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Planetary flybys are used to modify the orbital parameters of spacecraft.

For maximum gain, a large deflection angle is often desired. But the deflection angle is limited by the closest approach, which in turn is limited by the radius of the planet

flyby limit

However: what if a tunnel was dug along the flyby path? Then the closest point of the flyby could be even lower, and the deflection angle increased.

tunnel flyby

Or would it?

Considering the extreme case, going straight through the centre, a tunnel path offers no deflection at all, where a point mass flyby would in contrast give a full 180º.

no deflection

Sub-surface gravity can be found by applying the shell theorem, and assuming uniform density, the force of gravity then becomes proportional to distance $F(r) \propto r$, in contrast to the usual $F(r) \propto r^{-2}$ (as an aside, this law of gravity also happens to have stable orbits, the only other exponent for $r$ to do so. They are however ellipses with their geometric centre instead of a focus at the centre of mass, so it's not immediately clear how to patch them together with the outside hyperbola)

Writing a simple time stepping simulation, I found the deflection angle to indeed be increasing a little when digging tunnels some way into the planet, but then shrink again.

The question then arises: What is the optimal depth for flyby tunnels? Presumably it depends on $v_{\infty}$

SE - stop firing the good guys
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    (this does not appear to be a practical scheme for infrastructure) – SE - stop firing the good guys Sep 21 '22 at 21:07
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    Would you have to flare the exit of the tunnel to allow for planetary rotation? – Organic Marble Sep 21 '22 at 22:18
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    Presumably yes, but planetary rotation would not affect the trajectory in any way. – SE - stop firing the good guys Sep 21 '22 at 22:32
  • I'm sure a problem of flying through a globular cluster wes studied somewhere. In this case we don't need a tunnel. But it's not full equivalent of this problem, the mass distributions will be some different. – Heopps Sep 22 '22 at 09:47
  • Are you assuming a uniform density planet? For the Earth gravity actually increases with depth up to a point, which might increase the maximum achievable turn angle. – 2012rcampion Sep 22 '22 at 18:31
  • Yes, I'm assuming uniform density, and explicitly state that. Not that an analysis considering real planetary variable density would be unwelcome. – SE - stop firing the good guys Sep 22 '22 at 18:36
  • Short version: assuming uniform density sphere, optimal depth is 0. Gravity inside a shell (see e.g. http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html) is zero, so any depth below the surface >0 results in a spherical shell of >0 thickness contributing net-zero deflection. I have not considered the variable-density oblate spheroid. – Joshua Voskamp Sep 22 '22 at 19:10
  • Props for considering the extreme case; expanding on this, realize that the extreme case is the union between the 'accelerating directly toward point mass' and the 'no acceleration due to spherical shell' cases – Joshua Voskamp Sep 22 '22 at 19:14
  • Having a probe fly remotely through a curved tunnel during a flyby would be interesting. I'd like to see that! ;-) – Fred Sep 23 '22 at 04:00
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    As an after thought, the nature of the rock being flown through during such a "fly-by" could pose practicality challenges. For argument's sake, a planet with an internal structure similar to Earth might find such a tunnel eventually closing in on itself if the tunnel had to pass through the mantle. – Fred Sep 23 '22 at 04:15
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    This sounds like a perfect question for XKCD What If. – molnarm Sep 23 '22 at 12:42
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    Entertaining theoretical problem, but of course the wind resistance in the atmosphere and especially in the tunnel would probably stop the probe dead rather than increasing its deflection angle. – Christopher Hamkins Sep 23 '22 at 14:28
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    There was a hypothetical idea proposed of digging a parabolic tunnel through the crust of the Earth, equipping it with magnetic levitation, sealing it and sucking out the air, and then you could drop a passenger train from one end, which would give it just about enough momentum to come up the other side. They quoted New York to London in 1 hour, with virtually no energy used (other than to power the maglev). This seems similar, but in reverse... – Darrel Hoffman Sep 23 '22 at 21:46
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    @DarrelHoffman it seems so, except that the tunnel shape is recalculated for a nonzero (and in fact very very large) initial and final velocity! – uhoh Sep 24 '22 at 00:44
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    Perhaps if one wants a given exit angle and velocity, controlling the entry is best. Gravitational acceleration in a mineshaft (compared to the surface) would be an interesting study. – Robert DiGiovanni Sep 24 '22 at 06:39
  • Since the portion of the planet that is now "above" (farther from the center) the spacecraft would have a negative effect on the slingshot effect, I doubt deflection increases with depth. Rather, it probably decreases until it is zero when the path is through the center of the planet. This matches A McKelvy’s numerical analysis. – Todd Wilcox Sep 24 '22 at 11:52
  • @ToddWilcox using Newton's shell theorem for a radially symmetric (but varying with radius) mass distribution, it turns out that there is no net effect from the mass above. It integrates to zero and only the mass below the radius of the spacecraft contributes. I know it seems counterintuitive at first, but it's an incredibly useful result and makes these kinds of problems easier to solve. – uhoh Sep 25 '22 at 00:35
  • @uhoh Doesn’t that mean there’s less mass “below” and therefore the force of gravity (using the Newtonian approximation) is less? – Todd Wilcox Sep 25 '22 at 01:10
  • @ToddWilcox yes, but you're also closer to it (and in reality what's below you is also now higher density) so one has to do some actual math to find out what happens. I'm only saying that the "negative effect of the mass above" doesn't happen in the case of a radially symmetric mass distribution. – uhoh Sep 25 '22 at 02:00

