What produces gravitational waves with “periods between about 100 - 8000 seconds”? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Redshift for gravitational waves?Frequency of gravitational wave detectionWould humans hear gravity waves from a binary BH fusion nearby?What are gravitational waves actually?LIGO gravitational wave chirp signal frequencyWhich of the following statements about gravitational waves are true?What is the detection threshold of gravitational waves for LIGO?“Who saw” the binary neutron star merger first? What was the sequence of events? (GRB/GW170817)Do gravitational waves have distinct bands or parameters from which a source redshift can be inferred?Would an X-ray-based pulsar timing array in orbit or on the Moon to study GW background be effective?
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What produces gravitational waves with “periods between about 100 - 8000 seconds”?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Redshift for gravitational waves?Frequency of gravitational wave detectionWould humans hear gravity waves from a binary BH fusion nearby?What are gravitational waves actually?LIGO gravitational wave chirp signal frequencyWhich of the following statements about gravitational waves are true?What is the detection threshold of gravitational waves for LIGO?“Who saw” the binary neutron star merger first? What was the sequence of events? (GRB/GW170817)Do gravitational waves have distinct bands or parameters from which a source redshift can be inferred?Would an X-ray-based pulsar timing array in orbit or on the Moon to study GW background be effective?
$begingroup$
The Ulysses mission has a compelling story. I was sent to Jupiter to perform a gravitational assist shooting it out of the plane of the ecliptic in order to fly over the Sun's north and south poles to perform "fast latitude scans". Because of its design it was used for several important lines of scientific study.
Ulysses contained a pair of coherent transponders which received signals from Earth, shifted them in frequency in a coherent way using phase-locked loops and beamed them immediately back to Earth at two different frequencies.
From ESA's write up of the Ulysses Gravitational Wave Experiment:
In the spacecraft Doppler tracking method, the Earth and spacecraft constitute the two objects whose time-varying separation is monitored to detect a passing gravitational wave. The monitoring is accomplished with high-precision Doppler tracking in which a constant frequency microwave radio signal (S-band) is transmitted from the Earth to the spacecraft (uplink); the signal is transponded (received and coherently amplified) at the spacecraft; and then transmitted back to Earth (downlink) in both S- and X-band signals. This Dual frequency downlink is required in order to calibrate the interplanetary media which affects the two frequency bands differently. The downlink signal is recorded at Earth and its frequency is compared to the constant uplink frequency f0 to extract the Doppler signal, δf / f0.
The article goes on to say:
Since the optimum size of a gravitational wave detector is the wave length, interplanetary dimensions are needed for detecting gravitational waves in the mHz range. Doppler tracking of Ulysses provides sensitive detections of gravitational waves in this low frequency band. The driving noise source is the fluctuations in the refractive index of interplanetary plasma. This dictates the timing of the experiment to be near solar opposition and sets the target accuracy for the fractional frequency change at 3.0 × 10-14 for integration times of the order of 1000 seconds.
SUMMARY OF OBJECTIVES
The objective of the gravitational wave investigation on Ulysses is to search for low frequency gravitational waves crossing the Solar System. Because of the great distance to the spacecraft, this method is most sensitive to wave periods between about 100 - 8000 seconds, a band which is not accessible to ground-based experiments which are superior for periods below 1 second.
You can read more about Ulysses in eoPortal's Ulysses where I found both the link above and the following:
B. Bertotti, R. Ambrosini, S. W. Asmar, J. P. Brenkle, G. Comoretto, G. Giampieri, L. Iess, A. Messeri, H. D. Wahlquist, “The gravitational wave experiment,” Astronomy and Astrophysics Supplement Series, Ulysses Instruments Special Issue, Vol. 92, No. 2, pp. 431-440, Jan. 1992
Question: What produces gravitational waves with "periods between about 100 - 8000 seconds"?
gravitational-waves artificial-satellite nasa
asked Apr 20 at 4:16
uhohuhoh
7,70822275
$endgroup$
add a comment |
$begingroup$
The Ulysses mission has a compelling story. I was sent to Jupiter to perform a gravitational assist shooting it out of the plane of the ecliptic in order to fly over the Sun's north and south poles to perform "fast latitude scans". Because of its design it was used for several important lines of scientific study.
Ulysses contained a pair of coherent transponders which received signals from Earth, shifted them in frequency in a coherent way using phase-locked loops and beamed them immediately back to Earth at two different frequencies.
From ESA's write up of the Ulysses Gravitational Wave Experiment:
In the spacecraft Doppler tracking method, the Earth and spacecraft constitute the two objects whose time-varying separation is monitored to detect a passing gravitational wave. The monitoring is accomplished with high-precision Doppler tracking in which a constant frequency microwave radio signal (S-band) is transmitted from the Earth to the spacecraft (uplink); the signal is transponded (received and coherently amplified) at the spacecraft; and then transmitted back to Earth (downlink) in both S- and X-band signals. This Dual frequency downlink is required in order to calibrate the interplanetary media which affects the two frequency bands differently. The downlink signal is recorded at Earth and its frequency is compared to the constant uplink frequency f0 to extract the Doppler signal, δf / f0.
The article goes on to say:
Since the optimum size of a gravitational wave detector is the wave length, interplanetary dimensions are needed for detecting gravitational waves in the mHz range. Doppler tracking of Ulysses provides sensitive detections of gravitational waves in this low frequency band. The driving noise source is the fluctuations in the refractive index of interplanetary plasma. This dictates the timing of the experiment to be near solar opposition and sets the target accuracy for the fractional frequency change at 3.0 × 10-14 for integration times of the order of 1000 seconds.
SUMMARY OF OBJECTIVES
The objective of the gravitational wave investigation on Ulysses is to search for low frequency gravitational waves crossing the Solar System. Because of the great distance to the spacecraft, this method is most sensitive to wave periods between about 100 - 8000 seconds, a band which is not accessible to ground-based experiments which are superior for periods below 1 second.
You can read more about Ulysses in eoPortal's Ulysses where I found both the link above and the following:
B. Bertotti, R. Ambrosini, S. W. Asmar, J. P. Brenkle, G. Comoretto, G. Giampieri, L. Iess, A. Messeri, H. D. Wahlquist, “The gravitational wave experiment,” Astronomy and Astrophysics Supplement Series, Ulysses Instruments Special Issue, Vol. 92, No. 2, pp. 431-440, Jan. 1992
Question: What produces gravitational waves with "periods between about 100 - 8000 seconds"?
gravitational-waves artificial-satellite nasa
asked Apr 20 at 4:16
uhohuhoh
7,70822275
$endgroup$
$begingroup$
Binary black holes that are not in the final stage of their approach. The frequency of the GW increases as the BHs get closer together - look at the graphs from LIGO and you'll see frequency speeding up towards the end. In the early stages it could take them a very long time to complete an orbit - just like binary stars, really. But amplitude is quite a bit lower in the early stages, so hopefully this new detector is much more sensitive amplitude-wise.
$endgroup$
– Florin Andrei
2 days ago
add a comment |
$begingroup$
The Ulysses mission has a compelling story. I was sent to Jupiter to perform a gravitational assist shooting it out of the plane of the ecliptic in order to fly over the Sun's north and south poles to perform "fast latitude scans". Because of its design it was used for several important lines of scientific study.
Ulysses contained a pair of coherent transponders which received signals from Earth, shifted them in frequency in a coherent way using phase-locked loops and beamed them immediately back to Earth at two different frequencies.
