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(Excuse the pun in the title, couldn't resist) Paris and London are connected by a straight underground tunnel, as shown in the diagram below. A train travels between the two cities powered only by the gravitational force of the Earth. Find the maximum speed of the train and the time taken to travel from Paris to London. The distance between the two cities is $d$ and the radius of the Earth is $R$. Neglect friction. $x$ is the horizontal distance from Paris to the location of the train.

We are inside the Earth, so the gravitational force is not constant. The distance of the train from the centre is $r$ and the gravitational force is $-mg r/R$. The component of this that accelerates the train is $-mg r/R \sin \theta$ if $\theta$ is the angle between the line segment $h$ and the position vector of the train. This gives the differential equation $$\frac{dv}{dt} = v \frac{dv}{dr} = -g \frac{r}{R} \sin \theta$$ Now, $r^2 = h^2 + r^2 \sin^2 \theta$ from the diagram, which means $\sin \theta = \sqrt{1-h^2/r^2}$ . Plugging this into the integrand and simplifying yields $$v \frac{dv}{dr} = -\frac{g}{R} \sqrt{r^2 - h^2}$$ so that $$\int_0^{ v} \tilde v d \tilde v = -\frac{g}{R} \int_R^{\sqrt{(x-d/2)^2 + h^2}} \sqrt{r^2 - h^2} dr$$ which may be solved by a hyperbolic substitution.

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Do I have the set up right? When I try to put in the limits, the computation starts to get very messy. Thanks for any pointers.

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    $\begingroup$ I am sorry Kyle Kanos, I do not understand why you think such a question should be closed. I have taken the time to write out this post with correct latex and went to the bother of attaching a picture, the least I would like from you is an explanation. Furthermore, I have made an attempt to answer this question and explicitly demonstrated my problem. I am not new to SE and I don't believe I have violated any rules. Thanks for your cooperation. $\endgroup$
    – CAF
    Sep 8, 2014 at 18:30
  • $\begingroup$ Please see this meta post: meta.physics.stackexchange.com/q/6093. You are asking us to check your work, which is in violation of the rules. Thank you for your cooperation. $\endgroup$
    – Kyle Kanos
    Sep 8, 2014 at 19:07
  • $\begingroup$ Also, including the @ symbol (e.g., @KyleKanos) will ping a user directly so that they see your message and not come across it by happenstance. $\endgroup$
    – Kyle Kanos
    Sep 8, 2014 at 19:09

1 Answer 1

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You are massively overthinking the problem. The question was "find the maximum speed".

If you know the greatest depth of the tunnel, then compute the difference in potential energy between the surface and that point.

Now convert that to kinetic energy, and then to velocity.

Evaluating

$$U = \frac{mgr^2}{2R}$$

at both $r=R$ and $r=h$, and setting the difference equal to the kinetic energy, you obtain

$$v = \sqrt{gR\left(1-\frac{h^2}{R^2}\right)}$$

You may recognize that the latter part of this expression looks a lot like the sine of the half angle subtended by the start and end point seen from the center of the earth. You can compute that angle from the distance between the two cities. The rest should now be easy...

UPDATE as for doing this with differential equations - as I hinted in my comments, it's all about getting the right coordinates. If you measure $x$ from the center of the hole, you can write the following:

$$r^2 = x^2 + h^2\\ sin \theta = \frac{x}{r}\\ F_{gravity} = \frac{mgr}{R}\text{ (pointing to center)}\\ F_{horizontal} = F sin\theta = - F \frac{x}{r} = -\frac{mgx}{R}$$

And now integrating the equations of motion is really quite trivial...

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  • $\begingroup$ Thanks for your answer. So when I integrate the force equation, I obtain $U = mg r^2/2$ for the potential energy function. Then I can write generally, $$\frac{mgR}{2} = \frac{1}{2}mv(r)^2 + mg r^2/2$$ solve for $v(r)$, find the $r$ that maximizes it and hence $v_{\max}$? $\endgroup$
    – CAF
    Sep 8, 2014 at 19:27
  • $\begingroup$ Not sure why you need to solve and maximize. KE = gained by going from $r=R$ to $r=h$. So write expression for potential energy as function of $r$. If you are right about the form of $U$ (I did not check), then $\frac12 mv^2 = mg(R^2 - h^2)/2$ by substituting the values for $r$ at the surface, and at maximum depth, respectively. $\endgroup$
    – Floris
    Sep 8, 2014 at 19:32
  • $\begingroup$ Oops, I made an error in my expression, it should be $U = mg r^2/2R$ on dimensional grounds, so that $$\frac{mgR}{2} = \frac{1}{2}mv(r)^2 + mgr^2/2R $$ $\endgroup$
    – CAF
    Sep 8, 2014 at 19:49
  • $\begingroup$ The answer given is $$v^2 = \frac{g}{R} x(L-x)$$ where $x$ is defined in the question. I tried, but I can't quite see if the results are equivalent. It requires $R^2 - h^2 = x ( L-x )$. Thanks. Also, could you tell me whether the D.E I set up was in fact correct, just a more difficult method? $\endgroup$
    – CAF
    Sep 10, 2014 at 15:16
  • $\begingroup$ What is $L$ in your expression? Also - this looks like it attempts to be an expression for velocity as a function of $x$ where the question was for max velocity (i.e. at x = $d/2$). As for your DE - I really would want to think about the coordinate system in which it is easiest - inclined to write everything in terms of $\theta$. It doesn't look "obviously wrong" but as I said - you are going about this the hard way. $\endgroup$
    – Floris
    Sep 10, 2014 at 15:25

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