| |
|
 |
The Physics of Santa's Journey
Posted by Douglas, Dec 15, 2009. 186 views. ID = 3027
This post was written in 25 minutes.
|
 | I wrote this at 5:00 AM, before finishing my first cup of coffee, so if I made mistakes in my math or physics, please be forgiving. :) |  | This post has been awarded 37 stars by 9 readers. |
Dear Santa,
The world population is now over six billion people. That's a lot of people. If we figure that the average family size is four people, and that of those families, only 75% of them have people who are good enough to get presents from you, that means you have to visit about 1.1 billion homes on Christmas Eve.
Normally we think of a "night" as being about 8 hours, but of course, if you time things properly, you have much more time to do your rounds; when it's 6:00 AM in India, it's early evening in Maine. So say you've got 24 hours to do your visits to those 1.1 billion homes.
That means you need to visit about thirteen thousand homes every second, or forty-seven million visits every hour. Let's be generous and assume that (on average) the journey from one home to the next is one tenth of a mile.
From this we conclude that you need an average speed of 4.7 million miles per hour. Now, that's not the part that really interests me; others have done those calculations (though at the time they were done the world's population was significantly less, and you could comfortably travel at about a third the speed now required).
What interests me is this: in order to maximize your efficiency, and make this whole thing possible, I think you must constantly accelerate through half of each journey, then constantly decelerate through the second half of each journey.
Using my good ol' Physics equations,
d = vit + 1/2at2
.05 miles = 1/2a(1x10-18 sec2)
The resulting acceleration has too many zeroes for me to count, but it looks like it might be in the quadrillions of miles per second per second.
And if we assume you and your sleigh (with all those presents!) have a combined mass of 1,000 slugs (a very generous estimate, considering how many gifts you are lugging around) then the amount of force being applied to you, your sleigh, your reindeer and the gifts is in the...hmmm...what do we call that? Quintillions?
My point is this: even if you and your reindeer can magically withstand those kinds of forces, I'm pretty sure my new Van de Graff generator will not.
Please use UPS or the USPS this year.
With kind regards,
A Concerned Physics Teacher
Copyright 2009 Douglas. All rights reserved. FifteenMinutesOfFiction.com has been granted non-exclusive rights to display this work. For permission to reprint this item, please contact the author.
| This post has been awarded 37 stars by 9 readers. |  | This post is part of a writing prompt: Letter To Santa |
Comments
Jeff Howe
Dec 15, 2009 | LOL
~Posted by Jeff Howe, Dec 15, 2009 | 
Mathax
Dec 15, 2009 | Ok, this is really great...
~Posted by Mathax, Dec 15, 2009 | 
MrPhysics
Dec 15, 2009 | Seems to me that those speeds you would get considerable time dilation.
~Posted by MrPhysics, Dec 15, 2009 | 
Laura
Dec 15, 2009 | Somehow when I saw this title in my reading list, I KNEW it was by you... ;-)
~Posted by Laura, Dec 15, 2009 | 
'Chelle
Dec 16, 2009 | Thanks for this real life use of math! I have often wondered how fast Santa would have to travel...it is way worse for him than i imagined. Is there a point at which one would cease to be moving and would instead be considered omnipresent?
~Posted by 'Chelle, Dec 16, 2009 |
Douglas
Dec 16, 2009 | Ha ha...that question is way beyond my ability to answer!
~Posted by Douglas, Dec 16, 2009 | 
gabemay
Mar 1 | Hilarious! But are you serious? You're a Physics teacher? lol
~Posted by gabemay, Mar 1 |
|
|
|
|
