isotope is now after some time distributed on a spherical surface. As long as the observer is outside of the shell of escaping energy, to the observer the gravitational field looks like all the mass is concentrated at the center of the expanding shell of energy, exactly where the mass/energy originally was, and hence gravitationally undetectable. After a while the expanding surface has past the observer, and the observer is inside a spherically symmetric shell of mass/energy. Even if the observer is not at the center there will no gravitational field at all according to the observer. (consult any physicist or physics textbook, or do the exercise for the gravitational field in a 'hollow earth'). So to the observer this looks like a bit of disappearing mass at the location of fission event. It seems extremely difficult to me to actually measure such an event gravitationally though..

danbruc 9y ago

You are probably thinking of the planned Laser Interferometer Space Antenna (LISA) [1]. Launch is scheduled for 2034 but the technical demonstration mission LISA Pathfinder [2] was already done between 2015 and 2017 and exceeded its targets with regard to sensitivity. Other proposed space-based gravitational wave detectors are TianQin [3] and DECIGO [4]. [1] https://en.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna https://en.wikipedia.org/wiki/Laser_Interferometer_Space_Ant... [2] https://en.wikipedia.org/wiki/LISA_Pathfinder https://en.wikipedia.org/wiki/LISA_Pathfinder [3] https://en.wikipedia.org/wiki/TianQin https://en.wikipedia.org/wiki/TianQin [4] https://en.wikipedia.org/wiki/Deci-hertz_Interferometer_Gravitational_wave_Observatory https://en.wikipedia.org/wiki/Deci-hertz_Interferometer_Grav...