<p>Says <strong>@uhoh</strong> “<em>I'm curious how the strain scales with M</em>”</p>
<p>If <span class="math-container">$m_1$</span> and <span class="math-container">$m_2$</span> are the respective masses of an orbitally bound binary body, and we define <span class="math-container">$M_c$</span> as the <em>chirp mass</em>:
<span class="math-container">$$M_c=\dfrac{(m_1 m_2)^{3/5}}{(m_1+m_2)^{1/5}}$$</span>
Note that <span class="math-container">$M_c$</span> has dimensions of mass. We call <span class="math-container">$f$</span> the frequency of the gravitational wave: remember that, as <strong>@ProfRob</strong> correctly explains, <span class="math-container">$f$</span> is twice the orbital frequency of the binary pair.
<span class="math-container">$$f=2 f_{orb}$$</span>
<span class="math-container">$c$</span> and <span class="math-container">$G$</span> are respectively the speed of light and the Universal Gravitational Constant.
And <span class="math-container">$D$</span> is the luminosity distance between the binary pair and us.
Then, the expression for the strain <span class="math-container">$h$</span> is:
<span class="math-container">$$\boxed{h=\dfrac{4GM_c}{D \ c^2} \ \left ( \dfrac{\pi G}{c^3} \ f M_c \right )^{2/3}}$$</span>
For fixed <span class="math-container">$D$</span> and <span class="math-container">$f$</span>
<span class="math-container">$$h \propto M_c^{5/3}$$</span>
The source I have used are notes entitled &quot;<em>Relativistic Astrophysics - Lecture. Gravitational Waves</em>&quot; authored by John Wheeler. I have the notes on paper; unfortunately, I have not found the PDF freely available on the Internet.</p>
<p>Best regards.</p>

CC BY-SA 4.0
2023-04-19T06:16:07.167Z
OK @uhoh, best regards
Albert
CC BY-SA 4.0
2023-04-25T12:51:33.473Z
Done! https://astronomy.stackexchange.com/a/53471/31201
Albert
CC BY-SA 4.0
2023-04-25T20:46:21.660Z
Yes I appreciate it very much, thanks!
uhoh
CC BY-SA 4.0
2023-04-18T12:52:07.113Z
2023-04-18T14:08:48.500Z
2023-04-18T14:08:48.500Z
Albert
<p>A gravitational wave period between 100 and 8000 seconds corresponds to a frequency between 0.8 mHz and 60 mHz. This is roughly the frequency band that will be targeted by the LISA space-based gravitational wave observatory.</p>
<p>In this regime we expect three main classes of sources:</p>
<ol>
<li><p><strong>Massive black hole mergers</strong> Mergers of massive black holes after the merger of two galaxies.</p>
</li>
<li><p><strong>Extreme mass-ratio inspirals</strong> (EMRI) The merger of a stellar mass compact object (black hole or neutron star) at the center of a galaxy.</p>
</li>
<li><p><strong>Galactic compact binaries</strong> Binaries of two compact objects well before merger. These are the sources discussed in the other answers. We expect there to be millions of these in our own galaxy.</p>
</li>
</ol>
<p>However, none of these sources are expected to be loud enough to be (have been) detected by Ulysses.</p>

CC BY-SA 4.0
2023-04-23T21:39:14.880Z
2023-04-23T21:39:14.880Z
TimRias
<p>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 gravitational waves with periods of 100-800s.</p>
<p>We can use Kepler's third law to say something about these:
<span class="math-container">$$ a = \left(\frac{GM}{4\pi^2}\right)^{1/3} P^{2/3},$$</span>
where <span class="math-container">$P$</span> is the orbital period, <span class="math-container">$M$</span> is the total mass of the binary system and <span class="math-container">$a$</span> is the orbital separation.</p>
<p>For a binary with <span class="math-container">$M\sim 1M_{\odot}$</span> and <span class="math-container">$200&lt;P&lt;16000$</span>s, then <span class="math-container">$0.07 &lt; a/R_\odot &lt; 1.37$</span>. Since normal stars of mass <span class="math-container">$\sim 1M_{\odot}$</span> have radii that are similar to or larger 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 it might just be possible to observe W UMa (contact) binaries. More massive binaries have separations that increase as <span class="math-container">$M^{1/3}$</span>, but the radii of normal stars increases more like <span class="math-container">$M$</span>, so this conclusion is even firmer at larger masses.</p>
<p>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 &quot;double degenerates&quot;, the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts, with orbital periods of a few hours. Here is a prime example <a href="https://astronomy.stackexchange.com/questions/43435/time-domain-astronomy-and-fastest-eclipsing-binary-ztf-j15395027-20-mag-6-91">Time domain astronomy and fastest eclipsing binary ZTF J1539+5027 (+20 mag, 6.91 minutes): How to measure its minimum brightness?</a></p>
<p>Of course gravitational wave strain from a binary goes as something like <span class="math-container">$M^{5/3} P^{-2/3} d^{-1}$</span>, where <span class="math-container">$d$</span> is the distance. These binaries are much longer period than the (presumably rare) massive, merging black holes seen so far and so probably need to be close, in our own Galaxy, to be detected.</p>
<p>e.g. LIGO was capable of detecting <span class="math-container">$M \sim 30 M_{\odot}$</span> merging black holes, with <span class="math-container">$P \sim 0.02$</span> s at distances of a billion light years. A similar strain amplitude would be produced by a <span class="math-container">$M\sim 2M_{\odot}$</span> binary with <span class="math-container">$P= 200$</span> s at a distance of 20,000 light years.</p>