From ESA's write up of the Ulysses Gravitational Wave Experiment:
In the spacecraft Doppler tracking method, the Earth and spacecraft constitute the two objects whose time-varying separation is monitored to detect a passing gravitational wave. The monitoring is accomplished with high-precision Doppler tracking in which a constant frequency microwave radio signal (S-band) is transmitted from the Earth to the spacecraft (uplink); the signal is transponded (received and coherently amplified) at the spacecraft; and then transmitted back to Earth (downlink) in both S- and X-band signals. This Dual frequency downlink is required in order to calibrate the interplanetary media which affects the two frequency bands differently. The downlink signal is recorded at Earth and its frequency is compared to the constant uplink frequency f0 to extract the Doppler signal, δf / f0.
The article goes on to say:
Since the optimum size of a gravitational wave detector is the wave length, interplanetary dimensions are needed for detecting gravitational waves in the mHz range. Doppler tracking of Ulysses provides sensitive detections of gravitational waves in this low frequency band. The driving noise source is the fluctuations in the refractive index of interplanetary plasma. This dictates the timing of the experiment to be near solar opposition and sets the target accuracy for the fractional frequency change at 3.0 × 10-14 for integration times of the order of 1000 seconds.
SUMMARY OF OBJECTIVES
The objective of the gravitational wave investigation on Ulysses is to search for low frequency gravitational waves crossing the Solar System. Because of the great distance to the spacecraft, this method is most sensitive to wave periods between about 100 - 8000 seconds, a band which is not accessible to ground-based experiments which are superior for periods below 1 second.
You can read more about Ulysses in eoPortal's Ulysses where I found both the link above and the following:
B. Bertotti, R. Ambrosini, S. W. Asmar, J. P. Brenkle, G. Comoretto, G. Giampieri, L. Iess, A. Messeri, H. D. Wahlquist, “The gravitational wave experiment,” Astronomy and Astrophysics Supplement Series, Ulysses Instruments Special Issue, Vol. 92, No. 2, pp. 431-440, Jan. 1992
Question: What produces gravitational waves with "periods between about 100 - 8000 seconds"?
gravitational-waves artificial-satellite nasa
asked Apr 20 at 4:16
uhohuhoh
7,70822275
$endgroup$
The Ulysses mission has a compelling story. I was sent to Jupiter to perform a gravitational assist shooting it out of the plane of the ecliptic in order to fly over the Sun's north and south poles to perform "fast latitude scans". Because of its design it was used for several important lines of scientific study.
Ulysses contained a pair of coherent transponders which received signals from Earth, shifted them in frequency in a coherent way using phase-locked loops and beamed them immediately back to Earth at two different frequencies.
From ESA's write up of the Ulysses Gravitational Wave Experiment:
In the spacecraft Doppler tracking method, the Earth and spacecraft constitute the two objects whose time-varying separation is monitored to detect a passing gravitational wave. The monitoring is accomplished with high-precision Doppler tracking in which a constant frequency microwave radio signal (S-band) is transmitted from the Earth to the spacecraft (uplink); the signal is transponded (received and coherently amplified) at the spacecraft; and then transmitted back to Earth (downlink) in both S- and X-band signals. This Dual frequency downlink is required in order to calibrate the interplanetary media which affects the two frequency bands differently. The downlink signal is recorded at Earth and its frequency is compared to the constant uplink frequency f0 to extract the Doppler signal, δf / f0.
The article goes on to say:
Since the optimum size of a gravitational wave detector is the wave length, interplanetary dimensions are needed for detecting gravitational waves in the mHz range. Doppler tracking of Ulysses provides sensitive detections of gravitational waves in this low frequency band. The driving noise source is the fluctuations in the refractive index of interplanetary plasma. This dictates the timing of the experiment to be near solar opposition and sets the target accuracy for the fractional frequency change at 3.0 × 10-14 for integration times of the order of 1000 seconds.
SUMMARY OF OBJECTIVES
The objective of the gravitational wave investigation on Ulysses is to search for low frequency gravitational waves crossing the Solar System. Because of the great distance to the spacecraft, this method is most sensitive to wave periods between about 100 - 8000 seconds, a band which is not accessible to ground-based experiments which are superior for periods below 1 second.
