# Why are these periods the same: a low earth orbit and oscillations through the center of the earth?

I was learning about GPS satellite orbits and came across that Low Earth Orbits (LEO) have a period of about 88 minutes at an altitude of 160 km. When I took a mechanics course a couple of years ago, we were assigned a problem that assumed that if one could drill a hole through the middle of the Earth and then drop an object into it, what would your period of oscillation be? It just happens to be a number that I remembered and it was 84.5 minutes (see Hyperphysics). So if I fine-tuned the LEO orbit to a vanishing altitude, in theory, I could get its period to be 84.5 minutes as well. Of course, I am ignoring air drag.

My question is: why are these two periods (oscillating through the earth and a zero altitude LEO) the same? I am sure that there is some fundamental physical reason that I am missing here. Help.

• Wait a moment. You say "If I fine-tuned the LEO orbit to be 84.5 min" and then you wonder why it would be exactly 84.5 min? – ACuriousMind Jul 11 '14 at 15:33
• @ACuriousMind: I think my question is comparing two different oscillations: (1) the period of the person oscillating through a hole in the earth and (2) the period of a LEO orbit fine-tuned to an altitude that resulted in a period of 84.5 min? Is this clearer? I modified to question to reflect your comments. – Carlos Jul 11 '14 at 15:38
• Not really. You are comparing two oscillations, one of which you adjusted to be precisely of the same period as the other. I really don't get the question. – ACuriousMind Jul 11 '14 at 15:43
• Note that you aren't just ignoring air drag, you're ignoring resistance from the trees and stuff that the satellite would be crashing through, since it's orbiting at surface level ;-) – Steve Jessop Jul 11 '14 at 16:16
• I don't think this qualifies as an answer so: I think the intuition here is that, objects in orbit are free falling toward the centre of the earth. Much the same way the dropped ball is free falling toward the centre of the earth. So the time it takes an object to do a quarter turn around the earth, should be about as long as it takes the ball to reach the centre of the earth. Come to think of it, maybe the ball should be slower. Since half way to the core, it has the earth above it pulling it back up. The orbit is always free falling. – Cruncher Jul 11 '14 at 16:22

# Intuitive explanation

Suppose you drill two, perpendicular holes through the center of the Earth. You drop an object through one, then drop an object through the other at precisely the time the first object passes through the center.

What you have now are two objects oscillating in just one dimension, but they do so in quadrature. That is, if we were to plot the altitude of each object, one would be something like $\sin(t)$ and the other would be $\cos(t)$.

Now consider the motion of a circular orbit, but think about the left-right movement and the up-down movement separately. You will see it is doing the same thing as your two objects falling through the center of the Earth, but it is doing them simultaneously.

image source

caveat: an important assumption here is an Earth of uniform density and perfect spherical symmetry, and a frictionless orbit right at the surface. Of course all those things are significant deviations from reality.

# Mathematical proof

Let's consider just the vertical acceleration of two points, one inside the planet and another on the surface, at equal vertical distance ($h$) from the planet's center:

• $R$ is the radius of the planet
• $g$ is the gravitational acceleration at the surface
• $a_p$ and $a_q$ are just the vertical components of the acceleration on each point

If we can demonstrate that these vertical accelerations are equal, then we demonstrate that the differing horizontal positions have no relevance to the vertical motion of the points. Then we can free ourselves to think of vertical and horizontal motion independently, as in the intuitive explanation.

Calculating $a_q$ is simple trigonometry. It's at the surface, so the magnitude of its acceleration must be $g$. Just the vertical component is simply:

$$a_q = g (\sin \theta)$$

If you have worked through the "dropping an object through a tunnel in Earth" problem, then you already know that in the case of $p$, its acceleration linearly decreases with its distance from the center of the planet (this is why the "uniform density" assumption is important):

$$a_p = g \frac{h}{R}$$

$h$ is equal for our two points, and finding it is again simple trigonometry:

$$h = R (\sin \theta)$$

So:

$$\require{cancel} a_p = g \frac{\cancel{R} (\sin \theta)}{\cancel{R}} \\ a_p = g (\sin \theta) = a_q$$

Q.E.D.

This also gives some insight to an unfortunate consequence: this method can be applied only to orbits on or inside the surface of the planet. Outside of the planet, $p$ no longer experiences an acceleration proportional to the distance from the center of mass ($a_p \propto h$), but instead proportional to the inverse square of distance ($a_p \propto 1/h^2$), according to Newton's law of universal gravitation.

• To add to this: mathematically, the $z$ coordinate of the projectile is governed by the same equation of motion whether it is orbiting the earth or passing straight through (in the $z$ direction). – lemon Jul 11 '14 at 15:43
• Great answer Phil! I knew it was something fundamental. This really shows that circular motion and linear periodic motion are really one and the same. – Carlos Jul 11 '14 at 16:08
• This shows that the projection of a circular orbit is indeed sinusoidal. But it doesn't show that such a projection is a solution to an object oscillating through the center of the Earth. – BMS Jul 12 '14 at 13:31
• @BMS I've added a proof. – Phil Frost Jul 13 '14 at 11:37
• +1 but : "acceleration linearly decreases with its distance from the center of the planet" You should probably point out that this only holds for uniform densities, especially given that this assumption is fairly off. I realize you pointed this out right before starting your proof, but it'd be good for future readers to know where exactly this comes into play. – ticster Jul 13 '14 at 12:05

Phil's answer, while beautifully illustrated, is a little incomplete. It relies on the fact that in the case of the tunnel you're solving the one dimensional projection of the low earth orbit satellite, but doesn't prove this. I do this below. The force applied on the object, for a sphere of uniform density, is actually :

\begin{eqnarray} F &=& - \frac{4}{3} \pi \frac{G m \rho r^3}{r^2} \\ &=& - m g \frac{r}{R_{earth}} \\ &=& - k r \end{eqnarray}

Where $k = \frac{mg}{R_{earth}}$. This is equivalent to a spring problem, whose solution will indeed be sinusoidal with period $2 \pi \sqrt{\frac{R_{earth}}{g}}$, the same as a low earth orbit period. Again, while Phil's answer does provide an illustration of this, it doesn't actually prove it. In particular, it leaves out the crucial fact that this only holds for a sphere of uniform density.