CC BY-SA 4.0
2019-04-20T08:01:11.990Z
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?
uhoh
CC BY-SA 4.0
2019-04-24T00:02:20.863Z
I just ran across [this deleted answer](https://astronomy.stackexchange.com/a/30097/7982), saw that the supernovae fit within the 100-8000 limits and thought to add a supplemental answer. Then I saw it was *your deleted answer*. Do you think adding a reference to this phenomenon might be useful here?
uhoh
CC BY-SA 4.0
2019-04-24T06:40:33.190Z
@uhoh Look again. The x-axis is frequency, not period
ProfRob
CC BY-SA 4.0
2019-04-24T08:10:43.057Z
and so it is, thanks!
uhoh
CC BY-SA 4.0
2019-04-20T07:56:37.187Z
2023-04-18T13:46:27.267Z
2023-04-18T13:46:27.267Z
ProfRob
<p>The <a href="https://en.wikipedia.org/wiki/Ulysses_(spacecraft)" rel="nofollow noreferrer">Ulysses mission</a> has a compelling story. It 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 &quot;fast latitude scans&quot;. Because of its design it was used for several important lines of scientific study.</p>
<p>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.</p>
<p>From ESA's write up of the <a href="http://sci.esa.int/ulysses/40406-instruments/?fbodylongid=1940" rel="nofollow noreferrer">Ulysses Gravitational Wave Experiment</a>:</p>
<blockquote>
<p>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.</p>
</blockquote>
<p>The article goes on to say:</p>
<blockquote>
<p>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.</p>
<p>SUMMARY OF OBJECTIVES</p>
<p>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, <strong>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.</strong></p>
</blockquote>
<p>You can read more about Ulysses in <a href="https://earth.esa.int/web/eoportal/satellite-missions/u/ulysses#foot46%29" rel="nofollow noreferrer">eoPortal's Ulysses</a> where I found both the link above and the following:</p>
<p>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</p>
<hr />
<p><strong>Question:</strong> What produces gravitational waves with &quot;periods between about 100 - 8000 seconds&quot;?</p>

CC BY-SA 4.0
2019-04-20T07:11:53.543Z
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.
Florin Andrei
CC BY-SA 4.0
2019-08-08T18:03:36.403Z
"I was sent to Jupiter to perform a gravitational assist shooting it out of the plane of the ecliptic" That must have been an adventure :>) .
Keith McClary
CC BY-SA 4.0
2019-08-08T22:27:43.213Z
@KeithMcClary Oh rats! That was a Freudian slip; I'm not supposed to talk about it.
uhoh
CC BY-SA 4.0
2021-04-23T03:07:34.773Z
related, but at the kilo-Hertz rather than the milli-Hertz end of the spectrum: [What would a kHz gravitational wave detector look like? (mountains on millisecond magnetars) How would it differ from LIGO/Virgo?](https://astronomy.stackexchange.com/q/43518/7982)
uhoh
<orbit><general-relativity><gravitational-waves><artificial-satellite><nasa>
What produces gravitational waves with "periods between about 100 - 8000 seconds"?
uhoh
# What produces gravitational waves with "periods between about 100 - 8000 seconds"?

(Asked by: username_0 on 2019-04-20)

The [Ulysses mission](https://en.wikipedia.org/wiki/Ulysses_(spacecraft)) has a compelling story. It 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](http://sci.esa.int/ulysses/40406-instruments/?fbodylongid=1940):

> 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](https://earth.esa.int/web/eoportal/satellite-missions/u/ulysses#foot46%29) 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"?