You can read more about Ulysses in eoPortal's Ulysses where I found both the link above and the following:
B. Bertotti, R. Ambrosini, S. W. Asmar, J. P. Brenkle, G. Comoretto, G. Giampieri, L. Iess, A. Messeri, H. D. Wahlquist, “The gravitational wave experiment,” Astronomy and Astrophysics Supplement Series, Ulysses Instruments Special Issue, Vol. 92, No. 2, pp. 431-440, Jan. 1992
Question: What produces gravitational waves with "periods between about 100 - 8000 seconds"?
gravitational-waves artificial-satellite nasa
gravitational-waves artificial-satellite nasa
asked Apr 20 at 4:16
uhohuhoh
7,70822275
asked Apr 20 at 4:16
uhohuhoh
7,70822275
asked Apr 20 at 4:16
uhohuhoh
7,70822275
asked Apr 20 at 4:16
uhohuhoh
7,70822275
asked Apr 20 at 4:16
uhohuhoh
7,70822275
7,70822275
$begingroup$
Binary black holes that are not in the final stage of their approach. The frequency of the GW increases as the BHs get closer together - look at the graphs from LIGO and you'll see frequency speeding up towards the end. In the early stages it could take them a very long time to complete an orbit - just like binary stars, really. But amplitude is quite a bit lower in the early stages, so hopefully this new detector is much more sensitive amplitude-wise.
$endgroup$
– Florin Andrei
2 days ago
add a comment |
$begingroup$
Binary black holes that are not in the final stage of their approach. The frequency of the GW increases as the BHs get closer together - look at the graphs from LIGO and you'll see frequency speeding up towards the end. In the early stages it could take them a very long time to complete an orbit - just like binary stars, really. But amplitude is quite a bit lower in the early stages, so hopefully this new detector is much more sensitive amplitude-wise.
$endgroup$
– Florin Andrei
2 days ago
$begingroup$
Binary black holes that are not in the final stage of their approach. The frequency of the GW increases as the BHs get closer together - look at the graphs from LIGO and you'll see frequency speeding up towards the end. In the early stages it could take them a very long time to complete an orbit - just like binary stars, really. But amplitude is quite a bit lower in the early stages, so hopefully this new detector is much more sensitive amplitude-wise.
$endgroup$
– Florin Andrei
2 days ago
$begingroup$
Binary black holes that are not in the final stage of their approach. The frequency of the GW increases as the BHs get closer together - look at the graphs from LIGO and you'll see frequency speeding up towards the end. In the early stages it could take them a very long time to complete an orbit - just like binary stars, really. But amplitude is quite a bit lower in the early stages, so hopefully this new detector is much more sensitive amplitude-wise.
$endgroup$
– Florin Andrei
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any binary system produces gravitational waves at twice it's orbital frequency, i.e. with periods of half it's orbital period. So binary systems with periods between 200s and 16000s will produce such waves.
We can use Kepler's third law to say something about these:
$$ a = left(fracGM4piright)^1/3 P^2/3,$$
where $P$ is the orbital period, $M$ is the total mass of the binary system and $a$ is the orbital separation.
For a binary with $Msim 1M_odot$ and $200<P<1600$s, then $0.11 < a < 2.00 R_odot$. Since normal stars of mass $sim 0.5M_odot$ have radii that are similar to this, then the stars would probably need to be stellar remnants (white dwarfs, neutron stars or black holes) except right at the longest period end, where contact W Uma binaries would be. More massive binaries only have separations that increase as $M^1/3$, but the radii of normal stars increases more like $M$, so this conclusion is even firmer.
It could be possible to have a compact binary involving a low mass star plus a compact object - perhaps a Roche lobe filling one, so as well as "double degenerates", the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts.
