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It is extremely difficult to fetch accurate value of gravitational constant. How about we have a black hole toy and laboratory for it? I am (currently) not a university student and need help for it.

Tentacles3587
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No. To measure G, you must have independent measurements of mass and gravitational field. The mass of the black hole can only be inferred from its gravity.

John Doty
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  • +1. Could you have a planet and the black hole orbit their mutual center of mass, and infer the black hole mass from the radius of its orbit? – mmesser314 Sep 05 '22 at 16:02
  • @mmesser314 How, then, would you measure the mass of the planet? – John Doty Sep 05 '22 at 16:57
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    @mmesser314 No, because there is no analytic solution for a two-body system in GR. So the result would not be of high accuracy. – safesphere Sep 05 '22 at 20:45
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    @safesphere No analytic solution, does not imply no highly accurate solution. Some aspects of orbital dynamics around a black hole have been calculated with thousands (!) of digits of accuracy. – TimRias Sep 06 '22 at 09:56
  • I agree with @TimRias (https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic) but how to find that data? – Tentacles3587 Sep 06 '22 at 09:58
  • @TimRias Are you saying that numerical gravity can produce a high accuracy result for a realistic two body problem? Even if this were the case, you would not know the density distribution of the real planet and therefore the actual stress-energy tensor. Plus a real black hole would be spinning with an accretion disk that would is described by the vacuum Kerr solution. So calculating a real binary system with numerical gravity to a high precision is a just pipe dream. – safesphere Sep 06 '22 at 15:18
  • @safesphere Perhaps you're right for close orbits, but gravity has long range. See https://ssd.jpl.nasa.gov/orbits.html for extremely accurate orbital simulations. – John Doty Sep 06 '22 at 15:29
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    @safesphere How much accuracy can be achieved (and through what method) depends the orbital separation of the binary, the mass-ratio, compactness of the object orbitting the black hole. For example, of the object is compact and small enough, its internal composition would not be relevant for high accuracy calculation. All this is moot though, since no matter how accurate the simulation data, you would still not be able to measure G. – TimRias Sep 06 '22 at 16:15
  • @safesphere You don't actually need to do a 2 body calculation. Photon trajectories around a Schwarzschild black hole can be computed using $\phi=\int\frac{du}{\sqrt{b^{-2}-u^2+u^3}}$, where $u=r_s/r$ and $b\cdot r_s$ is the impact parameter. That's an elliptic integral of the 1st kind, which can be rapidly calculated to high precision using the arithmetic-geometric mean. See Carlson (1994) for details. – PM 2Ring Sep 06 '22 at 16:51
  • @PM2Ring: to be hyper-pedantic (because I know it doesn't really matter, you're neglecting the back-reaction of the photon onto the spacetime, so you're still taking the same geometric approximation you use when you treat schwarzschild as a background for computing planetary orbits. – Zo the Relativist Sep 07 '22 at 19:29
  • @Jerry Mea culpa, although I did mention "that critical impact parameter calculation ignores the spacetime curvature caused by the photon itself" in https://astronomy.stackexchange.com/questions/25552/can-light-be-trapped-in-orbit/50275#comment111574_50275 OTOH, I expect that the error in the measurement of the potential energy of the chunk of matter that we drop into the BH is larger than the errors due to the photon energy. – PM 2Ring Sep 07 '22 at 20:15
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This is fundamentally impossible.

In the metric for the black hole, $G$ only ever appears in the combination $GM$. Consequently, there is no way of interacting with a black hole that will allow you to independently measure the mass $M$ and $G$.

The same is true for the dynamics of a black hole binary. G will always be paired with one of the masses (and vice versa).

Not coincidentally, the same is true in Newtonian gravitational physics, which is why we know $GM_{\odot}$ very accurately, but not G.

John Doty
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TimRias
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  • I disagree that it's fundamentally impossible, but it's surely not possible in practice. You "simply" need to measure $r_s$ of the BH (eg by determining the photon sphere radius via photon deflection), drop a chunk of matter of known mass into the BH, then measure the new $r_s$. Of course, other sources of gravitational potential must be negligible, and to convert the observer's measurements to Schwarzschild coordinates you have to account for the spacetime curvature in their vicinity (and their velocity). – PM 2Ring Sep 06 '22 at 17:06
  • @PM2Ring Clever. – John Doty Sep 07 '22 at 21:18
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I think yes. In general relativity $G$ can be derived from the knowledge of Einstein's gravitational constant $\kappa$ and $c$, the propagation velocity of gravitation waves. See my answer to the question Is the Planck force a truly "Planck unit"?.

JanG
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  • While yes, energy loss due to gravitational waves is definitely proportional in magnitude to $G$, we can barely measure it as is, so I doubt you'll be able to beat cavendish-style experiments that way. – Zo the Relativist Sep 06 '22 at 15:31
  • I have rather thought of Planck force (~$c^4/G$) measurement. In principle it should be possible to determine $\kappa$ in the strong gravitation limit. Imagine Einstein living at the center of star on the verge to become a black hole. He would develop theory of spacetime (GR) without need or use of Newton's constant. – JanG Sep 06 '22 at 18:44