To the Sun!

No, dear website, I have not forgotten about you.

It has simply been a busy couple of weeks...

I am back in the classroom this year, teaching physics as an adjunct instructor for the University of Cincinnati. I have done this in the past, but not for these particular classes, and so it should be interesting. Last week saw me going to and from the University frequently in preparation for the upcoming semester, which began on August 21st. Do you know what else happened that day?

A total eclipse of the Sun, the path of which spanned the continental United States.

Image acquired from http://www.eclipse2017.org

I do not know who in Ohio's public university system decided that classes should start on that particular day, but they were clearly not engaged with popular astronomy or science in general. For if they were, then they would have known about this event. After all, astronomers have been accurately predicting eclipses for hundreds, if not thousands, of years.

Or perhaps they knew about it, but they just didn't think it would be a big deal. If so, they were wrong.

Therefore, my students' first assignment was to safely view the eclipse from wherever they happen to be (assuming they were under clear skies in North America), record multiple observations throughout the event and be ready to share their findings on Wednesday. Meanwhile, I went down to Dawson Springs, Kentucky, near the center of the path of totality, under clear skies to experience the greatest celestial spectacle.

The event was indeed life-changing, and no pictures, video, or written word can accurately describe what I saw. Nevertheless, I do have a lot of those and will be sharing them in the upcoming weeks.

But for now, I want to wrap up the thread I have been working on since the beginning of August. Surely those of you who follow my updates are aware of my blabberings about rockets and delta V, and perhaps you have wondered where I am going with this. Well, I think the best way to introduce the topic is by describing a question that someone asked me not long ago: what would I do if I knew that the world was ending tomorrow.

I thought for a moment and then replied, "I would try to find the most interesting way to die."

Note that I am not, by any means, saying that I want to die, at least not anytime soon. In fact, I can't die. The joke we told each other in graduate school, no matter how close to death we felt, was that you could not die until you finished all of your work, and I still have lots of work to do! (And lots of eclipses to see.) Still, there is no escaping the fact that everything alive will die someday, including me, and when that happens, I do not wish to fade away as a shadow of my former self. No—I want to go out in a blaze of glory at the top of my game! I want to go out in a way that is meaningful to me and my work and in a way which makes those around me jump and exclaim, "Holy s***, did you see that?!"

As a physicist, I can think of no better way to exit the Universe than by falling into a black hole. However, unfortunately, there are no black holes within present reach of Earth, meaning even though I can plot a course to one using present technology, I would be long-dead from heart failure, a stroke, or cancer, before even exiting our own Solar System. I must, therefore, seek realms within the Oort cloud, to serve as the stage of my last stand from which I take my final bow.

So then, why not the Sun?

Now we're talking! Image from NASA.

As a brief aside, one of my favorite facts about the Solar System is that one considers the distribution of its mass, around 99% of it is contained within the Sun. Then, most of the remaining 1% is in Jupiter. That means all of the other planets, and all of the moons, comets, and asteroids, and all of the dwarf planets and other minor planets, and all of the stuff crawling around on all of these things makes up a fraction of 1% of all of the "stuff" in the Solar System.

At this point, the common response I see from folks is that they declare, "Oh, we are so small and insignificant... woe are we!" to which I reply NONSENSE!!! Sure, space is big compared to us, but "judge me by my size, do you?!"

Well you should not, for I, and you, are sentient, and that makes us capable of wonderful things! To borrow a passage from An Atheist's Sermon,

That star of ours shining somewhere beyond the clouds…the one that’s been around for a couple billion years before you were born and will remain for millions of years after you die—does it dream? Does it hope? Does it love? Does it wake up every morning, face the world and try to make something of itself? Does it feel a single thing? ...Of course not.

Something that I have long wondered about, ever since I was a child, is what would it be like to fly into the Sun? Even back then I knew it would be hot, but since then I have learned the physics necessary to calculate exactly how long it would take to get there, assuming gravity is one's only motivating force:

DID YOU KNOW THAT: If starting from rest with respect to the "fixed" stars at Earth's average orbital radius, it will take just over 64 days to fall into the Sun? WELL, NOW YOU KNOW!!!

Posted by Aaron Eiben on Monday, April 24, 2017

In fact, as you approach the star, you might even be going fast enough to plow into it before you die of heat exhaustion, depending on how well your spacecraft protects you from that, as well as the ionizing radiation. The trouble is, setting up these initial conditions is actually quite hard if one's departure point is Earth!

