Hulse-Taylor versus Pluto-Charon

HarryWertM
If we could detect them all on Earth, what how would the magnitudes of gravitational waves from the Hulse-Taylor binary compare to the magnitudes of waves from our Pluto-Charon planetary system? I know the Pluto-Charon system is far below the Hz range for LIGO and even LISA, but if you could measure in that range, how would the magnitudes compare?

Staff Emeritus
Gold Member
Interesting question!

Here is a derivation that shows how the strength of gravitational waves scales: http://www.lightandmatter.com/html_books/genrel/ch09/ch09.html#Section9.2 [Broken] (see subsection 9.2.5). The result is that the radiated power is $\propto (m/r)^5$.

Charon is about 10^-9 solar masses, and the r is smaller by a factor of 10^-2. The result is that the power would be down by about a factor of 10^-35.

The distance to the Pluto-Charon system is smaller than the distance to the Hulse-Taylor system by a factor of 10^9, so 1/r2 provides an improvement of 10^18. But this is still way too small to make up for the 10^-35 in radiated power.

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Frame Dragger
If we could detect them all on Earth, what how would the magnitudes of gravitational waves from the Hulse-Taylor binary compare to the magnitudes of waves from our Pluto-Charon planetary system? I know the Pluto-Charon system is far below the Hz range for LIGO and even LISA, but if you could measure in that range, how would the magnitudes compare?

FAAAAAAR below... Pluto isn't terribly massive, and Charon is downright puny. I'm not sure that without some data of real gravitational waves, for reference, guesses at such a small scale would probably be orders of magnitude off. That's my guess, I could very well be wrong.

I would compare their magnitude by comparing relative masses of the binaries, distance from the barycenter of orbit, rotation about indiviual axis, and rotation about each other. A pulsar is a neutron star (or white dwarf, either way degenerate matter), which is massive as HELL, with ridiculous angular momentum... and that MIGHT be good enough for LIGO, and probably for LISA.

I think you'd need to run a simulation, and I don't think anyone has or is likely to anytime soon. It's just... too small, and gravity drops off too rapidly with distance. Keep in mind, that is a simulation with no practical value, taking up comp time. Interesting idea however, but even if Pluto and Charon collided, I don't think it would be too impressive. Hell, Jupiter and Titan are more impressive, closer, and Titan could easily be considered a planet next to Pluto.

EDIT: *looks and sees bcrowell has posted*. Then again, there could be a straightforword answer that I completely missed! :tongue: Damn!

EDIT2: Why did I bring up angular momentum?! Ahhhh... sometimes I really wish we were trained to destroy material. Ah... oh well

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Homework Helper
As a simple order of magnitude estimate, the strength of gravitational waves from a binary system is

$$h \approx \frac{G^2M^2}{rR}$$

So for the Pluto-Charon system I use

M = 1.3 * 10^22 kg
R = 18000 km
r = 28.7 AU

And for the Hulse-Taylor system I use

M = 1.4 Solar masses
R = 1.1 Solar radius
r = 21,000 light years

Putting in the numbers gives that the gravitational waves from the Hulse-Taylor system would be about 10 million times higher in amplitude.

Staff Emeritus
Gold Member
Putting in the numbers gives that the gravitational waves from the Hulse-Taylor system would be about 10 million times higher in amplitude.

It's a little tricky comparing your result with my estimate in #2, since yours is expressed in terms of amplitude, while mine is expressed in terms of power intercepted per unit area.

I could be wrong, but I believe the way to compare them is as follows. I think the thing that plays the role of a "field" is the time derivative of the amplitude, dh/dt, so that if you know h and want to find the intensity, you want to take $(\omega h)^2$, not just $h^2$. Since $\omega \propto m^{1/2}r^{-3/2}$, the factor of $\omega^2$ makes your form $h\propto m^2/R$ equivalent to my form $P\propto m^5/R^5$.

The ratio of the frequencies is 20, so using your estimate of h, we get $h\omega$ differing by a factor of about 10^8 or 10^9. Squaring that gives something like 10^17, which is consistent with my result.

Frame Dragger
It's a little tricky comparing your result with my estimate in #2, since yours is expressed in terms of amplitude, while mine is expressed in terms of power intercepted per unit area.

I could be wrong, but I believe the way to compare them is as follows. I think the thing that plays the role of a "field" is the time derivative of the amplitude, dh/dt, so that if you know h and want to find the intensity, you want to take $(\omega h)^2$, not just $h^2$. Since $\omega \propto m^{1/2}r^{-3/2}$, the factor of $\omega^2$ makes your form $h\propto m^2/R$ equivalent to my form $P\propto m^5/R^5$.

The ratio of the frequencies is 20, so using your estimate of h, we get $h\omega$ differing by a factor of about 10^8 or 10^9. Squaring that gives something like 10^17, which is consistent with my result.

I really love this site... you know that? I REALLY love it. Where the hell else will you chat about two approaches to calculating the intensity of g-waves produced by a binary in our own little sytem!

Anyway, for quick and dirty calculations that's pretty good, coming within a single order of maginitude.

Homework Helper
bcrowell,

Certainly you'll get different answers if you're looking for different thigns. I interpreted "magnitude of gravitational waves" to be the amplitude of the metric perturbation h. However, it may indeed be more appropriate to look at the power/intensity of the wave, as you say. At any rate, the conclusion is the same: waves from Hulse-Taylor are much stronger than those from Pluto/Charon.

Staff Emeritus