I am curious due to the course of the idle research that I am doing (my hobby), and I am curious for various reasons as to the answer of this question. Please forgive me in advance for possible convoluted language, as I occasionally have trouble explaining my questions. How could one go about calculating a Roche limit for a black hole? I have asked this question in other places, and please leave off pedantic responses about black holes not existing. Basically my problem is thus: current Roche limit formulae are solved with respect to primary and secondary density, and, as we currently understand them, black holes have infinite density, due to them being zero-dimensional. I have seen some suggestions that mass can be replaced for density in the calculation, but this is often presented with no justification, or modifications in the formula. Currently, for the purposes of this, I am using Sag A* as the primary, and 3 different test particles (The moon, earth, and sun, because in my own little math world, I can destroy all 3) as the secondaries, and possibly other masses as well. I am aware of the limitations of the limit, that they only really hold if a body is held together only with gravity. So, if someone can give me a pointer or two, I'd be eternally grateful.


1 Answer 1


You'll sometimes see the expression for the Roche limit written as a ratio of densities, and sometimes as a ratio of masses. The two are obviously interchangable because the mass is equal to the density times the volume ($m = \rho\tfrac{4}{3}\pi r^3$). In the Wikipedia article you link the Roche limit is written as:

$$ d = R \left( 2 \frac{\rho_M}{\rho_m} \right) ^{1/3} $$

but $R$ is the radius of the primary and $\rho_M$ is the density of the primary, so you can rewrite the equation as:

$$\begin{align} d &= \left( 2 \frac{\rho_M R^3}{\rho_m} \right) ^{1/3} \\ &= \left( 2 \frac{\frac{3M}{4\pi}}{\rho_m} \right) ^{1/3} \end{align}$$

where $M$ is the mass of the primary. This allows you to calculate the Roche limit for a black hole even though the density is undefined.

Actually this illustrates a very important point about black holes. As long as you stay outside the event horizon the gravity due to a black hole is exactly the same as any other spherical object with the same mass (and a radius smaller than your distance from it). As long as the primary body is much larger than the secondary (so the perturbation of the primary by the secondary can be ignored) you would not expect the size or density of the primary to appear in the expression for the Roche limit - only the mass matters.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.