Earth is in no danger, whatsoever, of falling into the Sun anytime soon, and we can thank our planet's angular momentum for that. As is the case in any orbit, Earth is moving so fast "off to the side" of the Sun, that by the time our star's gravity pulls our planet in along one axis, we are tens of millions of miles away long the other! What exactly do I mean by that? Hopefully, this animation will help make me clear:

Figure \((1)\): A parametric representation of Earth's orbital motion; distances measured in astronomical units (AU).

Figure \((1)\) shows the orbit of Earth broken down into a parametric plot of linear oscillations in two perpendicular directions. Each one of those oscillations passes through the coordinate origin, \((0,0)\), at which point lies the Sun. Therefore, Earth would be pulled in by the Sun's gravity and destroyed were it not for the fact that the planet's motion along one axis has carried it safely away by the time it reaches the center of the other line!

Since we are starting from planet Earth, the same fact applies to us. Thanks to momentum conservation, Earth's orbit is our orbit, and we will forever orbit the Sun until we change our momentum. But we must do that to get off of Earth's surface, anyway! In one of my first posts, I computed the escape velocity for an object on the surface of our planet and found it to be just over \(11\,\mathrm{km/s}\). That is pretty fast, but Earth's average orbital speed is just shy of \(30\,\mathrm{km/s}\); \(29.8\,\mathrm{km/s}\), to be a bit more precise. Therefore, even if we leave the Earth at escape velocity traveling in a direction that is exactly opposite that of Earth's orbit, we are still moving at a speed of about \(20\,\mathrm{km/s}\) tangentially with respect to the Sun, and our resulting orbit looks like this:

Figure \((2)\): The white dot represents the Sun (to scale!) The dashed arcs represent the orbits of Mercury, Venus, and Earth, respectively. Our starting point is marked by \(\times\). If we leave the planet, which is moving with velocity \(V_\oplus\), at escape velocity \(V_E\), then our net velocity with respect to the Sun is \(V_0 = V_\oplus - V_E\). That is enough to carry us around the Sun along the blue oval orbit, and so we have failed.

For our purpose, escape velocity is not fast enough! If we are going to fall directly into the Sun, then we must leave the Earth at its orbital speed of about \(30\,\mathrm{km/s}\) in a direction opposite the planet's motion. Only then will our velocity and Earth's velocity cancel, giving us a speed of zero with respect to our star. From there, gravity does the rest:

Figure \((3)\): Success!

So then, what on Earth can give us a velocity of \(30\,\mathrm{km/s}\)? Given today's technology, that would be a really big rocket, and so far, the biggest rocket that has ever flown is the mighty Saturn V:

A great white tower that ascends into heaven atop a pillar of smoke and fire as thunder shakes the Earth around it—where is your god, now?! Image Credit: NASA

It launched the Apollo missions to the Moon. It lifted America's first space station into orbit! It happens to be my own personal idol, but can it send me to the Sun? The answer is, unfortunately, no.

In the lead-up to this post, I calculated the airspeed velocity of an unladen Saturn V and found it to be just over \(16\,\mathrm{km/s}\). To completely cancel Earth's orbital motion, we need about twice that speed, and simply using two Saturn V's won't cut it.

Stacking one rocket on top of the other won't work because the whole thing will be too heavy to leave the launch pad. As I calculated last time, a fully fueled Saturn V is just able to lift \(1,246,780\,\mathrm{lb}\), and another fully fueled Saturn V weighs more than that. So even though it looks really cool, this rocket will not get off the ground. And it will probably collapse under its own weight and explode.

Burn, baby, burn!

My initial guess was, therefore, to strap two Saturn V's together and fire them simultaneously. Doing this would double the rocket's thrust, but I quickly realized that it would also double the vehicle's weight, and those two effects cancel each other out! This is easiest to see in the rocket equation:\begin{equation} \Delta v = I_{sp} \, g_0\ln \left(\frac{m_i}{m_f}\right). \label{082517TreIsp} \end{equation}Even though the initial and final masses both double, their ratio remains the same. Therefore, so too does the rocket's \(\Delta v\) and its final speed.

Congratulations; you have doubled the vehicle's cost and gained nothing. But at least it flies!

So that then made me wonder, what is the point of this:

Behold, the Delta IV Heavy, my favorite operational rocket. Image from Wikipedia.

Vehicles such as the United Launch Alliance Delta IV Heavy shown above as well as the upcoming SpaceX Falcon Heavy both use "strap-on" boosters that are identical copies of the first stage. Following my realization about the \(\Delta v\) of parallel rockets described above, this practice seemed, albeit counterintuitively, pointless. I mean, my simple calculation showed that the \(\Delta v\) of parallel rockets is no greater than that of a single rocket, but launch providers would not use parallel boosters if they did not provide some sort of benefit. Moreover, I know from experience in Kerbal Space Program that parallel boosters do provide an advantage! So I had to sit down and think, "what advantage would that be?"