3 Answers3

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The Optimal Depth is Precisely (Almost) No Depth

I solved this problem using a numerical differential equation solver to investigate a range of flyby approach angles. The results are fascinating!

I used the earth as a basis for the investigation. The planet radius is marked in red, and the starting point is near geostationary altitude. I implemented the "tunnel gravity" with a simple Boolean and then taking the mass under the position as the relevant mass.

These are my plots!

I plot here 3 different starting velocities and a range of starting angles for each. If you look carefully, you will see that for all three cases the sharpest turning is provided by the trajectory that passes closest to the planet's surface.

Another fascinating result is that the planet acts like a lens! producing a focal length that is dependent on the initial velocity.

I will now admit that this is not a proof (though I think it is quite convincing), but every starting point I tried had the same result (and I tried many more than shown here). I am sure someone could mathematically prove this (and explain the neato lensing behavior), but that person will not be me.

Editing to address further work and comments discussion. On closer inspection, there is very marginal (~0.1 degree) extra turning to be had with a relatively shallow minimum depth (~100 km below ground)

Addendum: Variable Density and a Trench 200 km at Its Deepest

User Uhoh solicited a model for Earth's density model from this question following discussions about the effect this would cause. I integrated the equations to give mass and redid my analysis. more of my figures Despite the modern-art aesthetic of these plots, there was very minimal change. The optimal minimum depth did reduce (as expected) but only by about 100 km. The left figure is not useful, I just wanted to include it because I think it's pretty. The middle figure shows a small band of the most optimal trajectories. The best had a depth of 190 km (0.97 Earth radii). The right plot shows a zoom of the middle and reveals that a 190 km deep trench would be, in fact, pretty deep.