**Tags:** orbit, general-relativity, gravitational-waves, artificial-satellite, nasa

> **Comments:**
> **username_1**: 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.  
> **username_2**: "I was sent to Jupiter to perform a gravitational assist shooting it out of the plane of the ecliptic" That must have been an adventure :>) .  
> **username_0**: @username_2 Oh rats! That was a Freudian slip; I'm not supposed to talk about it.  
> **username_0**: related, but at the kilo-Hertz rather than the milli-Hertz end of the spectrum: [What would a kHz gravitational wave detector look like? (mountains on millisecond magnetars) How would it differ from LIGO/Virgo?](https://astronomy.stackexchange.com/q/43518/7982)  

---

## 3 answers

### Answer by username_5 on 2019-04-20. Score: 8 | Accepted answer

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 gravitational waves with periods of 100-800s.

We can use Kepler's third law to say something about these:
$$ a = \left(\frac{GM}{4\pi^2}\right)^{1/3} P^{2/3},$$
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 $M\sim 1M\_{\odot}$ and $200<P<16000$s, then $0.07 < a/R\_\odot < 1.37$. Since normal stars of mass $\sim 1M\_{\odot}$ have radii that are similar to or larger 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 it might just be possible to observe W UMa (contact) binaries. More massive binaries have separations that increase as $M^{1/3}$, but the radii of normal stars increases more like $M$, so this conclusion is even firmer at larger masses.

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, with orbital periods of a few hours. Here is a prime example [Time domain astronomy and fastest eclipsing binary ZTF J1539+5027 (+20 mag, 6.91 minutes): How to measure its minimum brightness?](https://astronomy.stackexchange.com/questions/43435/time-domain-astronomy-and-fastest-eclipsing-binary-ztf-j15395027-20-mag-6-91)

Of course gravitational wave strain from a binary goes as something like $M^{5/3} P^{-2/3} d^{-1}$, where $d$ is the distance. These binaries are much longer period than the (presumably rare) massive, merging black holes seen so far and so probably need to be close, in our own Galaxy, to be detected.

e.g. LIGO was capable of detecting $M \sim 30 M\_{\odot}$ merging black holes, with $P \sim 0.02$ s at distances of a billion light years. A similar strain amplitude would be produced by a $M\sim 2M\_{\odot}$ binary with $P= 200$ s at a distance of 20,000 light years.

> **Comments:**
> **username_0**: 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?  
> **username_0**: I just ran across [this deleted answer](https://astronomy.stackexchange.com/a/30097/7982), saw that the supernovae fit within the 100-8000 limits and thought to add a supplemental answer. Then I saw it was *your deleted answer*. Do you think adding a reference to this phenomenon might be useful here?  
> **username_5**: @username_0 Look again. The x-axis is frequency, not period  
> **username_0**: and so it is, thanks!  

---

### Answer by username_3 on 2023-04-18. Score: 4

Says **@username_0** “*I'm curious how the strain scales with M*”

If $m\_1$ and $m\_2$ are the respective masses of an orbitally bound binary body, and we define $M\_c$ as the *chirp mass*:
$$M\_c=\dfrac{(m\_1 m\_2)^{3/5}}{(m\_1+m\_2)^{1/5}}$$
Note that $M\_c$ has dimensions of mass. We call $f$ the frequency of the gravitational wave: remember that, as **@username_5** correctly explains, $f$ is twice the orbital frequency of the binary pair.
$$f=2 f\_{orb}$$
$c$ and $G$ are respectively the speed of light and the Universal Gravitational Constant.
And $D$ is the luminosity distance between the binary pair and us.
Then, the expression for the strain $h$ is:
$$\boxed{h=\dfrac{4GM\_c}{D \ c^2} \ \left ( \dfrac{\pi G}{c^3} \ f M\_c \right )^{2/3}}$$
For fixed $D$ and $f$
$$h \propto M\_c^{5/3}$$
The source I have used are notes entitled "*Relativistic Astrophysics - Lecture. Gravitational Waves*" authored by John Wheeler. I have the notes on paper; unfortunately, I have not found the PDF freely available on the Internet.

Best regards.

> **Comments:**
> **username_3**: OK @username_0, best regards  
> **username_3**: Done! https://astronomy.stackexchange.com/a/53471/31201  
> **username_0**: Yes I appreciate it very much, thanks!  

---

### Answer by username_4 on 2023-04-23. Score: 2

A gravitational wave period between 100 and 8000 seconds corresponds to a frequency between 0.8 mHz and 60 mHz. This is roughly the frequency band that will be targeted by the LISA space-based gravitational wave observatory.

In this regime we expect three main classes of sources:

1. **Massive black hole mergers** Mergers of massive black holes after the merger of two galaxies.
2. **Extreme mass-ratio inspirals** (EMRI) The merger of a stellar mass compact object (black hole or neutron star) at the center of a galaxy.
3. **Galactic compact binaries** Binaries of two compact objects well before merger. These are the sources discussed in the other answers. We expect there to be millions of these in our own galaxy.

However, none of these sources are expected to be loud enough to be (have been) detected by Ulysses.


---

