Skip to main content
Misplaced don't. Misspelled word.
Source Link
John Doty
  • 21.9k
  • 1
  • 27
  • 60

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

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

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.

Source Link
TimRias
  • 12.9k
  • 1
  • 26
  • 37

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