A McKelvy
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    I suspect a very shallow underground flyby might be minimally better than a flat 0. I expect $1/h^2$ over time (for h - distance from surface) should be the function to maximize and you might gain a bit by digging the periapsis slightly in, to keep the arrival and departure trajectory closer to the surface. Still, I don't expect more than ~1% percent gain vs tangent trajectory. – SF. Sep 22 '22 at 15:25
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    @SF. That matches my simulations. At unit mass, unit radius, I get a 0.147º improvement for a vinf of 1, by going down to a depth of 0.986 – SE - stop firing the good guys Sep 22 '22 at 15:27
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    So we need a ditch, not a tunnel! – Organic Marble Sep 22 '22 at 15:37
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    @SF Yes, interesting. I ran a sweep and find the maximized turning at about 140 km underground, but this provides only a 0.1 degree of extra turning as compared to a 0 altitude flyby – A McKelvy Sep 22 '22 at 15:44
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    For a formal proof that optimal surface depth is zero, consider elaborating on Newton's Shell Theorem (e.g. http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html) -- any apparent deflection to be gained as a result of some below-ground depth is an error in the numerical solver. – Joshua Voskamp Sep 22 '22 at 19:12
  • Though perhaps I've misunderstood some element of the problem if I follow "taking the mass under the position" as to mean "subtracting mass of spherical shell above radius R from the total mass M"? – Joshua Voskamp Sep 22 '22 at 19:18
  • I love your answer! What radial density profile $\rho(r)$ did you use for Earth? – uhoh Sep 22 '22 at 19:34
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    @JoshuaVoskamp yes.. uhoh I assumed constant density (sorry should have included that in my answer). I imagine the result will be deeper if a realistic gradient is used. – A McKelvy Sep 22 '22 at 21:53
  • @AMcKelvy I didn't receive a notification of your comment because you didn't @reply me (from FAQ), but I'm following your answer so luckily I did. Since Earth's density is less than half of the mean at the surface and more than double the mean about half-way in, results will be very very different if you include even a very rough approximation for radial density. – uhoh Sep 23 '22 at 01:32
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    @AMcKelvy So I've just asked An approximate, continuous piecewise analytical model for Earth's radial density profile useful for numerical integration? – uhoh Sep 23 '22 at 01:38
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    @JoshuaVoskamp The shell theorem ruins any gains from deeper trajectories, rapidly decreasing the "active" mass pulling the craft the deeper it flies through the tunnel/trench. There is some angle to be gained by pulling the arrival/departure branches closer to the surface though. Keep the part within the trench very short and shallow and the losses it causes won't outweigh gains outside the trench, on a much longer segment of the trajectory, even assuming uniform Earth density. – SF. Sep 23 '22 at 02:12
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    @uhoh apologies, I flew too close to the sun trying to reply to you and Joshua in a single comment and was notified that I can’t @ two users in a single comment. I will be interested to see what your question reveals! – A McKelvy Sep 23 '22 at 02:23
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    @JoshuaVoskamp I don't think that's a valid argument. Only the part of the planet above is a shell with net gravity 0. That is, planets are not hollow. – SE - stop firing the good guys Sep 23 '22 at 05:39
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    @uhoh In a way, assuming a uniform density is more mathematically interesting. When accounting for increasing density, it is quite obvious that a sub zero depth can be optimal. (Like, for an iron core with a low density fluff layer around). – SE - stop firing the good guys Sep 23 '22 at 05:43
  • @SE-stopfiringthegoodguys Disagree! One person can not define what "more mathematically interesting" is for everyone else. https://en.wikipedia.org/wiki/Luneburg_lens – uhoh Sep 23 '22 at 07:02
  • @uhoh Agreed that it is subjective. Non-uniform density is certainly more difficult computationally (and typically that means "more interesting") but if we pose a question "can the sub-surface trajectory be more optimal" the right choice of density distribution (heavy core) provides a trivial "yes" for the answer. With uniform distribution answering that question becomes quite tricky and nuanced as proximity and shell theorem pull the result in opposite directions. – SF. Sep 23 '22 at 10:29
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    @OrganicMarble A meridian trench, perhaps? https://starwars.fandom.com/wiki/Meridian_trench/Legends – Max Q Lagrange Sep 23 '22 at 19:52
  • @AMcKelvy wow thanks for the extra effort, and for typing all those coefficients! I'm just curious what is the maximum angle of deflection for the two cases? for comparison purposes it can be approximated by $\arccos(\hat{\mathbf{v}}_i \cdot \hat{\mathbf{v}}_f)$ even though your initial and final velocities are at finite distance. The asymptotic angle will be larger but outside of Earth the gradient doesn't matter so at least this will show which will produce the larger deflection. I guess it won't matter much since it's shallow and Newton's shell theorem and all. – uhoh Sep 23 '22 at 20:54
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Assuming a uniform planet with radius $1$ and gravitational parameter $1$, if $V_\infty$ is the speed at infinity and $r$ is the closest approach to the center, then the maximal speed is $V_m = \sqrt{V_\infty^2 + 3 - r^2}$, and the specific angular momentum is $L = V_mr$. If I haven't made a mistake, the deflection angle is then $$ 2\left( \arctan\frac{{r}\sqrt{1-r^2}}{V_m\sqrt{V_m^2-1}} + \arctan \frac{L(V_\infty^2+1)}{\sqrt{V_\infty^2+2-L^2}} - \arctan LV_\infty \right), $$ where the first summand is due to the elliptic part of the trajectory inside the planet, and the rest is due to the hyperbolic parts outside.

