Hulse-Taylor versus Pluto-Charon

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In summary: 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.
  • #1
HarryWertM
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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?
 
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  • #2
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 (see subsection 9.2.5). The result is that the radiated power is [itex]\propto (m/r)^5[/itex].

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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  • #3
HarryWertM said:
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! :-p Damn!

EDIT2: Why did I bring up angular momentum?! Ahhhh... sometimes I really wish we were trained to destroy material. Ah... oh well :cry:
 
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  • #4
As a simple order of magnitude estimate, the strength of gravitational waves from a binary system is

[tex]h \approx \frac{G^2M^2}{rR}[/tex]

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.
 
  • #5
nicksauce said:
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 [itex](\omega h)^2[/itex], not just [itex]h^2[/itex]. Since [itex]\omega \propto m^{1/2}r^{-3/2}[/itex], the factor of [itex]\omega^2[/itex] makes your form [itex]h\propto m^2/R[/itex] equivalent to my form [itex]P\propto m^5/R^5[/itex].

The ratio of the frequencies is 20, so using your estimate of h, we get [itex]h\omega[/itex] differing by a factor of about 10^8 or 10^9. Squaring that gives something like 10^17, which is consistent with my result.
 
  • #6
bcrowell said:
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 [itex](\omega h)^2[/itex], not just [itex]h^2[/itex]. Since [itex]\omega \propto m^{1/2}r^{-3/2}[/itex], the factor of [itex]\omega^2[/itex] makes your form [itex]h\propto m^2/R[/itex] equivalent to my form [itex]P\propto m^5/R^5[/itex].

The ratio of the frequencies is 20, so using your estimate of h, we get [itex]h\omega[/itex] 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.
 
  • #7
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.
 
  • #8
Frame Dragger said:
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!

Yeah, sometimes we forget how different life was before the internet. I grew up in a small town (high school with 600 students) in the 1970's. If you had an interest that wasn't shared with one of those other 599 kids, you were just not going to be able to talk to anyone about that interest.
 
  • #9
bcrowell said:
Yeah, sometimes we forget how different life was before the internet. I grew up in a small town (high school with 600 students) in the 1970's. If you had an interest that wasn't shared with one of those other 599 kids, you were just not going to be able to talk to anyone about that interest.

Oh I know exactly what you mean! The amount of information available to anyone who knows where to look for it is stunning, and delightful.

I only wish that had been the going thing when I was a kid myself. Ah well, on the bright side I read whole libraries... now if I had grown before books were common... hell man, we'd all have been burned as witches and warlocks probably! :smile:
 

FAQ: Hulse-Taylor versus Pluto-Charon

What is the difference between Hulse-Taylor and Pluto-Charon?

Hulse-Taylor and Pluto-Charon are two different astronomical systems. Hulse-Taylor refers to a binary pulsar system consisting of two neutron stars orbiting each other, while Pluto-Charon refers to the dwarf planet Pluto and its largest moon, Charon.

How were the Hulse-Taylor and Pluto-Charon systems discovered?

The Hulse-Taylor system was discovered in 1974 by astronomers Russell Hulse and Joseph Taylor, who noticed irregularities in the pulsating signals of a pulsar. The Pluto-Charon system was discovered in 1978 by astronomer James Christy, who noticed a bulge on one side of Pluto's disk, indicating the presence of a moon.

Which system is closer to Earth: Hulse-Taylor or Pluto-Charon?

The Hulse-Taylor system is much closer to Earth, with a distance of about 2,000 light years, while Pluto-Charon is about 4.67 billion miles away from Earth.

How do Hulse-Taylor and Pluto-Charon contribute to our understanding of the universe?

Hulse-Taylor helped confirm Einstein's theory of general relativity, as the system's orbit showed the predicted loss of energy through gravitational waves. Pluto-Charon has provided valuable information about the formation and evolution of our solar system, as well as the composition and geology of dwarf planets and their moons.

Are there any current or future missions to study Hulse-Taylor or Pluto-Charon?

There are currently no missions specifically targeting Hulse-Taylor, but the system is regularly monitored by astronomers to continue studying gravitational waves. NASA's New Horizons mission is currently the only spacecraft to have visited Pluto and Charon, and it is expected to continue sending data until at least the late 2030s.

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