In Hill’s mechanism, the gravitational tidal force of a single MBH disrupts an approaching binary. One star is captured on an eccentric orbit around the MBH and, by conservation of energy, the other star escapes with a final velocity equal to the geometric mean of the ∼$10^4$ km/s infall velocity and the ∼$10^2$ km/s binary orbital velocity.
Hypervelocity Stars and the Galactic Center, Warren R. Brown, p.3

I don't understand how the high energy/velocity can be transferred to the star.

(Hills' 1988 Nature paper is still paywalled.)


Consider two unit masses with velocity $V$. Give them a small push apart so they have velocities $V+v$ and $V-v$ with $v\ll V$ . Their kinetic energies are then $$\frac{(V+v)^2}{2} \approx \frac{(V)^2}{2} +vV$$ and $$\frac{(V-v)^2}{2} \approx \frac{(V)^2}{2} -vV$$ We have transferred energy $\approx vV$ from one mass to the other with the expenditure of a much smaller energy $v^2$.

Now suppose the masses are initially moving with some tiny velocity and then travel into a potential well where they have velocity $V$. If we push while they are in the well then one will have enough kinetic energy to escape the well plus $vV$, while the other will be trapped in the well.

This is the same principle as the Oberth effect, where the ejected reaction mass plays the role of one of the masses.

  • $\begingroup$ just fyi it's always okay to accept your own answer. $\endgroup$
    – uhoh
    Feb 23 '20 at 3:21

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.