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Whilst perusing a question on WorldBuilding, a link was given to an article about Gravity Trains.

Some of the answers on the question, and the Wikipedia article, state a travel time of 42 minutes between any two points on Earth. The Wikipedia article also states that the maximum speed of a gravity train that goes directly through the center of the Earth is 7,900m/s.

It was this speed that made me think of orbits (it being close to the roughly 8km/s I have in my head) - and indeed another article confirms that "orbiting on the surface" of the Earth involves a speed of 7.9km/s, with the additional property of the period being 1 hour, 24 minutes - ie twice that of the theoretical travel time of a gravity train.

Are these coincidences? If not, how and why are they related?

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    $\begingroup$ The orbital movement is a superposition of the movement of two gravity trains perpendicular to each other. So there is no coincidence. $\endgroup$
    – M.Herzkamp
    Apr 29 '15 at 13:29
  • $\begingroup$ @M.Herzkamp: with some extra detail that would make a nice answer. As it is, I'm not sure it's clear to non-nerds why the (simple harmonic) motion inside the Earth is related to circular motion around it. $\endgroup$ Apr 29 '15 at 15:16
  • $\begingroup$ From the comment I think I can understand why the velocities match, but not why the period is twice that of the travel time between any two points. My last dealing with physics at this level was A Levels 15 years ago :) $\endgroup$ Apr 29 '15 at 15:19
  • $\begingroup$ I believe that the true answer is not quite as simple as the textbook answer, which assumes that the density of the planet is a constant, which would make the motion inside the Earth harmonic, which, coincidentally, is also the projection of a spherical orbit onto the plane of the gravity train... for elliptical orbits all of this falls apart, anyway, because that motion does not project on simple harmonic motions. $\endgroup$
    – CuriousOne
    Apr 29 '15 at 19:18
  • $\begingroup$ @James Thorpe: The period is twice the travel time, because you are back afterwards. It is equal to going to the other side and back again. $\endgroup$
    – M.Herzkamp
    Apr 30 '15 at 9:35
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To answer the velocity question: this is not a coincidence.

The acceleration due to gravity at the Earth's surface is $\frac{G}{r^2}$, where G is Newton's gravitational constant multiplied by the mass of the Earth, about $4 × 10^5$ km³ / sec² (Different sources use $G$, or $\gamma$, or practically anything else for this value), and $r$ is the Earth's radius (say 6371km) - which gives 0.0098-ish km/sec², which is about right.

The gravitational potential energy of a point on the Earth's surface is $\frac{-G}{r}$, following the usual convention that the potential energy at infinity is zero. The kinetic energy of an escape orbit exactly cancels the potential energy - therefore $\frac{G}{r}$ - and the kinetic energy of a circular orbit goes exactly half way towards cancelling it: hence $\frac{G}{2r}$. The velocities of the parabolic and circular orbits are therefore $\sqrt\frac{2G}{r}$ and $\sqrt\frac{G}{r}$ respectively: put the figures into a calculator and they will come out about right.

A cheerful fact about inverse-square-law gravity is that when you are inside a massive spherical object, the gravitational forces exerted by the layers of the object further away from the centre than you are (ie. by a hollow spherical shell) all cancel out. The gravitational force acting on you is therefore just that exerted by the imaginary sphere you are "standing on", however deep you are; and when you get to the centre, it is 0. At a distance $x$ from the centre of the earth, the gravitational acceleration is $\frac{(x/r)^3}{(x/r)^2}=x/r$, where the numerator of the big fraction accounts for the smaller gravitating mass and the denominator accounts for the fact that you are closer to it.

(All this depends on the Earth being of uniform density throughout: which it isn't).

When you are at a distance $x$ from the centre, the work required to climb a distance $\Delta{x}$ is therefore $\frac{G}{r^2}\frac{x}{r}\Delta{x}$, and the total work required to get from the centre to the surface is $\int_0^r \frac{G}{r^2}\frac{x}{r}dx$, or $\frac{G}{r^3}\int_0^r x dx$, or $\frac12 r^2\frac{G}{r^3}$, or $\frac{G}{2r}$. Thus falling from the surface to the centre of the Earth loses you half as much potential energy as you lost by falling to the surface from outer space.

If you fall to the centre of the Earth, your gain in kinetic energy will exactly cancel out your loss in potential energy. So you will have a kinetic energy of $\frac{G}{2r}$, which is exactly the kinetic energy of a circular orbit; and your velocity at the centre will be $\sqrt\frac{G}{r}$, just as it is on a circular orbit.

So this equality isn't a coincidence at all. It is true of every gravitating body (moon, planet, whatever) as long as its density is uniform.

Coincidence of period

As commenters have remarked, this is also not a coincidence, but a case of simple harmonic motion.

  1. Falling through the Earth from the North Pole to the South Pole has an acceleration proportional to the distance from the centre.
  2. "Falling" along a circular polar orbit, and looking purely at the distance north or south of the equatorial plane, a colloquial explanation is as follows. The Earth's gravity works at full throttle at the North Pole (negative if you look northwards) and South Pole (positive if you look northwards), zero at the equator (zero because it is there exactly perpendicular to the direction of motion) and with a strength proportional to the cosine of the latitude everywhere in between. But the distance of a satellite from the equatorial plane is also proportional to the cosine of the latitude. So again, the acceleration is proportional to the distance from the equatorial plane - and in addition, the acceleration at the end-points is the same as when falling through the Earth. So the acceleration is the same throughout; so the motion is the same throughout.

And this too will be true of all planets with uniform density.

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  • $\begingroup$ Thanks for all that! The explanations alongside the math has made it click how they are related $\endgroup$ Jun 6 '16 at 11:37
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It takes about 45 minutes for a low Earth circular orbit to go from one side of the Earth to the other say From New York, USA to Perth Australia. If a hole could be tunneled through a uniform density Earth and you were to jump in the hole gravity would act like a spring and it would take you about 45 minutes, (if no air or frictional resistance) to "fall" from New York to Perth.(the only problem is that if you didn't grab onto something you would oscillate between New York and Perth with a period of 90 minutes) This would work for any uniform density planet using Newton's law of gravity/Gauss's law of gravity but it wouldn't necessarily be 45 minutes.

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