# The Physics Philes, lesson 43: Challenge Accepted Part 1

*In which a challenge is issued, a challenge is accepted, and a challenge is…completed?*

Over the past several months I have collected physicist Facebook friends like people collected Beanie Babies in the ’90s. (Only my collection will prove much more valuable in the long run.) Recently, one such friend issued a challenge:

It reads:

Here’s a physics problem for y’all…

You’re studying black holes and are planning an expedition to view one up close and in person. You want to get as close as possible to get good measurements, but you know that you’ll have to worry about tidal forces causing “spaghettification.” So, you need to calculate a minimum safe distance to avoid being ripped apart. Assume you will approach along a radial line, with your feet toward the black hole and your head away. You know that the tidal force will act on each end of your body with the same magnitude, but in opposite directions, causing you to be pulled apart. The limiting factor in being ripped apart is the tensile strength of bone; you figure your body can withstand about 10^5 N of force before being pulled apart. The mass of the black hole you are studying is about 10^20 kg.

a. Estimate the minimum safe distance.

b. Compare your minimum safe distance to the “size” of the black hole, given by the Schwarzchild radius.

Mindy Townsend you might like this.

Ha! Nice try, buddy, but you’ve got to get up pretty early in the morning to trick me. Schwarzchild radius? Like that’s even a thing. (Turns out, it is a thing. More on that later.)

I’ll be the first to admit that his problem is a little above my pay grade. But, hey, the problem was dropped in my lap and now I have to try to figure it out. I’m going to have to take question b first and next week I’ll tackle question a. I need to do this for three reasons: 1. unexpectedly, it’s easy to calculate the Schwarzchild radius; 2. I sort of don’t know where to start with estimating a minimum safe distance; and 3. I have a test on Tuesday that I need to study for and I can only put off actual work for so long.

But before we dive into this problem, I think we need a little background. (OK, *I* needed a little background, but I know you appreciate it.) So let’s talk for a minute about stars and black holes.

Have you ever watched a rocket take off? It looks like it’s going so slowly, doesn’t it? Actually, it’s going pretty fast. It has to in order to hit its escape speed, or the speed at which an object must travel to break free from a gravitational field.

This concept not only comes in handy when we’re trying to put something into orbit. It applies to stars, as well. How do you think light gets from the sun to our eyes? The light is traveling at at least the sun’s escape velocity.

Escape speed can be calculated by using a handy little equation:

Whoa whoa whoa. Let’s stop right here. What is with all the capital letters? M stands for the spherical mass of the pulling object (like a star) and R stands for the radius of that object. But what about G? G – not to be confused with g, which is acceleration due to gravity – is Newton’s gravitational constant, otherwise known at 6.67 x 10^-11 N • m^2/kg^2. We can use the relationship between the gravitational constant, the masses of objects, and the distance between the objects to find the force of gravity anywhere in the universe.

But I digress. We can relate this escape speed equation to average density by replacing M with ρV, or

(The stuff in the parentheses is equal to V.) OK! Let’s replace M with all that stuff and see what it gets us.

Note the R in the far right side of the equation is not under the square root sign. Also notice that the escape speed has nothing to do with the mass of the object trying to get away. It’s all about the body trying to keep you from leaving.

Now, what does any of this have to do with black holes. Everything! Back in the late-1700s, an amateur astronomer called Rev. John Michell determined that if a body with the same average density of the the sun, but had 500 times the radius, the escape speed would be faster than the speed of light. Hmm…this is a problem because nothing can accelerate past c, the speed of light. Mitchell became the first person to suggest the existence of a black hole.

Black holes are objects that exert a gravitational force on other bodies but cannot emit any light of its own. There is a radius that a body has to meet to act as a black hole. That radius is called the Schwarzschild radius, named after Karl Schwarzschild (because why would you name it that otherwise?). He used the general relativity (which I will absolutely not explain here) to derive a handy little equation for finding that critical radius:

If a spherical, non-rotating body with a mass M has a radius less than that given by the above equation, then nothing can escape its gravitational pull. If a body happens to be within the Schwarzschild radius, it will not be able to escape. We’d say that the body has traversed the event horizon, or the surface of the sphere with a radius of Rs. Since no light can be emitted, we have no idea what’s going on in there. Like the windows of Robert Downey, Jr.’s limo; we can’t see what shenanigans are going on inside.

That’s not to say we don’t know anything. There are some things we can determine: mass by the gravitational effects on other bodies; electric charge by the electric forces it exerts on other charged bodies; and angular momentum because black holes tend to drag space.

So! With all that information shoved in our brain-cases, we can determine what the Schwarzschild radius of a black hole with the mass M = 10^20 kg. It’s pretty easy, actually. We know the gravitational constant and the speed of light. Now we just need to plug in the numbers:

Whoa! Is that right? I’m literally asking you if I did that correctly, because that is tiny!

Next week, if I can get my brain together, I’ll tackle the first problem of estimating how close a person can get to the black hole without turning into human pasta. Also, maybe you’ll get some more black hole facts. Until then, if you see anything I need to clarify or correct, please say so in the comments.

## 3 Comments

Double check the math as just taking the exponents, ( 10^20 * 10^-11 ) / 10^8 should leave you with something closer to 10^1 (+/- ^1). 🙂

But the denominator is (10^8)^2, which gives you 1 * 10^-7

Ah, yep. I didn’t square the denominator. 🙂