Of course gravitational wave strain goes as something like $M P^-4/3 d^-1$, where $d$ is the distance. These binaries are much longer period than the merging black holes seen so far and so probably need to be close, in our Galaxy, to be detected.
answered 2 days ago
Rob JeffriesRob Jeffries
55.1k4114177
$endgroup$
$begingroup$
Thanks for the clear answer! I'm curious how the strain scales with $M$. Perhaps some of that weakness could be made up by the pair being fairly massive?
$endgroup$
– uhoh
2 days ago
add a comment |
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$begingroup$
Any binary system produces gravitational waves at twice it's orbital frequency, i.e. with periods of half it's orbital period. So binary systems with periods between 200s and 16000s will produce such waves.
We can use Kepler's third law to say something about these:
$$ a = left(fracGM4piright)^1/3 P^2/3,$$
where $P$ is the orbital period, $M$ is the total mass of the binary system and $a$ is the orbital separation.
For a binary with $Msim 1M_odot$ and $200<P<1600$s, then $0.11 < a < 2.00 R_odot$. Since normal stars of mass $sim 0.5M_odot$ have radii that are similar to this, then the stars would probably need to be stellar remnants (white dwarfs, neutron stars or black holes) except right at the longest period end, where contact W Uma binaries would be. More massive binaries only have separations that increase as $M^1/3$, but the radii of normal stars increases more like $M$, so this conclusion is even firmer.
It could be possible to have a compact binary involving a low mass star plus a compact object - perhaps a Roche lobe filling one, so as well as "double degenerates", the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts.
Of course gravitational wave strain goes as something like $M P^-4/3 d^-1$, where $d$ is the distance. These binaries are much longer period than the merging black holes seen so far and so probably need to be close, in our Galaxy, to be detected.
answered 2 days ago
Rob JeffriesRob Jeffries
55.1k4114177
$endgroup$
$begingroup$
Thanks for the clear answer! I'm curious how the strain scales with $M$. Perhaps some of that weakness could be made up by the pair being fairly massive?
$endgroup$
– uhoh
2 days ago
add a comment |
$begingroup$
Any binary system produces gravitational waves at twice it's orbital frequency, i.e. with periods of half it's orbital period. So binary systems with periods between 200s and 16000s will produce such waves.
We can use Kepler's third law to say something about these:
$$ a = left(fracGM4piright)^1/3 P^2/3,$$
where $P$ is the orbital period, $M$ is the total mass of the binary system and $a$ is the orbital separation.
For a binary with $Msim 1M_odot$ and $200<P<1600$s, then $0.11 < a < 2.00 R_odot$. Since normal stars of mass $sim 0.5M_odot$ have radii that are similar to this, then the stars would probably need to be stellar remnants (white dwarfs, neutron stars or black holes) except right at the longest period end, where contact W Uma binaries would be. More massive binaries only have separations that increase as $M^1/3$, but the radii of normal stars increases more like $M$, so this conclusion is even firmer.
It could be possible to have a compact binary involving a low mass star plus a compact object - perhaps a Roche lobe filling one, so as well as "double degenerates", the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts.
Of course gravitational wave strain goes as something like $M P^-4/3 d^-1$, where $d$ is the distance. These binaries are much longer period than the merging black holes seen so far and so probably need to be close, in our Galaxy, to be detected.
answered 2 days ago
Rob JeffriesRob Jeffries
55.1k4114177
$endgroup$
$begingroup$
Thanks for the clear answer! I'm curious how the strain scales with $M$. Perhaps some of that weakness could be made up by the pair being fairly massive?
$endgroup$
– uhoh
2 days ago
add a comment |
$begingroup$
Any binary system produces gravitational waves at twice it's orbital frequency, i.e. with periods of half it's orbital period. So binary systems with periods between 200s and 16000s will produce such waves.
