# Hypervelocity planet

## Homework Statement

(a) Consider a planet orbiting a star of mass M?, on a circular orbit with circular
velocity vp. What is the planet's orbital radius, in terms of vp and M?

(b) Now suppose that the star (carrying the planet along with it) enters a nearly radial
orbit around a black hole of mass M, with total energy E = 0. What is the star's
velocity, v?, relative to the black hole, as a function of M and the star's distance R
from the black hole?

(c) At what black hole distance R is the planet tidally stripped from the star by the
black hole? That is, solve for the radius R where the star's Hill radius equals the
at this radius? Express this velocity in terms of M?, M, and vp. (2 points)

(d) Assume that the planet was moving at velocity vp relative to the star, in the same
direction as v?, at the time that it was tidally stripped. Assuming that the star and
planet are at roughly the same distance from the black hole, what is the planet's total
orbital energy? Show that the planet is now unbound from the black hole, and can
therefore escape to large distances.

(e) Using the total energy you calculated in part (d), what is the asymptotic velocity
of the planet at large distances from the black hole, v1? Again, express your answer
in terms of M?, M, and vp. Assuming that the planet's original orbital velocity was
vp 300 km/s, and that the black hole was roughly a million times more massive than
the star, M = 106M?, what is the maximum ejection velocity v1 that the planet can
achieve through this mechanism?

## Homework Equations

F=ma, binary self attraction: $$a_b=\frac {Gm_b}{r^2_b}$$
Tidal force pulling on binary: $$a_t=\frac {2GM}{R^3}r_b$$

## The Attempt at a Solution

The circular orbit is simple enough for part A:
$$\frac{v^2m_p}{r_p} = \frac {GM_\star m_p}{r^2_p}$$
$$r_p=\frac {GM_\star}{v^2_p}$$

And, if total energy is zero then kinetic and potential must be equal for part B:

$$\frac{1}{2}M_\star v^2_\star = \frac {GM_\bullet M_\star}{R}$$
$$v_\star=\sqrt{\frac {2GM_\bullet}{R}}$$

Now, more interesting is the radius at which the black hole pulls the planet and star apart for part C:

$$\frac {Gm_\star}{r^2_p}\leq\frac {2GM_\bullet}{R^3}r_p$$

$$R \geq r_p(\frac{2M_\bullet}{M_\star})^{\frac{1}{3}}$$

Recall that $$r_p=\frac{GM_\star}{v^2_p}$$

Part D has me a little flustered. If I understand it conceptually it is saying that now that the binary system has been disrupted by the black hole the planet should have enough energy to escape the black hole system. So I believe that:
$$\frac{v^2_p m_p}{r_{p\bullet}} > \frac{GM_\bullet m_p}{R^2_{p\bullet}}$$

where the velocity of the planet is now the velocity of the star plus the orbital velocity of the planet around the star.

I plugged everything in but it is one ugly looking equation. Once R and the two velocities are plugged in it is pretty unwieldy. I just want someone elses opinion on my methodology. Or perhaps I made a mistake in my algebra.

Thanks for any help.

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mfb
Mentor
I guess you can approximate ##r_{p\bullet} \approx R##.
##v_{p\bullet} = v_p+v_\star##. I did not check it in detail, but that does not look so ugly.

I'm not sure that I can simply say they are equivalent. I believe R can be very large depending on the mass of the blackhole.

mfb
Mentor
R is very large, that is the reason for the approximation (relative to the black hole, the planet is close to the star). Even better, you know that the planet in its orbit around the star is moving towards the black hole, so those 3 objects form a right-angled triangle (with the star at the right angle). The approximation is valid as long as the black hole is much more massive than the star.

Shoot, I think maybe the confusion comes from an accidental notation problem.

it is $$r_p$$ not $$r_{p \bullet}$$

the latter makes it look like I'm referring to the distance of the planet to the black hole. Rather it should just be the orbital radius of the planet to it's host star. So it could have a small orbital radius around the star say like our moon to Earth but the black hole R could be the distance from us to the Sun.

mfb
Mentor
Ah, that is the orbital radius around the star. Why do you use this in part d?

• 1 person
It's sad when someone else understands my equations better than I do. You were right. to begin with ##R≈r_{p \bullet}##

I wrote that correctly to begin with but then forgot exactly what I meant. I left it as little are but, yes they're equivalent. I expanded out everything using little r but in the end it canceled out anyway.

I turned it in today but my max velocity seemed low to me. I ended up with something like 1.5x105m/s. I expected velocities much higher than this given the question title. Though, I admit it is still quite fast.

Thanks for the assistance though. At the very least I've learned to take more care with my variables and to always look for equivalencies.