How to find the change in time over a given distance, without constant acceleration or velocity?

Question

I'm working on a problem that essentially is trying to calculate the maximum velocity and time it would take, for one to fall through a friction-less hole from Chicago to Rome.

I have derived a function with respect to distance that calculates the acceleration at any point x in the tube. Integrating this from x=0 to the center point of the hole, which in this case happens to be x=3633501, gives me my maximum velocity. However, I am at a loss for how to calculate the time it would take to get to the center of the tunnel, which I would then double to get the overall trip time.

The formula for acceleration with respect to position is as follows:

$$a(x) = 1.5404 \cdot 10^{-6}\cdot \sqrt{3.1272 \cdot 10^{13}+x^2-7267002x}\cdot \sin\left(\arctan(\frac{2968886}{\sqrt{2.24584 \cdot 10^{13}+x^2-7627002x}})\right)$$

If anyone has any idea where to go from here to calculate time, it would be greatly appreciated. I can supply any other info upon request, I'm just not sure what is important to include here and what isn't.

Thank you.

Answer 1

$\sin(\arctan(x))=\frac x{1+x^2}$ will get rid of the trig functions. After that I suspect you are in for numerical integration.

Answer 2

If you use conservation of energy, you can get the velocity as a function of distance along the tube. Suppose the length of the tube is $2L$, and we set the coordinate system such as $x=0$ is at the center of the tube. That point is at a distance $r_{min}=\sqrt{R^2-L^2}$ from the center of the Earth of radius $R$. Conservation of energy says: $$\frac{GMm}{R}=\frac{GM_im}{r}+\frac{mv^2}{2}$$ $M_i$ is the mass of the Earth that is inside the radius $r$ $$M_i=M\frac{r^3}{R^3}$$ All you need is now the relationship between $r$ and $x$ $$r^2=r_{min}^2+x^2=R^2-L^2+x^2$$

If I did not do any mistakes, you get $$v^2(x)=2GM\frac{L^2-x^2}{R^3}$$ To get time $v=\frac{dx}{dt}$ or $dt=\frac{dx}{v}$, so $$T=\int_{-L}^{L}\frac{dx}{v(x)}$$