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(Solved): (1.3) Suppose that GW170817 would have been seen with Einstein Telescope (ET) rather than the LIGO-V ...
(1.3) Suppose that GW170817 would have been seen with Einstein Telescope (ET)
rather than the LIGO-Virgo detector network. ET will have much better suppression of
seismic noise than the existing detectors, allowing it to be sensitive starting from a frequency
as low as 3 Hz . Given that the source's component masses were m_(1)?1.6M_(o.) and m_(2)?
1.2M_(o.), where M_(o.)?2\times 10^(30)kg, for how much time would the signal have been visible in
ET?
Since for the three V-shaped detectors that compose Einstein Telescope, the distance D
in exercise (1.1) above is essentially zero, one might think that localizing the source on the
sky will be impossible with ET. Can you think of a way in which ET might be able to do
sky localization after all? (1.4) Finally, consider a supermassive binary black hole inspiral with m_(1)=10^(7)M_(o.)
and m_(2)=2\times 10^(6)M_(o.). Calculating the ISCO frequency, you will see that it is far too low
for Earth-based detectors, but accessible to the space-based LISA, which is sensitive in the
range 10^(-5)Hz-0.1Hz. For how long will the signal be in LISA's frequency band?
How do you think LISA will perform sky localization of supermassive binary black holes?
Also, despite that the black holes themselves emit no light, can you think of a reason why
electromagnetic astronomers would want to have a look at this sky location? The two LIGO interferometers are located respectively in Hanford (Washing-
ton) and Livingston (Louisiana), at a distance of 3002 kilometers from each other. Suppose
a gravitational wave passes by, which first reaches LIGO Hanford, and 8.1 milliseconds later
it reaches LIGO Livingston. This information narrows down the possible location of the
source to points on a ring in the sky. What are the orientation and angular radius of that
ring?
Due to the finite accuracy in time-of-arrival measurements, the ring will not be infinitely
thin. If the typical timing uncertainty is 0.1 milliseconds, what is the width of the ring in
degrees? ^(1)
With a network of more than two interferometers, the sky position of the source can be
narrowed down to a patch in the sky that is in the intersection of the rings one gets from
each pair of detectors. Your result for the width of a ring gives the approximate angular
size of that patch. Using the internet, compare this to the angular size of the Moon as it
appears from Earth.
