 Solution

Equating stellar surface gravity to the differential black-hole acceleration givesso the tidal disruption radius isA nonrotating hole swallows the star without a visible disruption when this lies inside its capture scale, here approximated by the Schwarzschild radius $r_h=2G{M}_{\mathrm{BH}}/c^2$. Equating the two radii yields the Hills massFor a solar-type star this is of order $10^8{M}_{\odot }$.

The threshold does depend on spin. A Kerr black hole has spin- and inclination-dependent horizon, marginally bound, and capture radii. Prograde orbits around a rapidly rotating hole can approach more closely, allowing disruption by masses above the Schwarzschild Hills mass, whereas retrograde capture occurs farther out.

An intermediate-mass black hole lies well below this threshold for ordinary stars, so stars entering its loss cone are disrupted outside the horizon. The returning debris can grow the hole and produces a tidal disruption event that may reveal an otherwise quiescent cluster black hole through a flare. Dense clusters can supply repeated disruptions, although the rate depends on two-body relaxation, stellar collisions, binary interactions, and whether gravitational recoil or cluster dynamics ejects the hole.

Solved by gpt-5.6-sol high.