This expression is too complex for me to try and find the optimal $r$ analytically, but if $V \gg 1$, the angle is approximately equal to $$ \frac{2}{V_\infty^2}\frac{1-(1-r^2)^{3/2}} {r}, $$ which indeed takes the maximal value at $r = \sqrt{\frac{\sqrt 3}2}$, as this answer describes.

Litho
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    But inside the planet, the inverse square law no longer holds, so why should the path be elliptical? – TonyK Sep 25 '22 at 10:29
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    @TonyK In the linear law, which holds inside the planet, the motion along each axis is harmonic with the same period. It means that all trajectories are ellipses with the center at the center of the planet or straight lines through the center. – Litho Sep 25 '22 at 10:40
  • @Litho just curious, what is "the linear law"? I think TonyK has a point, can it really still be a conic section if the force is not $r^{-2}$? – uhoh Sep 25 '22 at 11:15
  • @uhoh: Inside a spherical body of radius R, the gravitational force on a particle at distance r < R from the centre is the same as if the spherical body had radius r (Newton's shell theorem). It would then have mass proportional to r^3, so the gravitational force would be proportional to r^3/r^2 = r, i.e. linear. – TonyK Sep 25 '22 at 11:50
  • @TonyK ah, yes that sounds familiar, and I guess that's for uniform density. Hmm... this sounds like a conversation from a year ago somewhere else about Hooke's law orbits. I'll go look. Anyway if it's that then the motion in x, y and z will indeed be sinusoidal and I guess that gives an ellipse, although the speeds along the ellipse will not behave the same as for a central force elliptical orbit. – uhoh Sep 25 '22 at 11:55
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    @uhoh I meant that $F\propto r$. Fun fact: $F\propto r$ and $F\propto r^{-2}$ are the only two power laws where all the bounded trajectories are closed. They also both give conic sections for trajectories. – Litho Sep 25 '22 at 11:56
  • @Litho got it. Oh this sounds familiar. https://space.stackexchange.com/a/56072/12102 and https://space.stackexchange.com/a/58120/12102 and somewhere a discussion with David Hammen a year or three ago... – uhoh Sep 25 '22 at 11:59
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    I am persuaded! – TonyK Sep 25 '22 at 12:03
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Adding my own numerical time-step simulations here.

The effect seems to be quite subtle, only increasing the deflection angle by a fraction of a degree at shallow depth.

For a unit mass, with unit radius, the optimal depth (found by hill climbing) seems to be dependent on the $v_\infty$ (velocity at infinity)

At sensible velocities, the depth tapers off with the velocity and is pretty close to 1.

small scale

At much faster velocities, the curve eventually flatlines, approaching $\sqrt{\frac{\sqrt{3}}{2}} \approx 0.93060486$

large scale

But until anyone actually does the math there doesn't seem to be much more insights to get from simulations.

SE - stop firing the good guys
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