Hence my post on the problem with payloads. Unless one is trying to test the rocket itself, they are never launched empty. After all, the purpose of a launch vehicle is to launch something into space! In rocket terminology, that something is called a payload, and as I demonstrated back then, adding a payload reduces a rocket's \(\Delta v\). Furthermore, as is the case with the stacked Saturn V, if the payload is too heavy, the rocket will not get off the ground! However, this is where the advantage of parallel boosters comes to play. Adding more boosters reduces the payload's impact on both the vehicle's \(\Delta v\) and its thrust-to-weight ratio, and this can be seen in the math.

Say we have a payload of mass \(m_{PL}\) and some number \(N\) of identical rockets, each with a wet mass of \(m_W\) and a dry mass of \(m_D\). In that case, the total initial mass of our vehicle is\begin{equation}m_i = N m_W + m_{PL}, \label{082517mi}\end{equation}and if we burn all of the rockets simultaneously, the vehicle's final mass will be \begin{equation}m_f = N m_D + m_{PL}. \label{082517mf}\end{equation}The ratio of the masses is therefore \begin{equation}\frac{m_i}{m_f} = \frac{N m_W + m_{PL}}{N m_D + m_{PL}} = \frac{m_W + m_{PL}/N}{m_D + m_{PL}/N}. \label{082517mr}\end{equation}In the last part of equation \eqref{082517mr}, I factored and canceled the contant \(N\) from the ratio's numerator and denominator, or, equivalently, multiplied both parts by \(1/N\). Doing this makes it easier to see what happens as \(N\) increases:\begin{equation}
\lim_{N \to \infty} \frac{m_{PL}}{N} = 0 \quad \Longrightarrow \quad \lim_{N \to \infty} \frac{m_W + m_{PL}/N}{m_D + m_{PL}/N} = \frac{m_W}{m_D}. \label{082517lim1}
\end{equation}But that's the mass fraction of an unladen rocket, meaning we can completely nullify the deletrious effects of our payload upon our \(\Delta v\) simply by strapping together an infinite number of boosters.

It's just that easy!

This design can also overcome our rocket's thrust-to-weight problem. If an unladen rocket on the launch pad weighs \(W_R\) and its engines at full throttle produce thrust \(T\), then its thrust-to-weight ratio is \begin{equation} \frac{T}{W_R}. \label{082517tw} \end{equation} As I described in the past if we want this rocket to fly as soon as we light its engines, then its thrust-to-weight ratio cannot be less than \(1\). Otherwise, the engines will burn while the rocket remains stationary on the ground. Adding a payload with weight \(W_{PL}\) does nothing to solve that conundrum:\begin{equation} \frac{T}{W_R + W_{PL}}. \label{082517twpl} \end{equation}In fact, if the payload weighs anything at all (\(W_{PL} > 0\)), the ratio just becomes smaller. On the other hand, just like before, strapping together \(N\) parallel boosters increases our vehicle's weight to \(N W_R + W_{PL}\) whilst simultaneously increasing its thrust to \(N T\) (assuming all engines are lit). In that case, the thrust-to-weight ratio becomes\begin{equation} \frac{N T}{N W_R + W_{PL}} = \frac{T}{W_R + W_{PL}/N}, \label{082517Ntwpl} \end{equation}after invoking the same arithmetic we used in the mass ratio above. What is more, just like before, as we increase the number of rockets,\begin{equation} \lim_{N \to \infty} \frac{T}{W_R + W_{PL}/N} = \frac{T}{W_R}. \label{082517Ntw} \end{equation}Note that after taking the limit, equation \eqref{082517Ntw} contains no reference to the weight of the payload. Therefore, it does not matter how much the thing weighs. We can always get it off the ground by adding more rockets!

So then, what is our solution? We are going to need a spacecraft with at least \(14\,\mathrm{km/s}\) \(\Delta v\) all on its own, but its weight is not an issue since we can always add more boosters. Still, I would aim for something with a high specific impulse \(I_{sp}\) to achieve that speed with the least amount of fuel possible, and I would also prefer an engine with a good thrust-to-weight ratio, so that I may get up to speed in the least amount of time possible. Therefore, according to the following chart, it looks like nuclear fission is the way to go.

Image from Wikipedia.

So now we have a nuclear-powered spacecraft with \(14\,\mathrm{km/s}\) \(\Delta v\) and enough consumables to keep one person alive for 65 days. Even though it's weight is not important, I think we can manage to keep it under the maximum lifting capacity of a single Saturn V, which again is just under \(1.3\) million pounds. (Of course, if our vehicle weighs more than an Apollo spacecraft at \(116,000\,\mathrm{lb}\), we will probably have to add more struts to the booster underneath to prevent the stack from collapsing in on itself—a well-known fact to students of KSP).

