Fastest Eclipsing Binary, a Valuable Target for Gravitational Wave Studies:
Each night, Caltech's Zwicky Transient Facility (ZTF), a survey that uses the 48-inch telescope at Palomar Observatory, scans the sky for objects that move, blink, or otherwise vary in brightness. Promising candidates are followed up with a new instrument, the Kitt Peak 84-inch Electron Multiplying Demonstrator (KPED), at the Kitt Peak 2.1-meter telescope to identify short period eclipsing binaries. KPED is designed to measure with speed and sensitivity the changing brightness of celestial sources.
This approach has led to the discovery of ZTF J1539+5027 (or J1539 for short), a white dwarf eclipsing binary with the shortest period known to date, a mere 6.91 minutes. The stars orbit so close together that the entire system could fit within the diameter of the planet Saturn.
"As the dimmer star passes in front of the brighter one, it blocks most of the light, resulting in the seven-minute blinking pattern we see in the ZTF data," explains Caltech graduate student Kevin Burdge, lead author of the paper reporting the discovery, which appears in today's issue of the journal Nature.
[...] J1539 is a rare gem. It is one of only a few known sources of gravitational waves—ripples in space and time—that will be detected by the future European space mission LISA (Laser Interferometer Space Antenna), which is expected to launch in 2034. LISA, in which NASA plays a role, will be similar to the National Science Foundation's ground-based LIGO (Laser Interferometer Gravitational-wave Observatory), which made history in 2015 by making the first direct detection of gravitational waves from a pair of colliding black holes. LISA will detect gravitational waves from space at lower frequencies. J1539 is well matched to LISA; the 4.8 mHz gravitational wave frequency of J1539 is close to the peak of LISA's sensitivity.
[...] As remarkable as it is, J1539 was discovered with only a small portion of the data expected from ZTF. It was found in the ZTF team's initial analysis of 10 million sources, whereas the project will eventually study more than a billion stars.
Very roughly speaking, imagine two geostationary satellites on opposite sides of the earth. Orbiting the Earth's center in under 7 minutes. And each satellite having a mass on the order of the mass of the Sun.
Correction: The article reports their orbit could fit within Saturn's diameter ~75,000 miles. Earth's equatorial radius is ~4,000 miles. Geostationary orbit is ~22,200 miles above the Earth's surface (or ~26,200 miles above Earth's center). That gives a distance between opposing geostationary satellites of ~52,400 miles. So, extend the location of each geostationary satellite another ~11,400 miles above the Earth (roughly 3 Earth radii). For another perspective, the radius of the Moon's orbit is ~230,000 miles which is approximately 3 times the distance between these two orbiting stars, each of which is about the size of the Earth.
Journal Reference:
Burdge et al. General relativistic orbital decay in a seven-minute-orbital-period eclipsing binary system. Nature, 2019 DOI: 10.1038/s41586-019-1403-0
Here is a video animation of the eclipsing binaries on YouTube.
(Score: -1, Troll) by Anonymous Coward on Tuesday July 30 2019, @12:17PM
All they see is flickering light, the rest is speculation.
(Score: 2) by Rupert Pupnick on Tuesday July 30 2019, @12:47PM (1 child)
Anyone have a sense of just how weak these gravitational waves are if you were express them as a ratio (wave amplitude)/(static value of gravitational field on Earth’s surface)? How many orders of magnitude are we talking about?
(Score: 3, Interesting) by bzipitidoo on Wednesday July 31 2019, @04:56AM
I recall that the problem LIGO faced is that even the strongest gravity waves would alter the distance of 1 km by about the width of a proton.
(Score: 0) by Anonymous Coward on Tuesday July 30 2019, @01:31PM (3 children)
I do not understand those miles. Can someone convert to really scientific distance measuring units, please? Original article costs $8.99 for me...
(Score: 1) by kiffer on Tuesday July 30 2019, @01:45PM
Sure, The distances listed are easily converted to more scientific units.
"""
The article reports their orbit could fit within Saturn's diameter ~8.07x10^-4 AU. Earth's equatorial radius is 4.30x10^-5 AU. Geostationary orbit is ~2.39x10^-4 AU above the Earth's surface (or ~2.82x10^-4 AU above Earth's center). That gives a distance between opposing geostationary satellites of ~5.64x10^-4 AU. So, extend the location of each geostationary satellite another ~1.23x10^-4 AU above the Earth (roughly 3 Earth radii). For another perspective, the radius of the Moon's orbit is ~2.47x10^-3 AU which is approximately 3 times the distance between these two orbiting stars, each of which is about the size of the Earth.
"""
(Score: 0) by Anonymous Coward on Tuesday July 30 2019, @08:08PM (1 child)
There is a reason the US doesn't use the units of eternal tyranny and globalism.
(Score: 2) by fido_dogstoyevsky on Tuesday July 30 2019, @10:46PM
Did I miss the sarcasm tag, or are you under the impression that the US uses metric instead of imperial measures?
It's NOT a conspiracy... it's a plot.
(Score: 3, Informative) by Immerman on Tuesday July 30 2019, @01:43PM (1 child)
> imagine two geostationary satellites on opposite sides of the earth. Orbiting the Earth's center in under 7 minutes.
I think someone doesn't quite get the concept of "geostationary orbit" The orbital period of any geostationary satellite is 24 hours - any other speed and it will more across the sky as seen from Earth aka NOT be geostationary
Of course if we had two suns orbitting at that distance the Earth would doubtless be tidally locked to them in short order, so I suppose whatever speed they were orbitting at would be geostationary. Of course they'd actually be orbiting each other rather than the Earth, and the Earth would almost certainly fall into one or the other* long before tidal locking could occur. (* or both, it'd probably be well within the Roche limit of both stars and be torn to shreds)
(Score: 4, Insightful) by maxwell demon on Tuesday July 30 2019, @06:10PM
No, the orbital period of any geostationary orbit is exactly the rotation period of the earth, which is about 23 hours and 56 minutes.
The Tao of math: The numbers you can count are not the real numbers.
(Score: 0) by Anonymous Coward on Wednesday July 31 2019, @04:58AM
This is how KITT Peak notifies the astronomers of what it found: https://media3.giphy.com/media/7GryQBpBTrdW8/giphy.gif?cid=790b76115d411dcd61477a3167960830&rid=giphy.gif [giphy.com]