# The Physics Philes, lesson 44: Challenge Accepted Part 2

*In which a challenge is issued, a challenge is accepted, and a challenge is completed…for real this time!*

If you will remember, last week I was issued a challenge involving black holes by this cheeky physicist I know. Even though I’m far too dumb to really get my head around it, I have vowed to answer his question – THE WHOLE QUESTION – eventually, at some point in my life. Who knows how long it will take.

For those of you who have forgotten, the question was presented to me as follows:

I answered question b last week, because it was easy and I had to study for a test. But this week I’m test-free, so I can devote heaps of time to figuring out question a. (Still, though. No guarantees.)

Before I attempt to find the minimum safe distance a person can be from this particular black hole, it might be worth figuring out what would even happen if someone tried to actually visit a black hole for real. Besides die, that is. Because you will totally die.

If you’re far away from a black hole, the effect of the black hole’s gravity is the same as it would be if it was another object with the same mass. (Get this, if the sun collapsed in on itself and formed a black hole, the planets would be unaffected. What.)

That’s not true if you’re close to a black hole, even a small one like the one in the question. Lots of really weird things would probably happen to you if you got too close.

For example, there is a thing called gravitational red shift. You probably have a pretty good idea what red shift is. It’s the phenomenon where, if something that is emitting light is moving away from you, the light wavelength will stretch and look more red to us. (If the thing is moving toward you, the wavelength might compress and thus look more blue.) It’s the Doppler Effect. The classic example is an ambulance speeding by. If you are standing still relative to the vehicle, you’ll notice that the frequency of the siren goes up as it travels toward you, and the frequency goes down as it travels away. A higher frequency corresponds to a higher pitch. That’s why, in the situation described, the siren sounds the way it does.

But what is *gravitational* redshift? That’s the phenomenon where electromagnetic radiation originating from a source that is in gravitational field is redshifted when observed in a region of a weaker gravitational field. So if you were to jump into a black hole and report back to your friends what you see via radio, your friends would have to continually adjust the receiver to lower frequencies. Assuming, of course, that they are trying to receive you signals from a place with a weaker gravitational field.

The friends would also never see our ill-fated adventurer actually cross the event horizon. This is because of time dilation, the difference of elapsed time between two events as measured by observers either moving relative to each other or differently situated from gravitational masses. The friends would see time move more slowly for our adventurer, while the adventurer would perceive his or her own time as moving along as usual.

Which is kind of too bad, because that means they are well on their way to being destroyed.

If our adventurer was falling feet first into the black hole, the gravitational pull would be greater on the feet than on the head. This difference is enough to stretch him or her along the direction toward the black hole and compress perpendicular to it. You know what this is called. Say it with me: SPAGHETTIFICATION. Spaghettification is caused by tidal forces, and it will rip you to shreds, right down to the atoms. (Although there is evidence to suggest that you’ll just burn to a crisp if you jump into a black hole which causes all kinds of problems in quantum mechanics. Please please PLEASE don’t make me explain it. My brain will fall out.)

From what I can piece together from the Google and clues my physicist friend has given me, it’s the tidal force that we need to worry about when trying to solve question a. (Remember the question? That’s kind of why I’m writing this.)

Before we get into the nuts and bolts of actually answering the question, I need to say that I am in no way sure I actually know how to do this. Worse than that, I’m not sure I can really explain why it has to be this way. Which sucks, because I don’t especially like to write these posts if I don’t have a good understanding of what’s going on. BUT. I have been challenged and my good name is now at stake. (Plus this particular physicist might be one of my professors someday and I don’t want him to think I’m stupid.) So…here we go.

First we need to find an equation. I spent a good day last week trying to figure this out so you don’t have to. I had actually dismissed this formula at first, because, well, I didn’t know anything about black holes. But I’m assured that this is indeed the formula we need to use to save our adventurer from certain doom:

OK, what does all of this mean? The left side of the equation obviously represents the tidal force discussed earlier. The two represents 2. I don’t know why. Just go with it for now. As we learned last week, big G is the gravitational constant. Big M is the mass of the black hole. The next two are variables I’m not sure about. I believe they stand for the mass and height of the thing being tugged upon; in this case, a person. For the purposes of our problem, I assumed that this person is one meter tall and weighs 100 kg. Why? Because those are easy numbers and I don’t need to muck this up any more than necessary. The denominator is the key variable for us in this equation. The r stands for how close our adventurer can get to the black hole without getting spaghettified. What it does *not* stand for is the Schwarzchild radius. Don’t use the Schwarzchild radius. You could waste half a day doing math and only coming up with nonsensical answers. Not that I know that from experience or anything…

Let’s first set out what we know:

Our target variable, then, is r. Let’s plug in those numbers!

Hey r^3! Get over to the left hand side where you belong!

Better. But we’re not done yet.

Don’t panic! This isn’t the answer! Remember, this is r^3. Now we need to take the cube root to find r.

Whoa! Last week we found that the Schwarzchild radius of a black hole with a mass of 10^20 kg is 1.48 x 10^-7 m. But our adventurer will be pulled to pieces if he or she gets closer than 237 m. That’s crazy-pants! If it’s true, that is. I retain the right to a do-over if it’s shown that I don’t know what I’m doing.

I had a really good time puzzling over these questions. But, unless I am challenged again, we’ll be back to regularly scheduled programing next week. Probably something about rotational motion.

Before I leave you this week, I need to give a BLAWG SHOUTOUT to a couple of people who have proven incredibly helpful: Carl Tracy and Chris Granade. Quick! Someone name the two most helpful Martian craters after these guys!

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There’s another way you can figure out the tidal force, which is more intuitive, but more difficult algebraically. The tidal force is essentially the difference between the force felt at your head and the force felt at your toes. So you can express this as:

Ftidal = abs(Fg(Rhead)-Fg(Rtoes)) = abs(Fg(R+d)-Fg(R-d))

Here, we’re well in the Newtonian limit, so Fg is the standard Newtonian expression for gravitational force. Assume for each of these terms that the mass is half the total mass of the body, and we get an equation of the form:

Ftidal = abs(GMm/(2(R+d)^2) – Gmm/(2(R-d)^2))

If you try to solve this algebraically, it turns into a fourth-order equation in d, which is no fun at all to solve. You basically have to plug it into Mathematica or Wolfram Alpha.

But if you want to do it by hand, we can make some approximations. There should be a college course on approximations in physics, but sadly there isn’t. So let’s go to the basics here. Let’s say you have an expression that looks like (A+B)^n (hey, we do!), and we expect B is much less than A (we do!). A little calculus (using a Taylor series) can show that a reasonable approximation for this is (A+B)^n = A^n*(1+n*(B/A)) It might look more complicated, but it actually simplifies things up a lot in many cases.

So, in this case, A=R, B=d/2 or -d/2, and n=-2 (since it’s in the denominator). This means that we have GMm/(2(R+d)^2) is approximately GMm/2 * R^-2 * (1-2d/R), and the other term is GMm * R^-2 * (1+2d/R). Since we’re subtracting these two, we get some nice cancellation, and we come out with Ftidal = 2GMmd/(R^3), which is the formula for tidal force that you found earlier.