On The Fuel Efficiency of Launching My Enemies Into The Sun

Published On: 6/13/2023

Have you ever been minding your own business, plotting to launch your enemies into the sun, and then BAM you come across this tweet:

I'm just going to assume yes, you have had this exact experience.

Let's talk about it:

Orbital mechanics are fun.^{[1]Well, ya know, if you're a nerd.} Comparatively little is going on, but the math gets real tricky real fast. There's only one force, gravity, and it's described by Newton's Law of Gravitation:^{[2]Real nerds use Einstein's General Relativity.}

$$F_{Gravity} = {GM_1M_2 \over r^2}$$

where \(M_1\) and \(M_2\) are the masses of the orbiting objects, \(r\) is the distance between them, and \(G\) is the gravitational constant. Gravity is a conservative force, so the total energy of an orbiting satellite doesn't change. Specifically^{[3]That's some wordplay that will make sense in about three words.} we use specific mechanical energy:

$$\varepsilon = \frac{v^2}{2} - \frac{\mu}{r}$$$$=-\frac{\mu}{2a}$$

where \(v\) is the satellite's velocity, \(a\) is the orbit's semi-major axis, and \(\mu\) is \(G(M_1+M_2)\). Also, because the gravitational force always points directly inward, specific angular momentum is conserved:

$$\vec{h}=\vec{r} \times \vec{v}$$

It takes a little calculus, but one can prove using Newton's Law and these conservation laws^{[4]Newton's Law can be used to prove the conservation laws, so it's really all you need.} that satellites move in conic sections around the body that they are orbiting. There are four conic sections, but only two that we really care about: circles and parabolas. By plugging in a circle's radius into both parts of the specific mechanical energy equation, the orbital velocity of a circular orbit is found:

$$v_{circle}=\sqrt{\frac{\mu}{r}}$$

The parabolic orbit is one in which the kinetic energy \(\frac{v^2}{2}\) is exactly equal to the gravitational potential energy \(\frac{\mu}{r}\), meaning our satellite has just enough speed to fly away from the sun forever. When Dr. Buckley talks about planning to "launch them out of the Solar System instead," this is what he means. Now, by setting the two aforementioned energies equal, we get what that speed is:

$$v_{escape}=\sqrt{\frac{2\mu}{r}}$$

There's one last piece we need to talk about to help make this make sense.^{[5]Okay, really a lot more pieces, but I'm trying to not write an entire novel to solve a silly problem.} Delta-v, often stylized \(\Delta v\), is a measure of how much faster or slower we make a spaceship go by running its engines for a period of time if we ignore all other forces. It's often a better metric to use than the total change in energy of a satellite during a burn. Partially because delta-v is a proxy for how much fuel, i.e. chemical energy, is used. And partially because a timed burn will result in the same delta-v at all points in the orbit, but will have drastically different impacts on the energy.^{[6]Foreshadowing.}

So now if we want to prove the validity of Dr. Buckley's statement, we just need to calculate the delta-v it would take to launch someone into the sun and compare that against the delta-v needed to launch them out of the Solar System. In both cases I'm going to assume our satellite is already flying in an orbit that matches Earth's, but is not orbiting Earth itself.^{[7]I can make this assumption because Earth's rotational speed would help us launch out of the Solar System but make it more difficult to launch into the sun. So if I didn't ignore this rotation, it would just make the sun launch comparatively even less efficient.} An object orbiting along Earth's orbital path will have Earth's orbital velocity:

$$v_{Earth}=\sqrt{\frac{\mu}{r_{Earth\ Orbit}}}$$

To launch it out the Solar System, we need to get it up to the escape velocity we talked about earlier:

$$v_{escape}=\sqrt{\frac{2\mu}{r_{Earth\ Orbit}}}$$

So we need to be going \(v_{escape}\) and we're currently going \(v_{Earth}\).^{[8]Fun side note about escape velocity: it does not care about direction. As long as you're not going to crash into something, you will get away from whatever you're flying away from. Remember kids, you can always get away from your problems, you just need to be moving really, really, really fast.} That makes delta-v pretty easy to calculate, we just need to subtract one from the other:

$$\Delta v_{escape} = v_{escape} - v_{Earth} $$$$= \sqrt{\frac{2\mu}{r_{Earth\ Orbit}}} - \sqrt{\frac{\mu}{r_{Earth\ Orbit}}} $$$$= [\sqrt{2} - 1] \sqrt{\frac{\mu}{r_{Earth\ Orbit}}} $$$$\approx 0.414 \sqrt{\frac{\mu}{r_{Earth\ Orbit}}}$$

