Falling through the Earth problem

1. Sep 17, 2011

Xyius

1. The problem statement, all variables and given/known data
I basically need to find the resulting motion of an object that is falling through the Earth. I also need to find how long it takes. I am ignoring friction and assuming the density of the Earth is constant.

2. The attempt at a solution
Here is an image of my work. (I used MathType program, its easier for me than using the LaTex code)
[PLAIN]http://img825.imageshack.us/img825/2844/earthc.gif [Broken]

This seems to have been the hard part, I am getting stumped on the easier part, knowing how to solve the differential equation! I know that this is of the same form for Hookes Law, but what is throwing me off is the fact that the formula is with respect to "r." I know that when solving the differential equation for simple harmonic motion, you use newtons second law which is is only in the x direction, but since this is "r" does that mean I have to do it for both x and y components and use trigonometry? Or can I simply write Newtons second law as..
$F=m\frac{d^2r}{dt^2}$

Last edited by a moderator: May 5, 2017
2. Sep 17, 2011

dynamicsolo

To keep this problem simple, it is assumed that the Earth is not rotating (often by having the object fall from pole to pole). So the motion is purely one-dimensional: the object goes from the surface to the center and on to the antipodal surface, and then back again. So r is treated as a single variable, allowing us to work out the result as simple harmonic motion. (You should find that the period of "oscillation" is a familiar value...)

3. Sep 17, 2011

Xyius

Okay so here is what I come up with..
[PLAIN]http://img705.imageshack.us/img705/1416/earth2q.gif [Broken]

What is an initial condition to solve for B? Also, my final answer depends on the persons mass still, is this correct? I recall reading that it takes a set time to fall through the earth regardless of mass or where the hole is drilled.

Also a conceptual question, if the hole is drilled side ways above the widest point of the earth. (Like towards the top, as it is shown in my problem.) Would the person not "scrape" against the side since the gravitational force is pulling them in radially? If their path is horizontal, would they not scrape against the side? (I know this problem says ignore friction but this is just a conceptual question.)

EDIT: Oops, I didn't take "m" into the factor for "g". That explains why it is there at the end. :p
EDIT2: Also the terms in the trig functions should be square roots! Doy :\

Last edited by a moderator: May 5, 2017
4. Sep 17, 2011

dynamicsolo

Don't forget that your SHM differential equation corresponds to ma = F , or

$$m \frac{d^{2}r}{dt^{2}} - F = 0 ,$$

so the m should "divide out".

The velocity $\frac{dr}{dt}$ at r = RE is zero.

This is why we make the assumption of a pole-to-pole excursion with no rotation; otherwise, you hit the wall in a straight tunnel. There is a considerable literature on "gravity trains", for which you would have to dig a curved tunnel through the Earth's interior, in part to avoid bumping, but also because it is the most effective path (OK, we just imagine this -- there are a few problems with actually doing it...). Interestingly, Hooke corresponded with Newton about this idea. You can read more, starting from, say,

http://www.damninteresting.com/the-gravity-express/

http://en.wikipedia.org/wiki/Gravity_train

(Frankly, it's "easier" to build a space elevator...)

Last edited by a moderator: May 5, 2017
5. Sep 17, 2011

Xyius

Okay so I got B=0 and my final result as..
$$r(t)=R_{E}cos(\sqrt(\frac{g}{R_{E}})t)$$

Now I want to find how long it takes, so setting it equal to "-RE" I get

$$\frac{\pi^2 R_{E}}{g}$$

Which is NOT right, considering the radius of the earth is 6378100m (Taken from google).

EDIT: Man I must be tired I am making mistakes everywhere! I forgot "t" is not under the square root. Lemme fix that now..

EDIT2: There we go! 42 minutes! Thats the magic number! Thank you VERY MUCH for your help :D

Last edited: Sep 17, 2011
6. Sep 17, 2011

dynamicsolo

OK, I guess you spotted it already: the angular frequency is $\omega = \sqrt{\frac{g}{R_{E}}}$, and the period is $T = \frac{2\pi}{\omega}$.
Recalling that $g = \frac{GM_{E}}{R_{E}^{2}}$, you can express your result for T purely in terms of the universal gravitational constant G and properties of Planet Earth. Does the number (in minutes) look at all familiar?

[EDIT: Hence, Douglas Adams' "42"... The problem does ask for the one-way trip time. The point I wanted to make is that this is half the "oscillation" period of 84 minutes, which is the orbital period for a satellite (very) close to the Earth's surface (for reasons of atmospheric drag, no stable satellite has a period much below about 90 minutes). The conditions of the problem remove the other dimensions of orbital motion, but still approximate the period reasonably. A satellite in orbit is a sort of three-dimensional gravitational oscillator.]

Last edited: Sep 17, 2011
7. Sep 17, 2011

Xyius

Wow very interesting!!

Now I am curious as to the approach to solve this problem in more than one dimension. (I am not looking to specifically do it at the moment.)

For example, say you want to examine a hole through the side above the widest point of the earth as I mentioned earlier. (Ignoring the friction force from sliding on the wall.) In this case would you need to use newtons law in the x and y direction?

EDIT: Could you switch the coordinate axis so the sum of the forces in the y direction is zero, and the only force propelling the mass would be the cosine of the gravitational force? (Kinda hard to explain without a picture ha!)

8. Sep 17, 2011

dynamicsolo

You would get to do something along those lines in a more advanced mechanics course...

I'm not entirely clear on the distinction you're making. (What do you mean by "the widest point of the Earth"?) If we treat the Earth as not rotating, you get the same result for any tunnel passing through a diameter of the Earth (that is, surface to center to antipodal surface). The two- or three-dimensional complications come in if your tunnel is along a chord through the Earth (surface to surface, but not through the center) or if you include the Earth's rotation.

9. Sep 17, 2011

Xyius

Yeah that's what I mean. How would I show it in the two dimensional case? We do not start motion in two or 3 dimensions until next week, so I am confused as to why the picture in this problem shows a path that goes surface to surface but not through the center.

I would just like an understanding of how the approach would be in the two dimensional case. It seems like the only thing propelling you would be the horizontal component of the gravitational force.

10. Sep 19, 2011

dynamicsolo

Because gravity is what is called a "central force", which means that the strength of an interaction between two (point) masses depends only on (radial) distance and not direction, it is more convenient to solve a two-dimensional problem in polar coordinates. Although his original derivation was not in modern notation, essentially Newton used his newly-developed techniques of calculation to show that a force which depends on the product of the masses and is inversely proportional to the square of the separation distance leads to trajectories in a plane which follow the form of the polar equation for a conic section (ellipse, parabola, hyperbola). This is a big part of what he was sitting on that led Halley to urge that Newton publish his work in gravitation and calculus, the major opening work in the field that became orbital mechanics.