# Orbital mechanics from Daniel Suarez's book “Delta-V”

In the book Delta-V, a climactic scene involves a high-acceleration return from the orbit of Ryugu to Earth. The book makes a big deal about the delta-v needed for the trajectory (hence the name) so I was curious how realistic the scenario is. Here's what gets presented:

• They are initially in Ryugu's orbit (Aphelion 1.4159 AU, Perihelion 0.9633 AU).
• They are delayed and depart a few weeks after the ideal transfer orbit.
• As a result, 24.42 km/s of delta-v are needed to intersect Earth's orbit.
• They reach Earth going 26.8 km/s.
• They use aerobreaking to shed 14 km/s and enter an elliptical orbit around Earth.

I would normally trust Suarez to use real orbital mechanics in his books. However, I was a bit suspicious that the ∆v to change orbit and the velocity at Earth were so close. Is this correct, or did Suarez fail to correct for gravitational acceleration during the journey?

Edit. I found some more information about the orbits which might provide info about the position of Ryugu relative to earth for this.

• Their outward trajectory from lunar orbit to Ryugu takes 48 days and requires ∆v = 1.7 km/s plus a postinjection burn of 1.9 km/s. It is dated Dec 13, 2033 to Jan 30, 2034.
• Closest approach for the return orbit occurs in May 2038 at 5.5e7 km
• Return was scheduled for Feb 10, 2038, but they actually left Feb 19.
• Rendezvous occurred around Mar 31, 2038
• A slow trajectory was possible Oct 2, 2036 arriving Jan 2038 (perhaps using gravity assists) using ∆v = 706 m/s

The more I look back at the hard numbers in this book the more I'm inclined to trust Suarez's calculations. I'd still be interested in the math behind them though.

• If they are operating under time constraints, instead of optimal transfer which would give only a couple km/s over the inevitable 11.2, they will burn extra for a shorter, more straight-line transfer and need to shed roughly as much at the destination, on top of the 'mandatory' parts. – SF. Dec 2 '20 at 0:03
• The afterward mentions he used KSP realism and SpaceEngine to simulate the trajectories, so I guess the velocities are correct. I'd still like to know more about the orbital parameters though. – Quantum7 Dec 8 '20 at 22:15

Gravitational acceleration works out to accelerating an object that starts out at a relative halt up to 11.2 km/s when it impacts Earth, because that's what the potential energy difference works out to when converted to kinetic energy. If you do the addition in terms of energy:

$$\sqrt{(11.19 km/s)^2 + (24.42 km/s)^2} = 26.86 km/s$$

Since kinetic energy scales with the square of velocity, if one of the component velocities is double the other, its contribution to the total energy is 4 times greater. The end result is going to be close to the greater of the component velocities (in the worst case about 1.4 times greater if they're equal).

There is an issue here, but it isn't Earth's gravitational acceleration, it's that the velocity relative to Earth at a distance is 24.42 km/s. If that's the delta-v they applied (not familiar with the book), it implies Ryugu was somehow at rest with respect to Earth and at the same distance from the sun, or the vectors and change within the sun's gravity well coincidentally added up in just the right way that their delta-v ended up equaling their earth-relative velocity in the end.

• Since it's a reasonably near earth asteroid it would be moving at a velocity similar to Earth. I have no problem with the arrival speed (and your calculations show that he got that right--my gut said it was reasonable, I didn't actually do the math) but just about everything else about that flight doesn't work. – Loren Pechtel Dec 2 '20 at 5:05
• @LorenPechtel its orbit extends out past that of Mars, and its orbital velocity varies from 22 km/s to 33 km/s, in directions that diverge significantly from a circular orbit. – Christopher James Huff Dec 2 '20 at 12:13
• As I said, reasonably near Earth. We don't know the path they are taking returning to Earth, I'll give him the benefit of the doubt on that. I listed a bunch of other things I find to be far more of an issue. – Loren Pechtel Dec 3 '20 at 21:23
• Can you give a bit more detail about how you got 11.2 km/s? That's the two-body potential energy due to attraction to earth (ignoring the sun)? What initial distance did you use? – Quantum7 Dec 8 '20 at 22:13
• Yes, that's from the fall into Earth's gravity well. I didn't use any particular initial distance, the spacecraft would have started out far enough that "infinity" is a very good approximation. – Christopher James Huff Dec 8 '20 at 22:33

I haven't worked out just how flawed his orbits are but they don't work.

1. 24 km/sec is an awful lot of rocket, especially considering the payload they're carrying. Note that this is way beyond the Apollo/Saturn V stack which I believe is the biggest ever built using chemical engines. (The total delta-v of the Dawn mission was higher but a big chunk of it came from ion engines.)

2. I can remotely conceive of bringing the second engine along for safety but there was absolutely no reason to burn both. They have plenty of time, they aren't taking advantage of a gravity assist, there is absolutely nothing gained by piling on the acceleration other than having to make all your support structures stronger and thus reducing the performance of your rocket.

3. The approach to Earth is not equatorial as evidenced by what parts of Earth they flew over. They endured a brutal aerocapture maneuver, even if their craft had enough lifting surface to do a plane change during the aerocapture it would have been even harder on the crew. How did they match orbital planes with the station that rescued them? Without coming pretty close to an orbital plane match there's no way they could have come over--rescue would have come from Earth, not a space station.

4. They had a tough hull but no heat shield. How did they survive an aerocapture far hotter than Apollo? The heat shield is not only to keep your hull from melting but to keep from baking your crew.

5. What in the world did they use to circularize after the aerocapture? You can't do that with cold gas and they wouldn't have had anything else after that bath of fire.

• Thanks for the careful read of the book! These are interesting additional points to look into. 1. I don't think specific impulse was given, but acceleration didn't become high until most of the fuel burned. 2. I think you're right. Better low acceleration for longer. 3. This is glossed over. Maybe the rescue came from a lunar cycler orbit? 4. This was waved away by claiming the special 3D-printed steel alloy was heat resistant, in combination with vacuum for eliminating convection. 5. I think they mentioned a second gentler aerobreak to reduce apagee. Maybe rendezvous on the second orbit? – Quantum7 Dec 8 '20 at 21:54
• @Quantum7 They didn't give an ISP but it couldn't have been better than hydrolox. There's no reason for the acceleration to ever become high. And you can't aerocapture purely with the atmosphere--orbital mechanics 101: Your orbit intersects the point where you completed your last maneuver. Their last maneuver was aerobraking--thus their periapsis is in the atmosphere. Additional aerobraking can't change that, only a rocket or a gravity maneuver off a moon can. They had neither. – Loren Pechtel Dec 9 '20 at 2:36
• I checked the book again. It mentions a second aerobrake on orbit 2 followed by thrusters to circularize into a orbit with "500 km, 38-degree inclination" despite having "very little gas". The plane change is not mentioned but I guess it could have happened during second aerobrake? He probably took some liberties here. – Quantum7 Dec 9 '20 at 12:42
• @Quantum7 I don't care about a second aerobrake, I'm just saying that there's no way for an aerobrake to put you in orbit. That requires a rocket--and there's no way they had a functional rocket at that point. I think he took a lot of liberties with that voyage. – Loren Pechtel Dec 10 '20 at 3:48