Calculating the delta-v needed to launch someone into the sun is easier mathematically but tougher conceptually because it actually requires the counterintuitive approach of firing your rockets retrograde until your orbital velocity becomes zero. Since your velocity is now zero, the sun is going to pull you straight down until you become a toasty, concerningly fleshy marshmellow. To truly grok why this is more efficient than firing your rockets directly at the sun requires spending way too long thinking about orbital mechanics^{[9]Thank you, Kerbal Space Program} but a good way to think about it is that the only reason you're not falling into the sun right now is because the Earth is moving as fast as it is being pulled downward. You don't need to boost towards the sun, you just need the Earth to stop.

The delta-v needed to stop moving entirely is simply the velocity we're going at right now. We're going {really fast}, we need to be going zero, so we need to get rid of {really fast}. Therefore our delta-v is just

$$\Delta v_{Sun\ Launch}=\sqrt{\frac{\mu}{r_{Earth\ Orbit}}}$$

There are some neat ideas that come out of this. Notably, if we wanted to launch someone into the sun, it would actually be easier to do the farther we started from the sun.^{[10]Those damn Plutonians have it so easy.} But, other than the coefficient out front, the two delta-v equations are identical. And from that we can quickly see that no matter which planet you start from, it will always only take 41.4% of the sun launch delta-v in order to launch someone out of the Solar System.

Drat, my revenge plans foiled by those meddling orbital mechanics....

But wait, there was something interesting in there about starting farther from the sun.^{[11]And I haven't resolved my foreshadowing yet.} What if we go out to one of the farther planets and then come back in towards the sun?

Let's go back to the equation for specific mechanical energy \(\varepsilon = \frac{v^2}{2} - \frac{\mu}{r}\). Because \(\varepsilon\) stays constant (at least while our rockets are quiet), our orbit is a constant tradeoff between velocity and distance from the sun. As one goes up, the other must go down. And remember what I said earlier about parabolic orbits, our velocity is just high enough to get us away from the sun. This means at the "top" of a parabolic orbit, we have zero velocity because all our kinetic energy has been converted into potential energy. Now a parabolic orbit doesn't have a top, but how 'bout instead of a fully parabolic orbit we stop dead short of it in an incredibly eccentric orbit. At the top of this giant oval we would have almost, but not quite, zero velocity because our orbit is almost, but not quite, parabolic. But remember, to launch ourselves into the sun requires us to get to zero velocity. So to launch ourselves into the sun from the highest point in our highly eccentric orbit requires almost no delta-v beyond what it took to get us to this high point in the first place.

In fact, we can make this delta-v arbitrarily small by getting closer and closer to a perfectly parabolic orbit. And this parabolic orbit was our goal when we were trying to launch our enemies out of the Solar System. So, if we plan our maneuvers right, it takes the same amount of delta-v to shoot someone into the sun as it takes to launch them out of the Solar System. In an attempt to get my name on a Wikipedia page, I am shamelessly calling this maneuver the Gier Transfer.^{[12]Though, to be fair, my transfer is just a special case of the bi-elliptic transfer.}

This counterintuitive result stems from an idea I touched on earlier, that a controlled burn will always add the same amount of delta-v, but will have different impacts on the specific mechanical energy depending on where you are in your orbit. The discrepency comes about because velocity is squared in our specific mechanical energy equation, so the same delta-v/fuel/chemical energy use translates to a greater change in mechanical energy the faster we're moving.^{[13]This is the same idea as the Oberth Effect, which states that it is more efficient to speed up when you're already going fast.} This also means that we generally get a larger change in mechanical energy by accelerating instead of decelerating, because we immediately get to take advantage of our higher speed. By accelerating in our first maneuver and letting gravity do the deceleration as we fly out toward the top of our orbit, we use our delta-v in a much more efficient way.

