How Long Would It Take To Get To The Moon?

“Dad?” my son asked while we were playing with his Legos. “How long would it take to get to the moon?”

“I think that depends on how fast you’re going,” I replied.

“No,” he says, sounding exasperated as only a 6-year-old can, “I mean, if you were going as fast as the Death Star!” Because that was entirely clear from the context, right?

“I don’t know,” I tell him. “I don’t know how fast the Death Star is.”

“It’s really fast,” he assures me.

Where to start?

There are a couple of things we need to know here, in order to answer the question. How far away is the moon? How fast do we have to go at minimum to make it? Oh, and how fast is the Death Star? So, let’s dig in.

How far is it to the moon?

The distance from the Earth to the Moon varies based on the time of the month, because the Moon orbits us in an ellipse – so it gets closer and then moves further away. At apogee (the farthest it gets from us), it’s 405,400 km away, while it gets as close as 362,600 km at perigee. So, clearly, how long it takes will really depend on how fast we’re going – just like any other trip we can take.

How fast do we need to go?

How fast you need to go to get to the moon will depend on the method you’re using to get there, and the amount of time you want to take. So, let’s start with the concept of escape velocity. This is the minimum speed required to “out-pull” gravity and leave an object behind. If you launch at that speed or greater, you fly away. If you don’t, you fall back to the surface. Eventually. Escape velocity varies with the gravity of the object and is approximately 11.2 km/s, or 40,320 kph on Earth. Assuming there is no friction, which is a popular physics assumption to keep equations simple. If you launch at that speed, you fly away from the earth – you slow down over time, as Earth’s gravity pulls on you, but you never actually stop moving. Ever.

There’s a down side to trying to get to the moon by launching at escape velocity (say, by using a variant of Project HARP’s big gun): Earth’s force of gravity is 9.807 m/s2, so you’re pulling around 1,142 gravities at the instant of launch. You would be a thin, wide smear on your pilot’s chair well before you reached the moon.

Clearly, we didn’t send a gelatinized melange of Neil Armstrong, Michael Collins and Edwin Aldrin to the moon on Apollo 11 – those three men made it to the moon and back with bones and organs intact, after all. So, how did they do it? Well, the important thing to remember is that escape velocity is only needed if you have an initial push and then add no additional thrust after that. This isn’t how the Saturn V – or any other rocket for that matter – works. They lift themselves at a slower pace, but apply a constant (or near-constant) thrust by carrying fuel. There’s a point of diminishing returns on this, because you have to lift your fuel as well as the ship (something described in the Tsiolkosky rocket equation, which I discussed when I tried to describe how to make a house fly).

The Saturn V was a multi-stage rocket, with the first stage burning for 2 minutes 41 seconds and pushing the rocket about 68 km into the air (hitting a velocity of 2,756 meters per second). Then it ditched the first stage and started the second stage burn. This pushed it another 107 km (for a total of 175 km) into the air over the course of 6 minutes, reaching a velocity of 6,995 meters per second). Stage 3 burned for about 2 minutes 30 seconds, reaching a velocity of 7,793 meters per second and putting it in orbit at an altitude of 191.1 km. Stage 4 burned for six minutes, pushing the ship to a velocity of 10,800 meters per second once it was time to head for the moon.

So, how long would it take?

How fast are you going?

Let’s say you just boosted off Earth with a canon, firing you straight up at escape velocity. Let’s also say you timed things so that you’d intersect with the moon at perigee. That’s 362,600 km, or 362,600,000 meters. At 11.2 meters per second, that’s 32,375,000 seconds to reach the moon. This translates into 8,993 days, or 24 years, 7 and one half months. Approximately. Your gelatanized corpse has a long trip ahead.

Apollo 11 was moving at 10.8 kilometers per second, which (mathematically) means you’d expect the trip to the moon to take 33,574.07 seconds. In theory, this means 9.326 hours. It actually took three days. Why? Well, there’s two reasons and they’re both gravity. See, the Apollo 11 wasn’t maintaining constant thrust. It had fuel that it used for course corrections and orbital insertions and the like, but it coasted most of the way. Earth’s gravity pulled on the ship the whole time, slowing it down. In addition, the ship didn’t fly in a straight line. It was in a long, figure-eight-shaped orbit with the Earth and the Moon – like so:

But what about the Death Star?

Ah, yes. That. Well, it still depends on the speed the ship can manage.

How fast is the Death star?

This is… questionable. According to the DS-1 Orbital Battle Station entry on Wookieepedia, the Death star had a speed of 10 megalight (MGLT).

So, what’s a megalight? Well, also according to Wookeepedia, a megalight “was a standard unit of distance in space”. Which is entirely unhelpful, although it does indicate that when it was used in the Star Wars: X-Wing Alliance instruction manual, it appeared to be a unit of distance and that when used as speed it should imply “megalights per hour”.

In all likelihood, “megalight” is a word that got made up because it sounded cool and had no actual meaning attached to it. But if we try to break it down, “mega” as a metric prefix means million. So, one megalight could be a million light seconds. However, this would mean that the Death star flies at 10 million light seconds per hour, or 2,777.7 times the speed of light – meaning that it could reach Alpha Centauri from earth in less than 14 hours of cruising on its “sublight” drives.  So I’m going to assume that this is not what was intended.

The Star Wars Technical Commentaries on speculate in “Standard Units” on what MGLT means in terms of real world [i]anything[/i]. The author of the article comes to the conclusion that 1 MGLT is “at least 400 m/s2” acceleration, which is roughly 40 gravities of acceleration.

One thing we also know about ships in Star Wars is that constant acceleration isn’t an issue – they have something close to the “massless, infinite fuel” I mentioned above. The Death Star isn’t fast, compared to the other ships in Star Wars, but it can accellerate at a constant 4 kilometers per second. Now Dummies.dom provides us with a simple formula for determining the distance (s) covered for a given time (t) at a particular acceleration (a), and that formula is s = 0.5at2. Which means we can reverse engineer, because all we need is the time. The equation looks like this:

362,600 = 0.5(4)t2
362,600 = 2t2
181,300 = t2
t = square root of 181,300 = 425.7933771208754 seconds

So, assuming that the Death Star didn’t engage it’s hyperdrive, it would take a little over 7 minutes to reach the Moon at a velocity of approximately 1,703.17 kilometers per second. And it would keep going, because it can only slow down at 4 kilometers per second. So, if the Death Star wanted to stop at the Moon, it would need to slow down about halfway there (yes, I know that orbital mechanics are a little more complex than this, but we’re talking about a 160 kilometer diameter ship that can accelerate at 4 kilometers per second. So cut me some slack, would you?). That it would have to accelerate to halfway to the moon, and then decelerate the rest of the way. So, that would look something like this:

2(181,300 = 0.5(4)t2)
2(181,300 = 2t2)
2(90,650 = t2)
2(t = square root of 90,650 = 301.0813843464919)
t = 602.1627686929838 seconds, or slightly over 10 minutes.

“All your tides are belong to us, now.”

Why Does The Sun Move So Slow?


I really wish I could remember what prompted this questions. I suspect it had to do with one of our conversations about time, and how “day” is from sunrise to sunset. There was probably something we were doing in the evening, something he was looking forward to doing, and the sun seemed to be just crawling through the sky. Whatever the reason, the question certainly seems to make sense. The sun takes all day to cross the sky, so it looks slow. It comes up gradually, takes four to six hours to reach noon, and then slowly sinks into the west.

Of course, appearances are deceiving.

How Fast Does The Sun Move Across The Sky?

This is actually sort of tricky, because the Sun isn’t actually moving around the Earth. To begin with, let’s refer back to the technical definition of sunrise and sunset:

Sunrise and sunset. For computational purposes, sunrise or sunset is defined to occur when the geometric zenith distance of center of the Sun is 90.8333 degrees. That is, the center of the Sun is geometrically 50 arcminutes below a horizontal plane. For an observer at sea level with a level, unobstructed horizon, under average atmospheric conditions, the upper limb of the Sun will then appear to be tangent to the horizon. The 50-arcminute geometric depression of the Sun’s center used for the computations is obtained by adding the average apparent radius of the Sun (16 arcminutes) to the average amount of atmospheric refraction at the horizon (34 arcminutes).

Now, an arcminute is 1/60th of a degree, so the day begins when the center of the sun is 5/6ths of a degree below the eastern horizon and ends when the center of the sun is 5/6ths of a degree below the western horizon. Assuming perfect viewing conditions, et cetera, et cetera. So, that means that the sun has to cover 181 2/3 degrees in a single day. Here in Cincinnati, sunrise on the day this article is published is(August 19, 2016) is 6:55 am, and sunset is 8:27 pm. So, it will require 13 hours and 33 minutes (813 minutes) to cover that distance. That works out to, let’s see… 181.6666 / 813 = 0.22345 degrees per minute, or 13.407 degrees per hour.

Now, let’s do some more math. Cincinnati is at 39.1031 degrees north. According to Ask Dr. Math, you “just multiply the equatorial circumference by the cosine of the latitude, and you will have the circumference at that latitude.” The equatorial circumference of the earth is 24,901 miles, so 24,901 * cos(39.1031) = 19323.48 miles. The Sun crosses 181.666/360 = 0.5046 of that distance in 813 minutes, so that’s 9751.197 miles in 813 minutes. That’s essentially 12 miles a minute or 719.65 miles an hour.

But that’s the speed at which the sun passes over the Earth at that latitude on August 19. It’s not how fast it appears to an observer on the ground! To an observer standing on the surface of the Earth, the distance to the horizon is approximately 2.9 miles. That means that my son, observing the motion of the Sun, is standing at the center of a perceptual circle with a circumference of 2(π)2.9 = 18.22 miles. Which means that he sees the Sun appear to take 813 minutes to cover 18.22 * 0.5046 = 9.19 miles. That works out to a perceived speed of 0.0113 miles per minute, or 0.678 miles an hour.

No wonder it looks so slow to him. When we’re out on walks, my son and I hit a pace almost four and a half times faster!

How Fast Does The Sun Move Through The Galaxy?

Of course, the speed of the sun gets even trickier. Because, although it doesn’t move through the sky (we move, creating the illusion), it still orbits Sagittarius A*, the supermassive black hole at the center of the Milky Way. This monster is some 26,000 light years from earth and weighs in at around 4,000,000 M☉. Our Sun travels in a roughly circular orbit around this distant behemoth at a speed between 217 amd 250 kilometers per second – let’s take the average of those five figures and call it 230.4 kilometers per second (143.16392 miles per second). That’s 829,440 kph (515,390.112 mph).

That sounds impressive, doesn’t it?

Here’s something to consider, though: the speed of light is 299,792,458 meters per second, or 299,792.458 kilometers per second. That means the Sun (and it’s attendant planets and dwarf planets and other detritus) are moving at 0.07685% of the speed of light. Remember that line above, the one that reads “this monster is some 26,000 light years from earth”? That means that our Sun orbits Sagittarius A* in a circle approximately 163,360 light years in circumference. As a result, it will take about 163.360/0.07685% = 212.5 million years to complete an orbit. (Actual calculations from real astronomers come in at between 225 and 250 million years, which makes sense – they have access to more accurate figures, and the sun would actually describe an ellipse instead of a circle.)

So, why is the sun moving so slow? It isn’t. It’s tearing through space at a pace five times faster than the New Horizons probe at it’s maximum velocity – the fastest ship ever built by humanity (although the Sun’s gravity had slowed it to ‘only’ 14 kilometers per second by the time it passed Pluto). We just don’t notice, because it’s a huge universe and we have a very small frame of reference.

What if the Sun turned into a Black Hole?

This week, I’ve been writing about the sun. I blame the summer solstice for this, because the news that Monday was the longest day of the year fired my son’s imagination and got him asking question after question about the sun, and about the stars, and about related astronomical phenomena. So far, I’ve answered his questions about whether or not the sun can melt (it can’t) and what the hottest star is (H1504+65). Now it’s time to move on to his next question, one which demonstrates that he’s learned some interesting things.

“What if the sun turned into a black hole?” he asked, as we walked up the stairs to the front door of our condominium building. “Would it swallow the earth and all the planets?”

That one took me off guard, because I’m pretty sure that when I was five I didn’t even know what a black hole was. But then, I also realized that the first black hole was discovered the year I was born, so it’s not surprising the term wasn’t in common usage when I was five.

It’s a chilling thought, isn’t it? “Nothing escapes a black hole,” science fiction tells us. “Not even light.” Black holes are the great white sharks of space – remorseless predators consuming everything in their path. And we’d never see them coming. But they have one other thing in common with sharks.


They have an exaggerated reputation.

Newton’s Laws of Motion and Universal Gravitation

Although aspects of his laws have been superceeded by Einstein and his General and Special Theories, Newton’s laws remain an excellent (if ever so slightly inaccurate) model of motion. In brief, his three laws of motion state:

  1. If no forces act upon it, a body in motion will remain in motion and a body at rest will remain at rest, and velocity will remain constant in either case.
  2. If force is applied to an object, there will be a change in velocity proportional to the magnitude to which the force is applied.
  3. If body A exerts force on body B, then body B will exert a force of equal strength but in opposite direction on body A. This is also stated as “for every action there is an equal and opposite reaction”.

In addition, Newton put forth a law of universal gravitation. This law states that “two particles having masses m1 and m2 and separated by a distance r are attracted to each other with equal and opposite forces directed along the line joining the particles. The common magnitude F of the two forces is


where G is an universal constant, called the constant of gravitation, and has the value 6.67259×10^-11 N-m^2/kg^2.”

Yeah? What does this have to do with black holes?

I’ll get to that. But first, let’s cover what a black hole actually is.

Fine. What’s a black hole?

Does it surprise you to know that NASA has some good resources about black holes? It really shouldn’t.

A black hole is a region in space where the pulling force of gravity is so strong that light is not able to escape. The strong gravity occurs because matter has been pressed into a tiny space. This compression can take place at the end of a star’s life. Some black holes are a result of dying stars.

Because no light can escape, black holes are invisible. However, space telescopes with special instruments can help find black holes. They can observe the behavior of material and stars that are very close to black holes.

Black holes come in four size categories, representing both their mass and their physixal size. There are:

  1. Micro black holes. These can run all the way up to about 7.342 x 10-8 M (the mass of our Moon), and can get as big as 0.1 millimeters. Yes, it would suck if one hit you.
  2. Stellar black holes. These range up to 10 M in mass, and can be up to about 30 kilometers in diameter (0.5 x 10-4 R).
  3. Intermediate-mass black holes, which can get up to 1,000 M and up to about the mass of the Earth itself.
  4. Supermassive black holes. These are the monsters that lurk at the center of most galaxies, massing up to 1010 M and up to 400 astronomical units in size.

Wow. So, why do you say they have an exaggerated rep?

It’s true that the escape velocity of a black hole exceeds the speed of light, which is what it means to say that “no light can escape”. However, no black hole will be larger or more massive than the sum of all of the mass that went into making the black hole in the first place. So, outside the event horizon (the point at which gravity becomes too powerful to escape), the black hole has the same effect as any other object of the same mass. With that in mind, Newton’s law of universal gravitation tells us that – if the sun were to be instantly replaced with a 1 M black hole – there would no impact on our solar system. the r2 figure in the equation is measured from center of m1 to center of m2, so nothing changes.


Well, all right. That’s not true. Black holes have no luminosity – no energy would be generated and nothing would reach the Earth. So, to quote Randall Munroe’s Sunless Earth article, “We would all freeze and die.”

How Do You Get To Space?

We’ve got another contextless question here. Just the question “how do you get to space?” listed in my notes. I strongly suspect, though, that it has to do with my son’s current obsession with Star Wars. There’s a lot of space ships in those movies, after all, and he’s got a toy Millennium Falcon that he abuses in true five-year-old fashion. So he thinks about space – or at least movie “space” – a lot.

What Is Space?

Generally speaking, “space” is “outside Earth’s atmosphere”. Specifically, though, this gets problematic. There’s six different layers of the atmosphere, after all:

  • The troposphere, which averages about 7 miles (11 km) thick, and ranges from 8 km thick at the poles to 16 km at the equator. This is what most of us think of as “the atmosphere”, with breathable air and weather and most of clouds. The temperature drops with altitude in the troposphere.
  • The stratosphere, which sits on top of the troposphere and extends upwards to 30 miles (45 km) above the Earth’s surface. You find the ozone layer here, as well as temperature increasing slightly with altitude – up to a high of 32 degrees Fahrenheit (0 degrees Celsius).
  • The mesosphere, which sits on top of the stratosphere and extends upwards to about 53 miles (85 kilometers) above the surface. Temperature begins dropping with altitude again in this layer, reaching a low of -130 degrees Fahrenheit (-90 degrees Celsius), and it’s the atmospheric layer in which meteors begin to burn up as they fall towards Earth.
  • The thermosphere, which sits on top of the mesosphere and extends upwards to about 372 miles (600 kilometers) above the surface of the Earth. Temperatures can reach thousands of degrees (Celsius or Fahrenheit), but it’s measured in the energy of the molecules of gas in this layer and there’s not a lot of molecules that high up (the average molecule would have to travel 0.62 miles/1 kilometer to collide with another molecule), so it doesn’t feel hot.
  • The exosphere, which sits on top of the thermosphere and extends upwards to about 6,200 miles (10,000 km).
  • The ionosphere, which starts 30 miles (48 km) above the surface and stretches to 600 miles (965 km) above the surface. That’s an average figure, though, as it grows and shrinks based on solar conditions. It’s also divided into sub-regions, based on what wavelength of solar radiation it absorbs. Note that the ionosphere, although considered a seperate layer of the atmosphere, overlaps the mesosphere, the thermosphere, and the exosphere.

So, where’s space? The definition “outside Earth’s atmosphere” could mean that you have to exit the exosphere to be “in space”, but this would put us in the ridiculous position of having to consider the International Space Station (which orbits at an average height of 249 miles) within the atmosphere. Clearly, that’s ridiculous.

The commonly accepted altitude definition for space is the Kármán line, which is 62 miles (100 km) above sea level. This line, sitting in the lower reaches of the thermosphere, is approximately the altitude above which wings no longer provide lift and below which atmospheric drag makes orbital paths fail without forward thrust.

Ways to get to space

Clearly, then, to get into space “all” we have to do is travel at least 62 miles (100 km) straight up. Simple, right? Well, maybe not. At present, we have only a handfull of ways to leave the surface of the Earth: balloons, really big guns, rotor wing aircraft, fixed-wing aircraft, and rockets. Each has some advantages and disadvantages.


For balloons, the current record-holder for unmanned flight is the Japanese BU60-1, which reached an altitude of 33 miles (53 km) and carried a 10 kg payload. The altitude record for a manned balloon was the StratEx, which reached an altitude of 25.7 miles (41.4 km). Both were impressive achievements, but neither made the Kármán line.

Balloons function because of Archimedes’ Principle:

When a body is fully or partially submerged in a fluid, a buoyant force Fb from the surrounding fluid acts on the body. The force is directed upwards has a magnitude equal to the weight, mfg, of the fluid that has been displaced by the body.

In other words, if an object is submerged in a fluid but is lighter than that fluid, the fluid pushes it up until the object it is the same weight as the surrounding fluid. And while we don’t tend to think of it in this fashion, our atmosphere behaves like a fluid. We don’t float in the air because we’re denser then the air. A balloon inflated with air doesn’t float, because it’s actually slightly more dense than the atmosphere (because you forced more air in to inflate the balloon). But balloons filled with hot air, or with a gas that is less dense than our atmosphere (hydrogen or helium, say) will float.

Because of this, you could theoretically float a balloon into space – all you need is for the balloon to be less dense than the surrounding environment, after all. There’s a catch, though. The material of the balloon needs to be strong enough to not burst if the interior is pressurized, strong enough not to be crushed by the denser exterior, and light – because the mass of the balloon is added to the mass of the interior to determine if Archimedes’ Principle will lift it. A vacuum would be the ideal interior, but so far we don’t have anything simultaneously strong enough to keep the atmosphere from crushing a vacuum balloon and light enough for the air to push it up.

Big Gun

Jules Verne proposed this in From the Earth to the Moon way back in 1865. All you need is enough power, and you can launch something (or someone) into space from the ground. How much power? Well, using a ballistic trajectory calculator and assuming that the projectile is fired straight up from sea level, you would need a muzzle velocity of 4,595.52 feet per second (1,400.715 meters per second) to launch something 100 kilometers into the air – ignoring the effects of atmospheric drag. If you’re curious, that’s 142.78 times the force of gravity.

This is clearly not a good way to put people into space, but it would work well for launching non-fragile items. And in 1966, Project HARP demonstrated that it would work. A 16-inch (and 119 feet long) gun constructed by the US Department of Defence in Yuma, Arizona fired a 165 pound shell 590,000 feet (179,832 meters) into the air on November 19, 1966 – that’s 111.74 miles (179.8 kilometers), putting it well past the Kármán line


Planes seem an obvious choice, right? We all know what they are, they take off and land, so why not fly into space? Well, there’s a good reason for that – the Kármán line (meaning that by the time you reach space your wings aren’t working) and the propulsion (both jets and propellers require air to function). So, any plane that could reach space would need to be a rocket as well as an airplane. Right now, the world altitude record for a airplane was set on August 31, 1977 by Alexander Fedotov, who reached 23.4 miles (37.66 kilometers) in a MiG-25.


Right now, this is the way we get to space. A rocket engine carries stored fuel and utilizes Newton’s third law by expelling that fuel in one fashion or another from one end of the craft to push the opposite direction. Most rockets in operation are combustion rockets, meaning that the fuel is ignited and burned in some fashion. These can be quite expensive, as the rocket has to lift all of its fuel at the time of launch. However, with sufficient fuel and engineering and money, you can build a rocket capable of reaching any altitude.

Other Strategies

Any number of launch methods that do not rely on rockets have been proposed – the space gun technically is counted in this category, but it differs from the others in the fact that it has actually been constructed. Most of the others have a US Department of Defense technology readiness level of 2, meaning that they are dependent on the invention of the materials and technologies needed to support them, and may not actually be feasable. They include:

  • Space tower: a tower that reaches above the Kármán line, possibly as far as geosynchronous orbit (22,369 miles or 36,000 km).
  • Skyhook: A satellite that lowers a lift cable (or the equivalent) and then reels in the payload.
  • Space elevator: A tether attached to the Earth at one end and a geosynchronous satellite at the other, which can then allow vehicles to climb into space.

How do you stay in space?

To quote Randall Monroe, “getting to space is easy. The problem is staying there.” Why? Well, gravity is still pulling you down. So, you have to go ‘sideways’ fast enough that you keep missing the Earth as you fall (which is pretty much what an orbit is). Using the Earth Orbit Velocity calculator on Hyperphysics, an orbit at the Kármán line would require an orbital speed of 25,745.41 feet per second (7847.2 meters per second), which translates to 17,549.1 mph (28,249.64 kph).

Fortunately, you don’t have to maintain constant acceleration at that speed – one of the defining features of space is that it’s pretty empty, meaning there isn’t a whole lot out there to slow you down once you get going. But still, you have to get going really fast to stay there – orbiting at the Kármán line means you circle the earth every 1.45 hours.

How do you get back down?

Oh, that’s easy. You fall. Whether or not you die is a matter of how you land.

Would a Helicopter Work On Mars?

This is one of those questions that I wish I could remember the context for. But I don’t. It’s just sitting there in the master list of the questions my son has asked, sandwiched between “What’s plankton?” and “What if the oceans froze?”, and I have no idea when or why he asked me this. Just “would a helicopter work on Mars?”

Usually I try to jot down something about the context. But not this time.

So, let’s get to the question. Would a helicopter work on Mars?

The simple answer appears to be “yes”. The Jet Propulsion Laboratory, on January 22, 2015, put out a press release on that very subject. They’ve designed a small proof-of-concept drone with the imaginitive name Mars Helicopter. It’s a 2.2 pound (1 kg) device, cubical in shape, with a 3.6 foot (1.1 meter0 rotor span. As designed, it would be deployed to work in conjunction with a future Mars rover.

The helicopter would fly ahead of the rover almost every day, checking out various possible points of interest and helping engineers back on Earth plan the best driving route.

Scientists could also use the helicopter images to look for features for the rover to study in further detail. Another part of the helicopter’s job would be to check out the best places for the rover to collect key samples and rocks for a cache, which a next-generation rover could pick up later.

If you prefer visuals, JPL also has a video about Mars Helicopter. It’s primarily CGI, since they haven’t constructed a fully-functional model, but it includes a few shots of a prototype trying to fly in a simulated Martian atmosphere.

But, what would it take to make a helicopter work on Mars? Not that I don’t believe JPL when they say it can be done – they are literally rocket scientists, and their entire job revolves around building things for space and other planets – but I’m curious. Fortunatly, NASA has me covered again. Let’s turn now to a paper describing the actual experiments NASA has done on rotary-wing aircraft for Mars, titled Experimental Investigation and demonstration of Rotary-Wing Technologies for Flight in the Atmosphere of Mars.

The paper begins by describing the challenges faced in constructing a rotary-wing aircraft for the Red Planet:

The Martian atmosphere is 95% CO2 with the remaining 5% comprised of N2 and other trace gases. Mars’ gravity is slightly greater than a third of Earth’s. The atmosphere of Mars is extremely cold and thin (approximately 1/100’th of Earth’s sea-level atmospheric density). Further, a seasonal variation of approximately 20% of the planetary atmospheric mass occurs on Mars (a consequence of polar CO2 condensation and sublimation). Given the thin, carbon-dioxide-based Martian atmosphere, developing a rotorcraft design that can fly in that planetary environment will be very challenging.

After that, the article goes into a lot of cool descriptions of various hypothetical rotorcraft designs and a whole lot of math that flies right over my head. It’s all stuff I’d love to understand, but it gets into some advanced-looking physics and aviation engineering concepts that I’ve never studied. Here’s what I could pick out:

* NASA is investigating both coaxial and quad-rotor configurations.
* Experiments have been done with battery power, electricity generated by fuel cells, and engines fueled by hydrazine. At present, some variety of electrical power seems to be the best option.
* It’s really hard to test an actual prototype under conditions that replicate Martian gravity, atmosphere pressure, and composition. At present, we’ve only managed it with the pressure.
* The rotors will have to be really big and spin really fast, and the rotorcraft will have to be really light.

A reasonable layman’s description of what a Martian helicopter can be found in this quote from the second page of the paper:

From an aeromechanics perspective, Mars rotorcraft will be very different from their terrestrial counterparts. Mars rotorcraft will have very large lifting-surfaces and will be required to have ultra-lightweight construction. For example, in order to lift ten kilograms of vehicle mass on Mars, a single main rotor (at a disk loading of 4 N/m2) would have to have a radius of approximately 1.7 meters.

For added fun, check out Interplanetary Cessna on Russel Monroe’s What If? site, which answers the question “What would happen if you tried to fly a normal Earth airplane above different Solar System bodies?” It’s not specifically on the topic, but it’s close.

Are There A Million Stormtroopers?

Let’s just start off with something important. We’re talking about these guys from Star Wars:


Not these guys from World War Two:


I’m completely fine with my five-year-old watching Star Wars. I’m really not fine with the idea of having to talk about something like the Holocaust with my five-year-old.

With that out of the way, my son is a huge Star Wars fanatic. He loves the movies (with Return of the Jedi being his favorite, because of Ewoks). Christmas was loaded down with Star Wars toys. He’s got a lightsaber. What can I say? He’s at the right age to enjoy the movies. Which is to say, he’s alive – I’ve got two lightsabers, myself, and we play Star Wars [i]a lot[/i]. And he also loves the idea of ‘a million’. I think it’s his go-to number for “lots and lots”, because I’m pretty sure he doesn’t really understand what ‘a million’ is.

“Are there a million Stormtroopers?” he asks from the back seat of the car, as he plays with a Stormtrooper and a toy dinosaur and rider that he’s just purchased with his allowance and has declared is the Emperor.

“Probably more,” I answer.

“How many more?” he asks, sounding intrigued by the idea that there can be more than a million.

“Lots more,” I answer.

So, clearly, this isn’t a serious sort of question. But I’m curious about the potential answer. Or answers, really, because there’s two different ways to answer this. We’ll start with the real answer, first. For that, we turn to the Internet Movie Database, and check the full cast lists for Episodes IV through VI.

Star Wars Episode IV names 8 people playing Stormtroopers (with 2 more added in the Special Edition), and four people voicing Stormtroopers. So, depending on how you count that, you get 8 to 14 Stormtroopers in the movie. Episode V credits 8 people, and Episode VI credits 5. StackExchange discusses this a little as well, pointing out that the live actors were padded out with matte paintings, cardboard cutouts, and blue-screen special effects.

So that’s the real-world answer. From here on out, I’ll just be indulging in naked nerdery. You are warned.

According to Wookieepedia, Stormtroopers “were the elite shock troops in the Imperial Army placed under the Stormtrooper Corps”. Their organization is described as follows:

  • Squad: 8 troopers plus a sergeant.
  • Platoon: 4 squads, plus a lieutenant and a sergeant major.
  • Company: 4 platoons, plus a captain.
  • Batallion: 4 companies, plus a major.
  • Regiment: 4 batallions, plus a Lieutenant Colonel.
  • Legion: 4 regiments, plus a High Colonel.

That gives a Legion 9,813 Stormtroopers, and brings us to the question of just how many legions there were. Wookiepedia doesn’t directly address this, but does point out that the Stormtroopers were the Imperial successors to the Clonetroopers of the Grand Army of the Republic. Assuming a similar structure, then, we add the following to the organizational list:

  • Corps: 4 legions, plus a General
  • Sector Army: 4 corps, plus a General.
  • Systems Army: 2 Sector Armies, plus a general.
  • Grand Army: 10 systems armies.

This brings us to 3,140,270 Stormtroopers, without considering the various specialists (such as the Scout Troopers, or the Snowtroopers, or the various and sundry other types created by the Expanded Universe). So, let’s try and compare this to the real world. Since Stormtroopers are usually the first into the breach and the first on the beach, let’s have a look at the Organization of the United States Marine Corps – specifically, at the Marine Air-Ground Task Force. This is a combined arms unit of Marines, comprising a Division-sized Ground Combat Element, a wing-sized Aviation Combat Element, and a Logistics Combat Element (which appears to be Division strength).

First, a few definitions: A wing is three or more groups of 4-6 squadrons of 6-15 aircraft. A Division is the next organizational step up from a Regiment. And the Marines follow (loosely) a “rule of three” – three squads to a platoon, three platoons to a company, and so on. So, for organizational comparison purposes, we can compare a Stormtrooper Legion to a Marine Division.

We will assume that there is a such entity as a Stormtrooper Aerospace-Ground Task Force.  Stormtroopers roughly follow a “rule of four” in their organization. So, an Aerospace Combat Element for a SAGTF would probably be set up as follows:

  • Squadron: 8 TIE Fighter pilots, plus an officer.
  • Group: 4 squadrons, plus an officer.
  • Wing: 4 squadrons, plus an officer.

That gives us an Aerospace Combat Element strength of 149 at the Wing size.

The Logistics Combat Element is a “catch-all”, handling logistics (transport, supply, maintenance), engineering (EOD, combat engineering, heavy equipment), medical, and other specialized units. The Marines get one at Division strength in a MAGTF, so you probably get a Legion-strength Logistics Combat Element with Stormtroopers. Oh, and don’t assume they’re noncombatants. They’re just not primary combatants.

At all of this, the next step up from a Legion should be the SAGTF, comprised of 19,975 Stormtroopers. Replacing the word “Legion” with the term “SAGTF” in the above estimates, we get 79,101 Stormtroopers in a Corps, 316,405 in a Sector Army, 632,811 in a Systems Army, and a total of 6,328,110 Stormtroopers.

This seems low to me, however. The Empire is enforcing its authority over a large swathe of a galaxy, after all. One which, (again according to Wookieepedia) contains 1,024 sectors. It seems reasonable that there should be a Sector Army per sector. With 316,405 Stormtroopers in a sector, that gives us 323,998,720 Stormtroopers. Sounds like a lot, doesn’t it?

NASA estimates that there are somewhere between 10,000,000,000 and 100,000,000,000 solar systems in our galaxy. The galaxy in Star Wars is a spiral galaxy similar to our own, so it’s not unreasonable to believe that there are similar figures – call it a rough average of 55,000,000,000 solar systems. Eyeballing the map of the Star wars galaxy, I’d say that the Empire occupies roughly 70% of that galaxy – call it 38,500,000,000 solar systems.

Princeton University has an article titled Known Planetary Systems, which states we have currently found 258 solar systems. One of them is inhabited. Now, clearly we’re not a starfaring civilization. But the Galactic Empire is, and so I think it’s safe to apply this ratio to the Empire to decide that there are 149,224,806 inhabited worlds – with “inhabited” ranging from subsistance-level populations like Tatooine to mining installations like Bespin or Kessel to ecumenopolises like Coruscant. So there are only 2 Stormtroopers for each inhabited world.

No wonder Leia was so confident that: