Fear, Flying, and Physics

[Still working on the next chapter, but hit a little snag. Shooting to have it up around May 16th. Link to my stories here.]

Flicker can’t fly.

That has been basic to her character from the beginning, for reasons of physics and psychology, but I’ve never gone into the details before, because they haven’t been relevant to the the story. Well, now they are, because she is learning.

Since details and inefficiencies matter, and Flicker has a preferred direction that is easier to accelerate in (straight down), I needed to take a look at just how hard she’s had to push to do the things she’s already done. I’d done a quick and dirty approximation once before, but I had somehow missed realizing there was already a formula for what I wanted. Warning: Physics ahead. And math. 8-)

Centripetal force is how hard you have to push to keep moving in a circle instead of flying off at a tangent, and Flicker has to worry about it all the time. In Newtonian physics it’s mv^2/r (not even a pi in there–it already got cancelled out by using radius instead of circumference). Flicker isn’t force limited, but she is acceleration limited. Centripetal acceleration is even simpler, just drop the mass, and you get v^2/r.

Note the exponent. It is not linear in velocity, which means I already screwed up Doc’s forced orbit a few chapters back, and the casual twelve million g figure Flicker told Donner about accelerating cameras down during practice for her video in Fall. Fortunately I hadn’t specified the velocity there, except that it was slower than the actual video, so I can adjust that to fit without contradiction.

So, how about the numbers? I want to get them right this time 8-).

For Earth surface orbital velocity of about 7,900 meters/second, and radius of 6,371 kilometers, we get

(7.9 * 10^3)^2 / (6.371 * 10^6) = 9.796 m/s^2

or just about exactly 1 g, as we would expect, since the 1 g is what makes it an orbit.

But Flicker goes quite a bit faster than that–enough faster that the puny 1 g of gravity Earth provides becomes negligible for keeping her on the planet. How fast would she need to be going to require 12 million g’s to stay on the surface? Well, we solve for v in

v^2 / (6.371 * 10^6) = 1.2 * 10^8 m/s^2

or v = 2.76 * 10^7 m/s = 27,600 km/s or about nine percent of the speed of light, which is the right magnitude. Whew.

Flicker’s canonical forward acceleration limit is ten billion g’s, or about 10^11 m/s^2 – that’s the point where side effects start to get serious enough that she wants to avoid them. She can decelerate a bit faster, and down (which is the direction of centripetal acceleration) is easier, but it’s still a good ballpark figure.

At speeds close to c, we get

(3 * 10^8)^2 / (6.371 * 10^6) = 1.41 * 10^10 or 1.4 billion g’s for Earth, and

(3 * 10^8)^2 / (1.737 * 10^6) = 5.18 * 10^10 or 5.2 billion g’s for the Moon. Still safely under 10 billion.

So much for Newton. What about Einstein? I finally tracked down the right formalism for analyzing Flicker using Special Relativity.

Flicker is accelerating herself, but still needs to stay on a non-moving surface, so what matters is her proper velocity and proper acceleration. Proper velocity is rest frame distance traveled in moving frame time, or the distance from A to B on the map divided by the time it took to get there, measured by a watch carried by the traveler. It has no upper limit, and is the velocity calculated by an inertial navigation system.

For Special Relativity, the important number is the Lorentz factor, or gamma, which is 1/sqrt(1 - v^2/c^2). Proper velocity w is just regular velocity v times gamma, and proper acceleration is regular acceleration times gamma squared.

At 0.8 c, the fastest Flicker goes on Earth, her gamma is 1.667, so the needed proper centripetal acceleration is

(1.667)^2 * (2.4 * 10^8)^2 / (6.371 * 10^6) = 2.51 * 10^10 or 2.5 billion g’s. Still fine.

At 0.96 c on the Moon during her video, her gamma was 3.571, so the needed acceleration was

(3.571)^2 * (2.88 * 10^8)^2 / (1.737 * 10^6) = 6.09 * 10^11 or 61 billion g’s. Starting to maybe be a problem, but still the same order of magnitude.

But that’s not Flicker’s top speed.

The very first thing I ever wrote about Flicker, almost four years ago, was just called ‘Speed’, and eventually became part of Chapter 12 of The Fall of Doc Future. On her first, disastrous trip to the Moon when she was fifteen, when Doc told her she could go as fast as she wanted in fifty times around the Moon, she topped out at 0.99999994 c and got into serious trouble.

At 0.99999994 c, gamma is 2,887. That gives a proper velocity of 866 billion m/s, so it would only take her 12.6 microseconds, her time, to go all the way around the Moon once. It would also take her 8.66 seconds to get up to that speed at 10 billion g’s, and she didn’t have time–it wasn’t even that long in the rest frame.

And there is another small difficulty. At a gamma of 2,887, the centripetal acceleration needed to stay on the Moon is

(2887)^2 * (3 * 10^8)^2 / (1.737 * 10^6) = 4.319 * 10^17 m/s^2.

432 quadrillion g’s. Houston, we have a problem. At least now I’m pretty sure how she got into trouble. Now that acceleration is not doing any work, not adding or subtracting any energy, theoretically, because it’s always perpendicular to her direction of motion, which is the 'that’s why down is different’ explanation I was thinking of using anyway.

But the small imperfections and inefficiencies, things like the Moon not being perfectly spherical–the source of her trouble–matter, because I’m using them as part of the story. At 432 quadrillion g’s, they aren’t small any more.

And a little voice is telling me 'At 432 quadrillion g’s and a gamma of 2,887, I don’t think it’s safe to ignore General Relativity any more.’

Specifically, Flicker was giving off gravitational waves. But my best attempt to figure out even a rough magnitude ended up in this bit from a book chapter on detecting them:

“Deriving equation (24) involves a lengthy and difficult calculation starting from Einstein’s field equations. The same is true of the derivation of the metric (1) for a gravity wave. These are two of only three equations in this chapter we simply quote from a more advanced treatment of general relativity.”

I don’t want to try to even approximate Einstein’s field equations. I really don’t. I have a pdf of Misner, Thorne and Wheeler’s Gravitation, and it is fear inducing. I have a PhD in math, but I only got a B in tensor analysis.

So… Any general relativists out there who would like to calculate the energy of the gravitational radiation given off by a 50 kilogram rest mass object traveling in a circle of radius of 1,737 kilometers at 0.99999994 c? Just, you know, for fun?

I think I can safely ignore the presence of the Moon, except possibly as a victim. Interestingly, the one thing I can calculate is the frequency–the speed of light divided by the Moon’s circumference, which works out to 27.5 hertz, almost exactly A0, the A below the C below low C.

So Flicker was playing bass. I just have no clue how loud.

Edit: Per the helpful comments from someone who knows more GR than me, it looks like Flicker was radiating gravitational waves somewhere in the 1-10 kilowatt range. So, detectable with a sensitive detector, but not damaging–especially not compared to the EM radiation she was giving off.

Oh, and the frequency is twice what I thought: 45 hertz, low A.