We can use Kepler's third law to say something about these:
$$ a = left(fracGM4piright)^1/3 P^2/3,$$
where $P$ is the orbital period, $M$ is the total mass of the binary system and $a$ is the orbital separation.
For a binary with $Msim 1M_odot$ and $200<P<1600$s, then $0.11 < a < 2.00 R_odot$. Since normal stars of mass $sim 0.5M_odot$ have radii that are similar to this, then the stars would probably need to be stellar remnants (white dwarfs, neutron stars or black holes) except right at the longest period end, where contact W Uma binaries would be. More massive binaries only have separations that increase as $M^1/3$, but the radii of normal stars increases more like $M$, so this conclusion is even firmer.
It could be possible to have a compact binary involving a low mass star plus a compact object - perhaps a Roche lobe filling one, so as well as "double degenerates", the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts.
Of course gravitational wave strain goes as something like $M P^-4/3 d^-1$, where $d$ is the distance. These binaries are much longer period than the merging black holes seen so far and so probably need to be close, in our Galaxy, to be detected.
answered 2 days ago
Rob JeffriesRob Jeffries
55.1k4114177
$endgroup$
Any binary system produces gravitational waves at twice it's orbital frequency, i.e. with periods of half it's orbital period. So binary systems with periods between 200s and 16000s will produce such waves.
We can use Kepler's third law to say something about these:
$$ a = left(fracGM4piright)^1/3 P^2/3,$$
where $P$ is the orbital period, $M$ is the total mass of the binary system and $a$ is the orbital separation.
For a binary with $Msim 1M_odot$ and $200<P<1600$s, then $0.11 < a < 2.00 R_odot$. Since normal stars of mass $sim 0.5M_odot$ have radii that are similar to this, then the stars would probably need to be stellar remnants (white dwarfs, neutron stars or black holes) except right at the longest period end, where contact W Uma binaries would be. More massive binaries only have separations that increase as $M^1/3$, but the radii of normal stars increases more like $M$, so this conclusion is even firmer.
It could be possible to have a compact binary involving a low mass star plus a compact object - perhaps a Roche lobe filling one, so as well as "double degenerates", the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts.
Of course gravitational wave strain goes as something like $M P^-4/3 d^-1$, where $d$ is the distance. These binaries are much longer period than the merging black holes seen so far and so probably need to be close, in our Galaxy, to be detected.
answered 2 days ago
Rob JeffriesRob Jeffries
55.1k4114177
edited 2 days ago
answered 2 days ago
Rob JeffriesRob Jeffries
55.1k4114177
answered 2 days ago
Rob JeffriesRob Jeffries
55.1k4114177
answered 2 days ago
Rob JeffriesRob Jeffries
55.1k4114177
55.1k4114177
$begingroup$
Thanks for the clear answer! I'm curious how the strain scales with $M$. Perhaps some of that weakness could be made up by the pair being fairly massive?
$endgroup$
– uhoh
2 days ago
add a comment |
$begingroup$
Thanks for the clear answer! I'm curious how the strain scales with $M$. Perhaps some of that weakness could be made up by the pair being fairly massive?
$endgroup$
– uhoh
2 days ago
$begingroup$
Thanks for the clear answer! I'm curious how the strain scales with $M$. Perhaps some of that weakness could be made up by the pair being fairly massive?
$endgroup$
– uhoh
2 days ago
$begingroup$
Thanks for the clear answer! I'm curious how the strain scales with $M$. Perhaps some of that weakness could be made up by the pair being fairly massive?
$endgroup$
– uhoh
2 days ago
add a comment |
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Binary black holes that are not in the final stage of their approach. The frequency of the GW increases as the BHs get closer together - look at the graphs from LIGO and you'll see frequency speeding up towards the end. In the early stages it could take them a very long time to complete an orbit - just like binary stars, really. But amplitude is quite a bit lower in the early stages, so hopefully this new detector is much more sensitive amplitude-wise.
$endgroup$
– Florin Andrei
2 days ago