Tuesday, January 30, 2007

Holes, Deeper Holes and Holes in Things

I have recently found myself falling into holes. Not actual holes (although it does happen on occasion), rather holes in the internet. You have all fallen into one of these holes before; it might have been on Wikipedia, or possibly Youtube, and it might even have happened during a Google search. You start out reading an article about Don Quijote de la Mancha (Don Quixote) and next thing you know its three hours later and you are watching a video about the adventures of the cartoon aardvark Arthur. Now you are beginning to understand what I am talking about. Of course while you where doing this you also discovered the new pointless evil of people in the world, the lovely Youtube chain curses. These go along the lines of some young girl gets killed for no apparent reason and now you must post this message you just read in three other places or she will haunt you for five years, makes sense to me. Now let’s talk about other kinds of holes, holes through things, to other places. For instance through the center of the Earth to China, now let all ignore the thousands of very good reasons why you cannot dig a hole through the center of the Earth (which I will not list, you are a geek you should know, if you seriously don’t just Google it). However, even if you where to dig a hole directly through the surface of the earth through anywhere in the USA or even North America you will end up nowhere near China. In fact to get to China you would have to dig a hole from somewhere in Southern South America, preferably somewhere near central Argentina. I also took the liberty of digging a hole in my back yard, which you can clearly see in the picture below.

However I ran into a bit of the problem when I got out to the other side, I had just dug a hole into the bottom of the Indian Ocean (somewhere off the south western coast of Australia), this had the adverse side effect of draining half of the Indian Ocean into New Jersey. Then of course there is the small problem of what to do with the 525,835,387 cubic feet of dirt I just dug up.


  1. If you were really a geek, you would know that it would take you the same exact amount of time to fall through the hole to the other side if you dug a straight line from here to china or from here to the other side of the earth. (In an earth that was uniform and frictionless and filled with a superfluid).

  2. I've just done my calculations. It will take about 42 minutes to fall through a hole to the other side no matter where you are. In fact, with the assumptions as stated by Jonathan, you will be moving with simple harmonic motion. Cool, eh!

    And I am not a geek.

  3. Well first of to get any number worthy of true relevance you have to include terminal velocity into your equation (appoximatly 55 m/s). In free fall you will reach this speed in approximatly 8 seconds, with a distance of appoximatly 154 meters. this means that if we are assuming that the force due to gravity will not change (even though it will) you get to fall the remaining 6399846 meters to the core of the earth at 55 m/s. Due to this you also will only go around 150 meters past the core before you fall back to the center where we now find a case of damped harmonic motion with the end result of you being stuck having to make your way 6400 Km to get back to the surface (this is also of course neglecting all the other things that would kill you before this actual happens). Also, assuming that there is no friction to air resistance and the equation d=Vi*t+.5*a*t^2 and a little algebra, sqrt(2*d/a)=t you then plug in d as 6400000 m and a as 9.8m/s^2 and you find that the time is equal to 1143 seconds. Now since there is no air resistance you would know that it would take the same amount of time to get back to the surface as it took to get down so we multiply this by 2 and we get 2286 seconds. When we divide this by 60 we find that the trip would take around 38 minutes. I am somewhat curious as to how you did your math. Also, the waste I produced from digging my hole is produced when my hole has a 4ft diameter.

  4. Aryeh, the thing is, the equation you quoted only holds close to the Earth's surface. Elsewhere you need to use a=GM/r^2. Figure out what the mass enclosed is in terms of radius...

  5. That is true Eli, but I didn't feel like spending the time to figure out the equation I would need to solve for a system that the acceleration due to gravity changes each second due to the distance I just fell. If you want to go ahead, I don't even know if I have the tools to properly solve this.

  6. Wimp. It's really easy for something falling directly through Earth's center. For any chord it needs a tad more work.

  7. First, let me say that I am just a high-school graduate, so my math may not be water-tight, but I am reasonably confident that my reasoning is correct.

    Below are the data that I use. I got this from Britannica.
    Radius = 6378140 m
    Mass = 5.976 * 10^24 kg
    Density = 5498.44 kg/m^3

    If we were to assume that the earth is a perfect sphere with uniform density and the person falling through the earth encounters no air resistance, an interesting result is observed. Let us first consider the simplest case whereby the person falls through the diameter of the earth.

    g = G M / r^2
    M = 4/3 * Pi * r^3 * rho, where r is the distance of that person falling from the centre of the earth. (Now, M is not the mass of the total mass of earth! If a person is at distance r from the centre of the earth, the gravitational effect of the earth will behave as if it originates a point mass with mass as given by the formula. This is in accordance to Newton's Shell Theory.)

    Hence g = 4/3 * G * Pi * rho * r, i.e. g is directly proportional to r, which means that the person will move with simple harmonic motion. (ignoring friction, etc) Simple physics will tell you that omega^2 = 4/3 * G * Pi * rho, where omega is equal to 2Pi / T with T being the period of the simple harmonic motion.

    Solve[ (2Pi /T)^2  4/3 Pi density (6.67 * 10^-11), T]

    T -> 5069.36 seconds, which is about 84 minutes
    But 84 minutes is the time for the person to fall to the other side and back, so divide by half and you get 42 minutes. The effect is that the person will move with simple harmonic motion with the centre of the earth as the equilibrium position, with period of 5096 seconds.

    It takes me some time to prove that T must still be 84 minutes for any chord, but Jonathan is right. Try working it out. Hint: resolve the forces.

  8. Well done, Daniel! If you keep it symbolic one step further, you can get a neat result. Rho = Mass / Vol = MassEarth/(4/3)*pi*RadiusEarth^3. Putting this into your expression for omega gives: omega^2=G MassEarth / RadiusEarth^3. Then substitute to find period, etc. Compare this to the result for period of orbit from Kepler's Laws...

  9. Wish I could comment, but I have to rewrite a report, and study electron correlation potentials for my research, so you're on your own.