An alternate explanation (which really is the same as the answer from @Phil): as per Kepler's laws, an orbit is an ellipse, and the orbiting period is proportional to the semi-major axis of the ellipse.

A satellite in the lowest orbit will try to follow a special kind of ellipse (namely, a circle), whose semi-major axis is really the Earth radius (this is the "lowest orbit" because the satellite grazes the ground -- we ignore the atmosphere here).

The oscillation in the hole is really another orbit -- it is a degenerate ellipse which has been flattened to a line. Yet its semi-major axis is still the Earth radius.

Same semi-major axis, hence same period.

Edit: as was pointed out, that expanation is bogus in two ways:

• The degenerate case for a "flattened" ellipse would be a half-diameter. If all the Earth's weight was concentrated at its center, the orbit, starting from "ground" level (6300 km or so from the center) with (almost) no lateral velocity would be an accelerated fall toward the center; when close to the center, the object would miss it "by mere inches" and quickly run around it, before speeding up back to the initial position at ground level. Furthermore, that "flattened ellipse" would have a semi-major axis of length about 3150 km (half the radius), for a period which would be eight times smaller than the low orbit.

• The Earth weight is not concentrated at its center. In fact you get an "oscillator" trajectory, that allows you to emerge in New Zealand if you started from England, precisely because the "Earth mass at a single point" model is not the one used in this thought experiment.

While it is understandable that the low orbit and the ocsillator end up with periods of the same magnitude (they both are kinds of "free fall" against an Earth with the same weight, and starting at ground level), that hand-waving remark would have been equally applicable with an oscillator period being twice or half that of the low orbit. They seem to end up quite close to each other and I now have no idea whether this is mere coincidence or for some fundamental reason.

• Don't Kepler's laws assume that the center of the potential is at one focus of the ellipse though? (I honestly don't remember.) The foci of a degenerate ellipse are at the ends. – David Z Jul 11 '14 at 23:57
• There's one very small fault with this explanation. Kepler's laws assume the mass of the larger body to be concentrated at its center. When you go through the hole, the effective mass of the earth decreases as you approach the center, so your orbit through the center of the earth is not a Keplerian ellipse. – Tristan Jul 12 '14 at 0:01
• Most of what I know about orbital mechanics I learned from Kerbal Space Program, but I think it's true that the "flattened ellipse" path is not Keplerian. There is a Keplerian orbit for a flattened ellipse: a radial elliptic trajectory, but it's a quite different thing. – Phil Frost Jul 13 '14 at 11:36

My question is: why are these two periods (oscillating through the earth and a LEO) the same? I am sure that there is some fundamental physical reason that I am missing here. Help.

It's a result of the (flawed) assumption of a uniform density Earth. The Earth is anything but a constant density object. The Earth's core is five times more dense than surface rock. Gravitational acceleration reaches a maximum of over 10 m/s2 at the core-mantle boundary, which is a bit less than halfway to the center of the Earth. A uniform density model implies that gravitational acceleration is about half the surface value at this depth.

A better model of the Earth is to assume that acceleration due to gravity is a constant 10 m/s2 from the surface to halfway to the center of the Earth and then drops linearly to zero at the center of the Earth. This yields a period of 76.41 minutes rather than the 84.3 minute period of a 6371 km orbit (obviously ignoring air drag).

An even better model is to use numerical integration with the Preliminary Reference Earth Model (A. Dziewonski and D. Anderson (1981), "Preliminary reference Earth model," Physics of the earth and planetary interiors 25:4, 297-356. (tabular data at http://geophysics.ou.edu/solid_earth/prem.html)). This yields a period of 76.38 minutes, which is very close to the simple model described above.

Phil Frost's argument in his answer (v4) is correct. Assuming a spherical Earth with constant density $\rho$ (and assuming for simplicity that the object for some reason can move freely$^1$ through Earth so that there is no air drag, and so that we can skip all the tunnel drilling and not worry about that the Earth's rotation could press the object up against the tunnel wall; and assuming that we use Earth-Centered Inertial (ECI) coordinate system, so that there are no fictitious forces; etc), then the governing 3D vector-valued ODE (derived from Newton's laws) is

$$\tag{1} \frac{d^2{\bf r}}{dt^2}~=~-\frac{4\pi G\rho}{3}{\bf r}, \qquad\qquad {\bf r}~\equiv~\left(\begin{array}{c}x\cr y\cr z\end{array}\right).$$

This ODE (1) separates in three independent SHOs for the $x$, $y$ and $z$ coordinates with common characteristic period

$$\tag{2} T~=~\sqrt{\frac{3\pi}{G\rho}}~\approx~ 84~ {\rm min}.$$

In particular, for an arbitrary trajectory with $|{\bf r}|\leq R$ (= radius of spherical Earth), the period is independent of initial position and initial velocity.

--

$^1$More precisely: move freely apart from gravity.