Monday, February 11, 2013

Numbers slumber

Over at Down the Road, Jim has posted about ancient math skills. Unlike him, my stamina for math ran out early. From being on the math team that won city-wide when I was in grade 8, by grade 13 I was getting a D-for-effort 51% pity pass in relations and functions. Numbers that weren't tied to something concrete that I could manipulate in my mind have always given me trouble. The only math I ever really shined at in high school was trigonometry, because I could see it, experience it, show it, use it. It was the one trick I could really show off with my math yo yo. The rest of them? Forget it. I can manage quadratic equations provided they're simple enough, but don't bet the mortgage I'll get the right answer.

Every once in a while, though, I get this hankering to dust off the math skills I do have and see if I still have any chops. Recently I wound up in a kind of a slow-motion debate online with a guy who pointed out that the Andromeda galaxy is due to, eventually, collide with ours. He was essentially characterizing the event as something like gigantic car crash; something that would end life on Earth. And I just shook my head.

Galaxies are nothing like as solid as objects in the common world. Scaled up, they behave much more akin to liquids or even gasses. Stars seem huge, but that's only in relation to the scales we're used to here on Earth. In reality, they're specks unimaginably far apart in space so vast it's really beyond our ability to really grasp. This "collision" will take place over millions of years, and will be something more like two swarms of bees passing through each other than two solid objects colliding. The shapes of the galaxies will be massively disrupted, that's true. But collisions between stars, or disruptions to their planetary systems? Extremely unlikely for any given star due to the vastness of the distances involved.

Consider: it takes light between three and four seconds to cross the diameter of the Sun. It takes light roughly seven hours to reach the orbit of Pluto, which traditionally we've thought of as the "edge" of the solar system, whatever we might call it today. After that... four and a half years to reach the nearest next star, Alpha Centauri. Seven hours, then nothing to speak off for over four years, and that's if you head in exactly the right direction. Other "nearby" stars are decades or centuries away at the speed of light.

So I set out to try to enlighten this guy with an illustration. And this was my reasoning.

What affects the Earth in a noticeable way? Well... the Moon. You can see the tides. Okay, there's my standard.

What's an impressive distance, but one that sounds dangerously close on the scale we're talking here? A light year. Over four times closer than Alpha Centauri.

Okay. So how massive would a star have to be, passing a light year from the Sun, to have the same gravitational influence on the Earth as the Moon? That's the mathematical task I set for myself.

The starting premise I have from my old days watching Cosmos. I know that gravity, like light, propagates by the inverse square rule; that is, a body twice as far away must have four times the mass to have the same influence as a body half the distance away. I did some scribbling on my cream-coloured desk in pencil and decided the following:

a/b = √c/d, where

a = the distance from the Earth to this hypothetical star (9,460,730,472,581 km)
b = the distance from the Earth to the Moon (370,000 km)
c = the mass of the hypothetical star (unknown)
d = the mass of the Sun in lunar masses (81.3 [lunar masses in the Earth]*332,000 [terrestrial masses in the Sun]) = (26,991,869.9)

My calculations had to be rough, because the numbers I was inputting were, but I knew I should get something approximate. Essentially, if you square the result of the first calculation, it's equal to the mass of this star divided by the mass of the Moon, as expressed in solar masses. The number I came up with at the end of it all was that for a star passing one light year away to have the same gravitational influence on the Earth as the Moon does, it would have to be 23,207,735.6 times the mass of the Sun. That's very rough, of course, but if my math is right, then the answer has to be somewhere around that figure. I don't think there are any stars that massive. VV Cephei, I star I've marveled about since I was a kid, is thousands of times the radius of the Sun, but even so, probably has a mass only about 20-25 times that of the Sun. The long and the short of it is, even a very massive star would have to come improbably close to the Sun to even mess with the tides, let alone chuck the Earth out of its orbit or suck the atmosphere away, or any of the other disasters this guy seemed to have expected for the planet... When Galaxies Collide! And this was me trying to put what little math I can apply to some kind of practical use to demonstrate a point.

Even if I'm off, I'm still kind of proud I could marshal this much of a proof without getting stymied. :)

P.S. I forgot to mention that I scaled down the example in order to provide the guy with something he might be able to relate to. If you imagine stars to be the size of the average car... say, 10 feet across... then at that scale, Alpha Centauri, the nearest "car", would be 58,000 miles away. There's a whole lot of space in space, and even at that scale, the likelihood that a passing car, or 18-wheeler, or even a B-52 is going to cause you much trouble in terms of collision is, obviously, perishingly small.


Rudy Limeback said...

brilliant -- i'd say you still got it

i like relating things to everyday objects, and that car analogy at the end was great

jim said...

See, you're a scientist. Scientists use math only to further, or explain, the science. Mathematicians do math just cause.