I suppose I shouldn't. Once I get out of bed on Friday morning, it'll be 30 hours before I can get into another bed at my destination. It's going to be a long trip, and I'm going to be reading, drawing, and trying to sleep (not all at once). The longest layover is 6 hours and it's at LAX, where I'll be waiting for the 747 that will take me to the Philippines.
Somewhere in the back of my head is a little, distant worry about the recent uptick in terror attacks. I'm hoping that nothing will happen so that my trip can actually go as scheduled; it would annoy me considerably if some islamic fascist fundamentalist dickweed screwed up my trip.
After 9/11, air travel took--what, three days? Two days?--to resume after all the planes were grounded. I would hate staying in Los Angeles for two days.
Anyway, my last day of work ends at 6 AM on Thursday morning. I don't know how much sleep I'll manage to get.
* * *
I was thinking about physics again not long ago, and I began to wonder how much mass the sun loses due to its energy expenditure.
I was going to look up all sorts of facts--the masses of the particles involved in the solar phoenix, the rate of reaction, etc, etc--and then it hit me: doofus, Einstein already did all the hard work. E=m*c*c. All you need to know to find how much mass the sun loses is its total energy output; the rest is just algebra. (The simple kind.)
It's basic physics. Well, it wasn't basic 100 years ago, but it is now: if the sun is a big ball of fusing hydrogen, the energy that comes from it must have come from the conversion of some of its mass to energy. It's therefore relatively simple to apply Einstein's most famous formula and find the answer to my question.
Hmm, here is a figure: 420 trillion kilowatt-hours of solar energy reaches the Earth, and that's one billionth of the total output.
So the sun generates 420,000,000,000,000,000,000,000 kilowatt-hours. Fine. Running that through my handy unit convertor yields...erk.
A kilowatt-hour is 36 million joules. So 3.6 x 10e7 times 4.2 x 10e23 yields 1.512 x 10e30 joules.
Now to divide that by the "speed of light squared" in m/sec. (3 x 10e8) e2 is 9 x 10e16. So now we divide, and get 1.68 x 10e14.
...yeesh. That's got to be wrong, because that's 1.68 x 10e14 kilograms: 168,000,000,000,000 kilograms of mass. One hundred sixty-eight trillion kilograms.
Of course the figure I cite above doesn't say over what period of time the Earth receives those 420 trillion kilowatt-hours. And the sun masses about 2 x 10e30 kilograms:
___________________168,000,000,000,000 kilograms per [time unit]
11,910,000,000,000,000 [time unit]s.
If it's seconds, it comes to 377 million years. If it's days, it's thirty-two trillion years. If it's hours--the most reasonable unit since the energy figure I used is given in kilowatt-hours--it's still fifty-six billion years.
Since we know the sun's total lifespan is about 10 billion years, I know there is an error somewhere in my figures. For one thing, my figures don't take into account the sun's gravitational energy, which is a non-trivial factor. Also, the sun will run out of hydrogen before it loses enough mass to stop fusing. But fifty-six billion years is at least on the correct order of magnitude, even if it is off by a factor of five, so for a "back of the envelope" calculation it's really not all that bad. I'm sure I could refine it down to the well-known lifespan if I really cared all that much, but the number I got is close enough to satisfy me: it is an approximate assload of mass by any terrestrial measure, but not all that much on the astronomical scale.
Still, that's kind of sobering, isn't it? The sun is shrinking!
* * *
What really makes me feel good about that is knowing that I'm smart enough to a) figure out the problem, and b) understand that while my answer is in the right ballpark, it's not gospel. And how many people can ask themselves, "How much mass does the sun lose?" and get a semi-useful answer?
Heck, how many people would even think to wonder about something like that?