Saturday, March 4, 2017

Underground from Portland to New York (Part 1)

Portland to New York Using only Gravity (Part 1)

The Brachistochrone

The brachistochrone is a classic problem put forth by Johann Bernoulli and solved by some of the mathematical heavyweights of the day. The basic premise of the problem is that given two points connected by a curve, find the shape of the curve such that a particle moving only under the influence of gravity would travel along the curve in the minimum time. Somewhat surprisingly, this curve is not a straight line. Even Galileo figured this out in 1638 though mistakenly thought the curve was a circular arc. It turns out that the curve is a cycloid but that's not what we're going to prove here.

Going Underground

Lets say we want to build a transportation tunnel from Portland, Oregon to New York, New York. We've invented a frictionless surface and we can evacuate the tunnel so there won't be any air resistance. We want to go completely green and only use gravity for travel. Oh, and to add to our perfectly realistic assumptions, we live on an earth of uniform density and of perfectly spherical shape.

What is the shape of the tunnel that will get us across in the shortest amount of time?

In similar fashion to solving the Brachistochrone problem, we must minimize the functional $$ I=\int_{t0}^{t1}dt = \int_a^b \frac{ds}{v} $$

To get started we'll employ some basic physics. Because there is no energy added to the system, the sum of kinetic and potential energies will be a constant throughout the trip. The kinetic energy, \(T\), is familiar and we can write it directly as $$ T = \frac{1}{2}m_Tv^2 $$ where \(m_T\) is the mass of the train. The potential energy, \(V\), is not quite as straightforward as it is not the familiar \(V=mgh\) we've seen in our entry level physics class. To derive it we first need to find the forces involved. The force of gravity is $$ \vec{F} = -G\frac{m_Em_T}{r^2} \uvec{r} $$ where \(m_E\) is the effective mass of the earth when we are somewhere inside of it at a distance \(r\) from the center. This mass is all the mass that is located less than or equal to a distance \(r\) from the center. So $$ m_E = \frac{4}{3}\rho\pi r^3 $$ where \(\rho\) is the (uniform!) density of the earth. Plugging that back into our force equation gives $$ \vec{F} = -G\frac{\left(\frac{4}{3}\rho\pi r^3\right)m_T}{r^2} \uvec{r} = -\frac{4}{3}G\rho\pi m_T r \uvec{r} $$ (Note the negative sign since the force is directed opposite the radial direction.) Since we know the acceleration due to gravity is \(g = 9.8 \frac{m}{s^2} \) at the surface, \(r = R\) we make the substutition of $$ \eqalign{ g &= \frac{4}{3}G\pi\rho R \cr \vec{F} &= -\frac{m_T g}{R}r\uvec{r} } $$ As an aside, here it is interesting to note that inside the earth, the force actually increases linearly with the distance \(r\). This is because the mass increases as \(r^3\) even though the gravity is an inverse-square law.

Now to compute the integral for potential energy, we will choose our reference (\(V=0\)) point to be \(r = R\). The line integral is along the radius to a distance \(r\) from the center. $$ V = -\int_R^r -\frac{m_T g}{R} x dx = \frac{m_Tg}{2R}(r^2 - R^2) $$ Since the total energy \(T+V=0\) at the start and no energy is added to the system we can set the total energy to 0 and solve for the speed as a function of \(r\). $$ \eqalign{ T+V &= \frac{1}{2}m_Tv^2 + \frac{m_Tg}{2R}(r^2 - R^2) = 0 \cr v &= \sqrt{\frac{g}{R}(R^2-r^2)} } $$ We are now at a point where we can write the functional we wish to minimize as $$ \eqalign{ I &= \int_{t0}^{t1}dt = \int_a^b \frac{ds}{v} \cr &= \int_a^b\frac{\sqrt{(dr)^2+(rd\theta)^2}}{\sqrt{\frac{g}{R}(R^2-r^2)}} } $$ To be continued in part deux.

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