【人社】离职当月是不给缴纳社保是“惯例”？

Question

There are three formulae for the Roche limit distance. They are all simple, but the second one doesn't require knowing the densities, just the ratio of their masses and the size of the neutron star.

$d= R_m(2$${M_M}\over{M_ m}$$)^{1/3}$

where $R$ is the radius of the neutron star, $M_M$ is the mass of the black hole, and $M_m$ is the mass of the star.

If the mass ratio is a billion, and the neutron star radius is 5 miles, the formula says that a neutron star could orbit a black hole until its surface was 1,260 miles above the event horizon.

Right?

So if a neutron star orbits lower than this, the tidal force will tear it apart even though its surface gravity is 200 million Gs.

Is this correct?

Answer 1

Wrong. It means the "distance" between the centre of the black hole and the Roche breakup limit is 1260 miles.

Since the Schwarzschild radius of the black hole is of order 2 miles per solar mass, neutron stars are about 1.5 solar masses and you assumed $M_M/M_m \sim 10^9$, then the Schwarzschild radius of the black hole is billions of miles.

The neutron star will hardly notice crossing the event horizon; the tidal forces are negligible there. It is not until it (inevitably) gets much closer to the singularity that it breaks up.

Notes for the pedantic:

I put "distance" in quotes because the radial coordinate in Schwarzschild spacetime is not a Euclidean distance; it is a circumference divided by $2\pi$.

According to any external observer, the neutron star cannot pass through the event horizon. It will simply redshift and disappear (without breaking up).

There is no stable orbit within 3 Schwarzschild radii of a non-spinning black hole.

The quoted formulae uses Newtonian physics, but by a quirk of nature this turns out to be approximately correct for a free-falling body in General Relativity.