Needing less delta-v to accomplish the same end result feels like we must be getting something for nothing. But by plugging our orbital radii into the equation for \(\varepsilon\), we actually see that it takes the same change in energy to launch someone out of the solar system as it takes to launch them into the sun if you start from a circular orbit:

$$|\Delta \varepsilon_{out}|=|-\frac{\mu}{\infty}-[-\frac{\mu}{2r}]|$$$$=\frac{\mu}{2r}$$$$|\Delta \varepsilon_{in}|=|-\frac{\mu}{r}-[-\frac{\mu}{2 r}]|$$$$=\frac{\mu}{2 r}$$

Launching someone directly into the sun is just an inefficient way to do it. Our maneuver takes advantage of that weird chemical versus mechanical energy discrepency to instead launch someone into the sun as efficiently as possible. We're metaphorically^{[14]and technically literally} raising our miles per gallon.

There is a caveat here however, it will take much longer to do the efficient sun launch. Not only are we falling into the sun from a farther distance, we're also taking forever to get out there! If you want to see your enemies destroyed in your lifetime this might be where you want to focus your attention.

Quick set up: a little bit before Newton claimed only geniuses get hit on the head by apples, Kepler came up with his own laws. The third one states that the square of an orbital time period is proportional to the cube of it's semi-major axis, or, in a readable form:

$$T=\frac{2\pi}{\sqrt{\mu}}a^{\frac{3}{2}}$$

With a little bit of elbow grease we should be able to calculate the time it takes to fly out to progressively higher orbits for the purpose of launching our enemies into the sun and compare that against the efficiency of our delta-v usage.

To give just enough detail here that someone can check my work if they were so inclined, \(a\) going out is Earth's orbital radius plus our highest height divided by two, and \(a\) coming back is simply our highest height^{[15]I realize it has a name but I've spent this long not calling it the aphelion, I don't intend to start now.} divided by two. Note that our orbital time period would be half of what Kepler would give us because we only achieve a half orbit each time. Our delta-v can be calculated using the math of Hohmann transfers:

$$\Delta v_{NQP}=\sqrt{\frac{2\mu r_{T}}{r_{S}(r_{S}+r_{T})}} - v_{Earth}$$$$\Delta v_{Into\ Sun}=\sqrt{2\varepsilon_{NQP}+\frac{2\mu}{r_{T}}}$$$$=\sqrt{-\frac{2\mu}{r_{S}+r_{T}}+\frac{2\mu}{r_{T}}}$$$$=\sqrt{\frac{2\mu r_{S}}{r_{T}(r_{S}+r_{T})}}$$

where \(r_S\) is our starting orbital radius (i.e. Earth's orbital radius), \(r_T\) is our top most point, NQP is "Not Quite Parabolic," and the "Into Sun" delta-v is calculated by solving for the velocity at the top of our extremely eccentric orbit.

Plugging this all into your programming language of choice results in this neat looking graph:

You can see that we get quite an increase in efficiency for not too much extra time. If you're willing to wait somewhere 10ish years, you'll half the amount of fuel you need. But Dr. Buckley laid down the gauntlet, so if you want your delta-v use to be within a couple of percent of a simple escape orbit, then you're going to need to wait a while. Even stopping at Pluto results in a wait time of over one hundred years, so you better have invested in some cryotech if you want to be around to watch.

While technically it will always be just slightly more efficient to launch your enemies out of the Solar System, I think that fraction of a meter per second of delta-v is really not that much of an issue, especially compared against the numerous other inefficiencies associated with space travel. What you really need to be worried about is how you're going to live long enough to see your sweet revenge be enacted. Because really, if you take the right orbital path, launching your enemies into the sun becomes not a question of fuel economy, but one of time.

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