Then, take the spacecraft and place it atop a Saturn V—strut as needed. To recover our original \(\Delta v\) and thrust-to-weight ratio, we shall have to add more boosters. Many more boosters. In fact, according to the following chart, I recommend using no fewer than 500 parallel rockets:

Figure \((4\)): We want both lines to converge to zero. And don't worry—500 is much smaller than infinity.

Therefore, I call this launch vehicle the Saturn D (that's 500 in Roman numerals).

Having assembled our rocket, we must now figure out the right trajectory to nullify our initial angular momentum with respect to the Sun, that trajectory once again being directly opposite Earth's direction of motion. A moment's reflection tells me that one solution is to launch from the point on Earth's equator where the apparent solar time is 1800 hours (6 PM) at the moment of a solstice and go straight up. Hopefully, the following diagrams will illustrate why:

When viewed from high above the northern hemisphere, Earth is seen to both rotate and orbit the Sun counter-clockwise. Image from Wikipedia.

The two images above depict situations in which the solstices occur exactly at solar noon on the prime meridian, which is not always the case, meaning we will not always launch from Indonesia. What is true is that in both diagrams, Earth's orbital motion is directly into the page, and so we must therefore launch directly out of it! The apparent solar time at that point, no matter how Earth might be rotated, is 6 PM.

And I just realized something else: Earth is spinning, and that contributes to our initial velocity! We must compensate for that to achieve zero motion relative to the Sun, and at our planet's equator, this rotation carries us eastward "at a speed of 460 meters per second—or roughly 1,000 miles per hour." Thus, we shall have to adjust our launch angle westward and sacrifice a little less than \(0.5\,\mathrm{km/s}\) of our precious \(\Delta v\). Fortunately, I feel that is a small task for the mighty Saturn D.

Finally, the stage is set, and all of the details have been worked out, so let us proceed with a description of how this experiment would unwind.

At the end of a long and prosperous life, I, Aaron Gustav Parker Eiben, embark on one final mission: Project Icarus. It's late December, just before the solstice, and somewhere on Earth's equator, I board my nuclear spacecraft, Apollo's Chariot, which sits atop a glistening Saturn D. As the Sun sets, at 5:05 PM apparent solar time, hundreds of loudspeakers begin blaring Europe's "The Final Countdown," only to soon be drowned out by the Earth-cracking roar of no fewer than 2,500 F-1 rocket engines. The sound is incredible. It shatters windows, triggers landslides, and collapses unstable structures for hundreds, if not thousands, of miles. It probably also liquefies the organs of anyone who happens to be nearby, and eventually, it will be heard everywhere around the world, possibly more than once. At 6 PM apparent solar time, the launch clamps, which have been anchored deep into the subterranean bedrock, release and the Saturn D rises triumphantly from its pad. As the vehicle clears its launch towers, it gracefully pitches westward and climbs higher, and higher, and higher into the sky.

SO LONG, JERKS!! (Yes, Apollo's Chariot is another Saturn V, but it has nuclear engines.)

Though the Sun is no longer visible from the pad, having set at the moment of launch (or shortly thereafter due to atmospheric refraction), the rocket remains in complete illumination as it climbs, standing out as an incredibly bright point of light as it eventually disappears over the western horizon. Stages are dropped as their fuel is expended and at some point they all rain back down to Earth. (Don't worry, most of the equator is ocean, and so they probably do not land on anyone's house.) Meanwhile, I continue my ascent. When the launch is finally complete, all that is left of the spacecraft is me and my habitation module, speeding away from Earth at just under \(30\,\mathrm{km/s}\); \(0\,\mathrm{km/s}\) with respect to the Sun!

From there, gravity does the rest, and \(64.6\) days later, I find myself screaming headfirst into our star, traveling at a speed of just over \(616\,\mathrm{km/s}\)—about \(0.2\%\) the speed of light.*

THAT would be an interesting way to die.

Thanks for reading all the way through! I hope you enjoyed this, and I will see you next time.

—Aaron

*When I calculated the fall time to the Sun, I used Newtonian mechanics because I do not yet know enough general relativity. Therefore both it and my impact speed are probably a bit off and will be refined eventually.

P.S. Hitting the Sun is hard. Really hard:

P.P.S. After developing the concept of the Saturn D, I considered reaching out to my favorite YouTube rocket surgeon, Scott Manley to see if he would take on the challenge of building the Kerbal equivalent of this mighty launch vehicle for a direct-ascent mission to Kerbol. As far as I know, he has not completed such a task exactly, though he has taken the easy route: