# Too fast for a black hole to consume?

1. Aug 16, 2012

### Boliver

I read an article yesterday about the fastest moving pulsar yet detected. The Chandra X-ray observatory spied a pulsar in SNR MSH 11-61A and IGR J11014-6103 moving at approximately 6 million miles an hour. This raised the question in my mind of what would happen if this pulsar were to collide with a stellar sized black hole moving at such rate with respect to the black hole. Does a black hole have a "processing" rate where it converts hadrons into whatever the form of matter or energy is inside a black hole? Is the conversion instantaneous? Is it possible for a fast moving pulsar to blow right through a black hole because it is too big and moving too fast to eat? Or how about two black holes moving toward each other with a combined speed approaching the speed of light? Would the black holes collide like billiard balls? I hope these questions are not too naive but I am merely an arm-chair astro-physicist wannabe.

2. Aug 16, 2012

### Staff: Mentor

Once something has crossed the event horizon there is no getting out, even at the speed of light. Note that a black hole does not "eat" or "process" anything. The event horizon of the black hole is not a physical surface, it is simply a boundary that marks the point where the escape velocity exceeds the speed of light.

So if a pulsar was to collide with a black hole any part of it that is smaller than the event horizon is lost for good, the parts that are outside will either fly off very fast as they slingshot round the BH or fall in or go into orbits (perhaps falling in later).

3. Aug 16, 2012

### hubble_bubble

The frame dragging would also have an effect depending upon the angle and direction of collision.

4. Aug 18, 2012

### stevebd1

According to 'Exploring Black Holes' by Wheeler & Taylor, there are three types of in-falling radial plunger for a static black hole-

Drip (dropped from rest at $r_o$)

Rain (dropped from rest at infinity)

Hail (hurled inward at speed $v_{far}$ from a great distance)

The shell velocities ($v_{shell}$) (i.e. the velocity measured at a specific radius) for each plunger are-

Drip

$$\left(1-\frac{2M}{r_o}\right)^{-1/2}\left(\frac{2M}{r}-\frac{2M}{r_o}\right)^{1/2}$$

Rain

$$\left(\frac{2M}{r}\right)^{1/2}$$

Hail

$$\left[\frac{2M}{r}+v_{far}^2\left(1-\frac{2M}{r}\right)\right]^{1/2}$$

where $M=Gm/c^2$ (multiply by c for SI units). If you use the above equations, you'll see that regardless of whether an object is dropped from rest close to the event horizon or whether it approaches from a great distance at close to c, all objects cross the event horizon, relative to the EH, at c. These equations only apply to radii greater than 2M as there are no stable shells (or stable r) inside the event horizon.

For the velocity of an in-falling object as observed from infinity, multiply the above equations by $(1-2M/r)$.

Last edited: Aug 18, 2012
5. Aug 21, 2012