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So You Want To / Write a Hard Science-Fiction Story with Space Travel

"Did you ever notice when you're jacking off, that it's more of a turn-on to fantasize about the girl next door than it is to fantasize about a supermodel? Because with the girl next door, you're thinking, hey, this could really happen!"
Carlin's observation evokes one of the core appeals of hard science fiction to some. A sufficiently hard science fiction story, even if it's set in another star system, could really happen — or at least, it should be very difficult for the readers to come up with ways that it couldn't happen. Writing hard science fiction set in space carries with it all of the baggage of writing any other genre of literature. Your characters must be believable, your plots and descriptions must not be boring, the story must be satisfying to the reader in some way, etc.. Every piece of advice in the Write a Story article is just as sound when writing hard SF as it is when writing a western, a modern romance, a historical naval drama, or any other literary genre you can imagine. However, to work as a piece of hard SF with space travel, the writer must go one big step farther: The technology, the mechanics of space travel, the planets, the aliens (if there are aliens), all the details of that futuristic setting must be realistic. The author must take pains to follow the known laws of physics, chemistry, biology, astronomy, and planetology, and how they apply to any areas of engineering that will appear in the story. This means the author must know the laws of physics, chemistry, etc., or have good access to someone who does. While the laws of the author's fictional universe are allowed to deviate from the laws of Real Lifeon occasion, the author must be consciously aware of each of those deviations, must have an excuse for them (even if he never tells the reader this excuse, he must have it in his own head), and above all must take pains to limit the damage that such departures from reality can potentially do to the story. Since space travel is involved, it's important to remember that human beings have travelled in space for over five decades now. We know what is involved in getting from the Earth's surface to low Earth orbit. We know what's involved in landing on a rocky world 400,000 kilometers away. We know what effect microgravity has on human bones and muscles. A realistic story involving space travel must take all this accumulated human knowledge into account. The cartoonish world of 1950s B-movie astronauts having a "navigational error" that sends them to an "uncharted planet" with an Earthlike ecosystem inhabited by alien women who speak English is, and should be, a Discredited Trope — but so should portraying space travel like anything other than space travel just because it looks neater that way in your head. One of the best resources out there for realistic future space travel is the Atomic Rockets page, which covers everything from "what designs are on the drawing board for spacecraft capable of crossing interstellar distances within a human lifetime?" to "why should my female crew members not wear skirts?"

Heartbreak Hotel

Sadly, the rules of writing about realistic future space travel — like the rules of writing about anything realistic — are primarily a set of rules about what you can't do. The more ideas you have about what you'd like to have your characters do, the more ways reality will step in and say "No." First off, here is a list of tropes that are frowned upon in a realistic universe: There are sound reasons why all of the above tropes will probably not work in Real Life. It's not impossible to have them work in a way that doesn't violate the laws of science or good sense, but it will requite extra painstaking labor from the author to make that happen.

Doing the Research

This should go without saying, but: If you're going to set your story someplace we already know something about — like Mars or Alpha Centauri — for goodness' sake, read up on what we know about the place before you start writing! We've sent space probes to every planet in the solar system, we've accrued reams of data on just about every star that has a name, we've even mapped out the interstellar medium in our neck of the galaxy. The data are out there, and thanks to the Internet they're not even hard to acquire any more. You wouldn't set a story in the Sahara Desert and have your hero go swimming in one of the "numerous lakes" there. You wouldn't set a car chase in downtown Florence, Italy and then make up the street names and city layout. Similarly, don't set your story on Mars and have your hero swelter in the unbearable heatnote I'm looking at you, Babylon 5 novel #1 "Voices" by John Vornholt!, or put an Earthlike planet in orbit around Alpha Centauri without at least mentioning the bright "B" star that should be visible from time to time in the sky. Making up details about places we don't have strong data about is one thing, but making up details about places where our existing data would make those details flat-out impossible is quite another.

Doing the math

If doing an arithmetic problem like "A train leaves Chicago at 8 AM going 60 miles per hour" taxes the limits of your skills, putting realistic space travel into your story is probably not for you. Space travel is all about doing the math, and the math can get hairy — especially when dealing with speeds above 5-10% of the speed of light, where special relativity starts to rear its ugly head. But if you're up for the challenge, it's definitely worth doing. Even if you don't show your work to your readers, getting the numbers right (or nearly right) will go a long way toward your story's sense of realism. Let's take as our example a rocket trip from Earth to Saturn. How long will it take to make the flight? You could go the easy route, and look up how long it took for the Voyager 1 space probe to fly from Earth to Saturn, and just assume that you space ship flies about as fast as the Voyager probes did. But when you do look it up, you balk — it took nearly three years for Voyager 1 to make this trip! You can't have your intrepid space cadets waiting around for three years just to get to Saturn, they've got important space adventures to have, space wars to fight, and space women to woo. You want them to get to Saturn a lot quicker. So, you give them a space ship that never runs out of fuel — or uses fuel that's so efficient that it won't run out even if it runs its engines continuously between Earth and Saturn. So, with the ability to accelerate indefinitely, now how long will it take to get to Saturn? Let's say you decide to limit your space ship to a cruising acceleration of 1g, so that the crew will experience thrust equal to Earth's surface gravity. After all, this is a hard science fiction story, right? You can't just go slinging Inertial Dampening around. Any acceleration your ship undergoes will be experienced by your crew as G forces. So, if they're going to be cruising under engine power for anything longer than a few minutes, you'll probably want to keep your acceleration down to 1g. 1g works out to an acceleration of 9.8 meters per second per second — at the end of one second, you'll be going 9.8 m/s, at the end of two seconds, you'll be going 19.6 m/s, and so forth. At speeds much less than the speed of light, which is the speed we're dealing with in the Earth-to-Saturn example, the formulas relating speed, acceleration, time, and distance travelled while undergoing constant acceleration are pretty straightforward. Ignoring the sun's gravity (which can indeed be neglected for a ship that accelerates at one g) we have:
  1. v = a * t + v0
  2. d = 0.5 * a * t2 + v0 * t
  3. 2 * a * d = v2 - v02
... where d is the distance travelled in meters, v is your current velocity in m/sec, a is your acceleration in m/sec2, t is the elapsed time in seconds, and v0 is your initial velocity at the start of your trip. So, what value should we plug in for d? The distance from Earth to Saturn, right? Well, unfortunately, Earth and Saturn are both in orbit around the sun, which means they move. Their movements are more complicated than the simple equations I've written above, but for a first approximation, let's say you plan your trip to happen when Earth and Saturn are closest together, i.e. both on the same straight line from the sun (what astronomers call "opposition"). Looking up their orbital parameters on Wikipedia, we see that the average Earth-sun distance is 1.0 A.U., and the average Saturn-sun distance is 9.582 A.U., so the two will be about 8.5 A.U. apart at closest approach. (They could actually be closer. Saturn's orbit is rather eccentric, varying between 9 AU and 10 AU from the sun; but 9.5 A.U. is close enough for a first approximation. And waiting for Earth and Saturn to be at opposition when Saturn happens to be at perihelion will mean your space cadets will be sitting around tapping their toes for decades.) 1 A.U. is about 150 million kilometers, so 8.5 A.U. works out to around 1,280,000,000 kilometers. But wait — our equations require d to be in meters, not kilometers. We have to inflate that number by a factor of a thousand. 1,280,000,000 km is 1,280,000,000,000 m. Let's hope your calculator has enough spaces for all those digits. (If not, you might need to enter it in scientific notation, as 1.28 x 1012.) Now, how long will the trip take? Let's plug our numbers into the second equation above. We know d, and we know a. We can assume v0 = 0, if we're taking off from Earth at a standing stop. (We won't actually be; our space ship will be taking off from Earth orbit, which means it'll be moving at about 7800 m/s, and our zero-velocity reference point here is our destination — Saturn — which is orbiting at a different speed as the Earth. But our overall speed is going to be so great that these little speed differences shouldn't matter much.) This is what we get:
d = 0.5 * a * t2 + v0 * t
1,280,000,000,000 m = 0.5 * 9.8 m/sec2 * t2 + 0 * t
1,280,000,000,000 m = 4.9 m/sec2 * t2
Unfortunately, we're trying to solve for t here, so we need to do a little algebra. Let's divide both sides by 4.9 m/s2 :
1,280,000,000,000 m / 4.9 m/sec2 = t2
260,000,000,000 sec2 = t2
Taking the square root of both sides:
So the trip will take 510,000 seconds. Since there are 86,400 seconds in a day, this means it'll take about 5.9 days. What will be our final velocity (v) when we reach Saturn? For that, we need the first equation above; luckily, since we now know t, it'll be a lot easier:
v = 9.8 m/sec2 * 510,000 sec + 0
Woops! We arrived at Saturn in less than 6 days, but now we're whooshing past it at 5 million meters per second! (That's 5000 kilometers per second, about 1.6% of the speed of light.) At that speed, we'll only have a few seconds to have adventures on Saturn before we speed right out of the Solar system. What we'd like to do instead is arrive at Saturn with a velocity of zero, or nearly zero. That's going to require us to turn over and decelerate: We'll accelerate to the half-way point between the Earth and Saturn, then do a "skew flip" so that our engines are pointed in the opposite direction, then decelerate for the rest of the trip. This adds an extra step or two to the math we'll need to do, but thankfully each step should be just as easy. First, we need to change d from the full Earth-Saturn distance (1,280,000,000,000 meters) to half that distance, and see how long it'll take to get there:
640,000,000,000 m = 0.5 * 9.8 m/sec2 * t2
640,000,000,000 m = 4.9 m/sec2 * t2
640,000,000,000 m / 4.9 m/sec2 = t2
130,000,000,000 sec2 = t2
Taking the square root of both sides:
So, it'll take 360,000 seconds — a hair under 4.2 days — to get to the half-way point. Our velocity v at this point will be:
v = 9.8 m/sec2 * 360,000 sec
3500 kilometers per second. Slightly more than 1% of the speed of light, but that's okay, because we're in deep space nowhere near Earth or Saturn. Now, how long will it take to decelerate from this speed to a speed of 0 at 1g (9.8 m/sec2)? For this, let's plug the numbers back into the first equation. Remember, this time, our acceleration is negative (a negative acceleration is the same thing as a deceleration), and we do have an initial velocity:
0 = -9.8 m/sec2 * t + 3,500,000 m/sec
Subtracting 3,500,000 m/sec from both sides:
-3,500,000 m/sec = -9.8 m/sec2 * t
Dividing both sides by -9.8 m/sec2:
-3,500,000 m/sec / -9.8 m/sec2 = t
Hmmm! It takes 360,000 seconds to decelerate from 3500 km/sec to 0 — exactly the same amount of time it took to accelerate from 0 to 3500 km/sec! And if we ran the numbers, we'd see that on this second leg, we covered exactly the same 640,000,000 km we covered in the first half of the trip. The two legs of the journey are mirror images of one another. Once we've figured out how long it takes to get to the half way point, we know that it'll take exactly as long to go the rest of the way (assuming our deceleration during the second leg is equal in magnitude to our acceleration during the first leg). So ... using our constant-acceleration trajectory, it takes 360,000 seconds to reach the half way point, and another 360,000 seconds to go the rest of the way, for a total of 720,000 seconds to get from Earth to Saturn. That's 8.3 days, a little over a week. It's still not something your space cadets can do on their day off, but it's a heck of a lot quicker than the 3 years Voyager 1 took to make the trip — and unlike Voyager 1, your space ship ends up at rest relative to Saturn so it can stay there as long as you need it to. By the way, that constant-acceleration trajectory may be fast, but it is ferociously inefficient in terms of fuel use, and no real spacecraft actually use it. (For just how inefficient it would be, see the "Why Rockets Are So Big" section farther below.)

Special Relativity

Now ... what if our heroes can go faster than this? Let's say we've decided we need shorter flight times, so screw it, we're introducing Inertial Dampening technology into our story, like in the Honor Harrington or Star Trek universe. Now our space ships can accelerate at 1,000 g, or 9,800 m/sec2, and we should be able to get to Saturn much faster. In fact, using the first equation above, it looks like we should be able to accelerate to 30,000,000 m/sec in less than an hour — that's a tenth of the speed of light. And there's where we run into problems. Because above about a tenth of the speed of light, acceleration doesn't affect velocity in a straightforward manner any more. Your clock runs a little slower to a fixed observer than it does to you. Your momentum is slightly higher to a fixed observer than your acceleration history says it should be. The universe shrinks slightly in the direction you're moving. In short, you run smack-dab into special relativity, and now the math gets a lot more complicated. For one, the change in your velocity per second, given a constant acceleration from your space ship's reference frame, now depends on your current velocity — which turns the relationship between the two into a first-order differential equation. The basic relativistic equation for determining your "gamma factor" — the amount by which your mass goes up, time slows down, or distances in the direction of motion shrink — is as follows:
  • γ = 1 / sqrt (1 - v2/c2)
... where sqrt means square root, and c is the speed of light in a vacuum (about 300,000,000 m/sec). As you can see, it varies according to how fast you're going (that pesky v2 term in the denominator). Here are some typical gamma factors:
At restγ = 1
At 10% of cγ = 1.005
At 50% of cγ = 1.15
At 86.6% of cγ = 2
At 90% of cγ = 2.29
At 99% of cγ = 7.09
At 99.9% of cγ = 22.37
At 100% of cγ = ∞
(As you can see, a massive object such as a space ship can never achieve 100% of the speed of light, because its momentum would be infinite — it would take an infinite amount of energy to accelerate to c.) The Atomic Rockets website lists many, but not all, of the useful equations that come into play for a rocket travelling at these relativistic speeds. These include:
  1. T = (c/a) * ArcCosh[a*d/(c2) + 1]
  2. t = sqrt[(d/c)2 + (2*d/a)]
  3. v = c * Tanh[a*T/c]
  4. v = (a*t) / sqrt[1 + (a*t/c)2]
  5. γ = Cosh[a*T/c]
  6. γ = a*d/(c2) + 1
... where T is "proper time" (the number of seconds elapsed in your space ship's frame of reference), t is the number of seconds elapsed from a fixed observer's frame of reference, d is distance travelled from a fixed observer's frame of reference, a is acceleration in the space ship's frame of reference (which is assumed to be held constant), Tanh is hyperbolic tangent (there's a button for this on most scientific calculators), and ArcCosh is inverse hyperbolic cosine. So, let's say you want your inertial-dampener-equipped space ship to accelerate at 1,000g, or 9800 m/sec2, and you want to follow the same course you did with your piddling little 1g space ship — accelerate to the half-way point between Earth and Saturn, 640 million km away, and then decelerate for the same distance to arrive at Saturn with v = 0. How much time will that take, as far as the folks back on Earth are concerned? For that, we can use equation 2 above:
t = sqrt[(d/c)2 + (2*d/a)]
t = sqrt[(640,000,000,000 m / 300,000,000 m/sec)2 + (2 * 640,000,000,000 m / 9800 m/sec2)]
t = sqrt[4,550,000 sec2 + 130,600,000 sec2]
The folks back on Earth measure 11,600 seconds, or about 3.2 hours, for the space ship to reach the half way point. Meanwhile, how much time has elapsed for the folks on board the space ship? For that, we need equation 1 above:
T = (c/a) * ArcCosh[a*d/(c2) + 1]
T = (300,000,000 m/sec / 9800 m/sec2) * ArcCosh[9800 m/sec2 * 640,000,000,000 m / ((300,000,000 m/sec)2) + 1]
T = (30,600 sec) * ArcCosh[0.06969 + 1]
Your intrepid spaceship crew measures the elapsed time as 11,360 seconds, which is only slightly less than the time t measured by the folks back on Earth. How fast will you be going at this half-way point? For that, we'll need equation 3 above:
v = 300,000,000 m/sec * Tanh[9800 m/sec2 * 11,360 sec / 300,000,000 m/sec]
v = 300,000,000 m/sec * 0.3549
... or a tad over 1/3 of the speed of light. This explains why our proper time T and our Earth time t are so close together: At 1/3 of light speed, the gamma factor is only about 1.07, and our space ship was only going this fast near the end of this leg anyway. If we'd taken a longer trip — say, from Earth to Sedna in the Kuiper belt, or from Earth to Alpha Centauri — we'd have more distance in which to accelerate, which would let us get closer to the speed of light, and the relativistic effects would have been more pronounced. Unfortunately, these equations only address the situation when your space ship starts out at 0 velocity. If you want to apply these equations to a situation where you start out already moving at (say) 1/5 of light speed, they get even more complicated, and often times will not have already been derived for you. For example, the equation for the amount of time a fixed observer measures that it takes your accelerating space ship to cross a given distance, assuming you started out with a velocity that gave you an initial gamma factor of γ0, is this hairy beast:
t = c/a * (sqrt [(a*d/c2 + γ0)2-1] - sqrt[γ02-1])

Orbits

If your space ship is close to a large gravity source like a planet or a star, and is moving at a low enough speed, it isn't going to travel in a straight line. The gravity of the big object will cause your spacecraft to follow an orbital trajectory. The equations for an orbit are more complicated than the equations for straight-line movement, because your acceleration is always changing; it depends on how far you are from the big object at that particular instant. Whole volumes have been written on how to calculate an orbit precisely, but there are some simple straightforward cases that are at least somewhat easier to calculate. All orbits are shaped like conic sections. If your space ship is moving slower than the "escape velocity" — or more precisely, if its total kinetic energy and gravitational potential energy is less than the gravitational potential energy it would have at an infinite altitude — its orbit will be shaped like an ellipse. If it's moving precisely at the escape velocity, its orbit will be shaped like a parabola. If it's moving faster than the escape velocity, its orbit will be shaped like a hyperbola. This is assuming, however, that your space ship spends all its time from this moment forward in free-fall, without firing its engines. Parabolic and hyperbolic orbits are basically escape trajectories. The spacecraft leaves the big object in question and never comes back. An elliptical orbit, on the other hand, is stable, and allows your space ship to go around and around the big object over and over again. It's what most people think of when they hear the word "orbit." An ellipse looks an oval-ish shape, with two points inside it called the "foci" (plural of focus). In an elliptical orbit, the center of the big object you're orbiting is going to be at one focus, while the other focus won't contain anything at all. The formula for how long it takes to make one complete orbit was first deduced by Johannes Kepler when studying the motions of the planets around the sun. Sir Isaac Newton expanded on this formula so that it applied when orbiting any object. The formula is: .. where P is the period of the orbit in years, M is the total mass of all objects involved in the orbit in solar masses, and A is the semi-major axis of the orbit's ellipse in Astronomical Units. The simplest kind of elliptical orbit is one with no eccentricity at all. We call this a circle. In a perfecly circular orbit, both foci are at the same point in space, which is at the center of the circle. The semi-major axis A of a circle is just its radius (half its diameter). Suppose you want your space ship to orbit the Earth in a perfect circle 200 kilometers above the surface, like the Space Shuttle does. What will the period (P) of that orbit be? If we want to use the equation above — P2M = A3 — we'll first have to find the combined mass of the Earth and your space ship in solar masses. The Sun's mass is about 2 x 1030 kg, while the Earth's mass is about 6 x 1024 kg. Let's say your space ship weighs in at 1000 tonnes, i.e. a million kg. Here's the mass of the Earth in solar masses:
3.003740720000000000 x 10-6
... and here's the combined mass of Earth and your space ship in solar masses:
3.003740720000000001 x 10-6
As you can see, unless the two objects involved in the orbit are of comparable mass, the mass of the teensy tinsy little orbiting object really doesn't matter for this equation. We can just use the Earth's mass by itself, and be done with it. We don't even need all those decimal digits — the real Earth isn't a perfect sphere, so its gravity is a tiny bit "lumpy", which means that our answer for how long the orbit will take is only going to be approximate anyway. The value we need for M is 3 x 10-6 solar masses. Now, we need the semi-major axis, that is, the radius of the orbit. We're at 200 kilometers altitude, so that means the orbital radius is 200 kilometers, right? Wrong. The radius is the distance from our space ship to the center of the Earth. The Earth is 6371 kilometers in radius, on average, so the orbit's radius is 6571 km. For our formula, we need this expressed in Astronomical Units. One A.U. is 149,598,000 km, so the value we need for A is:
Plugging those two numbers into the formula, we get:
P2 * (3 x 10-6) = (4.39 x 10-5)3
P2 * (3 x 10-6) = 8.47 x 10-14
P2 = (8.47 x 10-14) / (3 x 10-6)
... which works out to 88.4 minutes. And, indeed, the space shuttle does take about this long to orbit the Earth once. Now, let's try it with a 100 kilometer orbit above the surface of Mars:
Mass of Mars = 6.4 x 1023 kg = 3.2 x 10-7 solar masses
Radius of orbit = 100 km (orbital altitude) + 3397 km (radius of Mars) = 3497 km = 2.3 x 10-5 A.U.
P2 * (3.2 x 10-7) = (2.3 x 10-5 A.U.)3
P2 = (2.3 x 10-5 A.U.)3 / (3.2 x 10-7) = 4 x 10-8
P = 2 x 10-4 years = 105 minutes.
Not surprising that this is similar to the period for our low-Earth orbit above. Mars is only about 1/10 the mass of the Earth, but our orbital radius was about half as high as before (and 1/2 cubed is 1/8), so the two factors nearly cancel each other out. Note that you don't have to do these calculations in solar masses, A.U.s, and years. If you want to use other units, you can introduce a constant k that adjusts for the differences in your unit system, like so: This is usually done with a k that converts everything into kilograms, meters, and seconds; but in the olden days, you might have picked a different k that converts your units into pound-masses, feet, and seconds. There's even a video about orbital mechanics here that uses furlongs!

Why rockets are so big

You've doubtlessly seen the footage of the Apollo moon mission launches. An enormous Saturn V rocket, hundreds of feet tall, lumbered off the launch pad on an enormous column of flame. Yet the actual Apollo spacecraft was just a tiny cylinder perched atop it, with an even smaller cone on top of that where the crew actually lived. Why was the rocket so big, and the actual usable space so small? Cars can pull themselves along the ground by spinning rubber tires. Boats can push water through their propellers. Airplanes can push air through their propellers or turbines. Rockets, on the other hand, have none of these options. Every ounce of thrust their engines produce has to come through the expenditure of onboard propellant — in other words, they accelerate forward by throwing material backward. (Boats and airplanes also accelerate forward by throwing material backward, but they get this material from the environment around them. Rockets have to carry all this "reaction mass" on board.) This severely limits the efficiency of a rocket engine when compared with a fluid-breathing or surface-friction engine, even moreso than the need to carry their own oxygen to combust with their fuel. Even worse, it means that at the start of your flight, you have to produce that much more thrust just to push all your unburned fuel along with you, so each kilogram of fuel you add provides progressively less and less total acceleration. This cascade effect can add up very quickly. The equation for how much total acceleration your rocket can undergo before it runs out of fuel — the total "delta-v budget" of your rocket — was derived by Tsielkovsky over a century ago:
  • Total delta-v = ve * ln(M/Me)
... where ve is the exhaust velocity of your engines, ln means "natural log" (there's a button for this on all scientific calculators), M is the mass of the rocket with fuel, and Me is the mass of the rocket without fuel (the empty weight). M/Me is sometimes called the "mass ratio" of the rocket. The ve for, say, the Space Shuttle's main engines is about 4500 meters per second. In order to orbit the Earth, the Space Shuttle must travel at 7800 meters/sec, and it must be about 300 kilometers above the Earth's surface (it requires at least another 1600 m/s of delta-v to lift it that high and overcome atmospheric drag along the way). This means it needs a total delta-v of around 9400 m/s, which is over twice its own exhaust velocity. From the rocket equation above, this means its M/Me ratio must be more than e2, or 7.39. The shuttle's weight with fuel must be over seven times as high as its weight without fuel! Discarding its spent solid rocket boosters in mid-flight (a trick similar to staging) can help a little, but not much. With such a stultifying mass ratio just to get into Earth orbit, you can see why flying to other planets in a matter of days — or worse, flying to another star within a human lifetime — just isn't practical for modern chemically-propelled rockets. How impractical is it? Well, let's take the example of the trip to Saturn discussed above, where our space cadets undergo a continuous 1g acceleration to the half way point, and a continuous 1g deceleration for the rest of the trip. We established that their velocity is 3,500,000 m/s at the half way point, so we need 3,500,000 m/s of delta-V to get that far, and another 3,500,000 m/s of delta-V to brake to a halt, for a total delta-V requirement of 7,000,000 m/s. If their space ship's engines had an exhaust velocity of 4,500 m/s, the same as the Space Shuttle's main engines, what mass ratio (M/Me) would be required to attain 7,000,000 m/s of delta-V?
7,000,000 m/s = 4500 m/s * ln(M/Me)
Dividing both sides by the exhaust velocity:
(7,000,000 m/s) / (4500 m/s) = ln(M/Me)
To get rid of the natural log, we need to take the natural exponential of both sides:
e1555 works out to an absolutely gargantuan 3.7 x 10675. In other words, your space ship must carry 3.7 x 10675 times as much fuel as its own empty mass! To put that into perspective, the estimated mass of the entire observable universe (excluding exotic forms of mass such as dark matter) is only some 1053 kilograms. A space ship with a very modest 1000 kg empty mass would have to carry 3 x 10625 observable universes' worth of fuel. Most hard SF authors will solve this problem by using more exotic forms of rocket propulsion which have much much higher exhaust velocities, or which can derive their propellant from someplace other than the rocket's fuel tanks. These include:
  • Nuclear fission (NERVA) engines
  • Ion engines, such as those on the Dawn and Deep Space One spacecraft
  • The Orion Drive
  • Controlled nuclear fusion engines
  • Ground-based laser pushers
  • Ramscoops
With the exception of ion engines, all of these are mere drawing-board designs at present, and all of them have practical problems. NERVA engines don't have that much better an exhaust velocity than chemical engines, and require shielding to protect the crew (and the ship's more delicate electronics) from their radioactivity. Ion engines have extremely low thrust levels (the engines on the Dawn spacecraft can, at max throttle, produce about 1/3 of an ounce of thrust). Orion's nuclear putt-putt motor requires an enormous pusher plate that dramatically increases the dead weight the spacecraft has to carry. Controlled nuclear fusion has never been accomplished, at least not in a way that produces more energy than it consumes. Laser pushers would require a ground-based installation to keep a constant lock on the spacecraft as it dwindles away into interstellar space, would consume an enormous abount of power even at 100% efficiency, and could only accelerate the spacecraft away (i.e. they can't be used for braking, unless the destination already has its own laser pusher installation). Ramscoops rely not only on the controlled nuclear fusion of light hydrogen (which is even trickier than the controlled nuclear fusion of heavy hydrogen), but also on the ability to collect the extremely rarefied interstellar gas without inducing significant drag, which might not even be possible. But even if controlled nuclear fusion does become a reality (allowing what Robert A. Heinlein called a torch), that still won't eliminate the need for big rockets if you want to get anywhere in a reasonable amount of time. Sure, your exhaust velocity might now be on the order of (say) 2% of light speed, but the rocket equation still applies. Let's try the trip to Saturn under a continuous 1g acceleration again, only this time let's give the space ship "torchship" engines with an exhaust velocity of 2% of light speed, or 6,000,000 m/s. The trip still requires 7,000,000 m/s of delta-V, so:
7,000,000 m/s = 6,000,000 m/s * ln(M/Me)
Dividing both sides by the exhaust velocity:
(7,000,000 m/s) / (6,000,000 m/s) = ln(M/Me)
To get rid of the natural log, we need to take the natural exponential of both sides:
e1.17 works out to 3.22. So the space ship's fuelled weight will need to be 3.22 times its empty weight. This means it's still going to have to carry 2.22 times as much fuel as its empty mass. You're still stuck with a big rocket. And when you get to Saturn, you'll be out of fuel. You'll need to completely refill your fuel tanks if you want to make the return trip to Earth. Forget about the notion of a "ship" patrolling the "seas of interplanetary space" for months on end, hopping from planet to planet without refuelling.

Realistic World Building

We humans evolved on, and (so far) all grew up on, Earth. We instinctively expect the air to be breatheable, the temperature to be liveable, the gravity to be 9.8 m/s2, the days to last 24 hours, trees and grass, animals and plants and fungi, et cetera, et cetera. The sad fact is, though, that no other planet we've detected thus far is even remotely habitable by human standards. The bigger ones are Jupiter-like balls of gas, while the smaller ones are almost universally airless. The few worlds we've found that do have both an atmosphere and a solid surface have been blanketed in gases that no human can breathe, at pressures anywhere from near-vacuum to 90 times Earth's sea level. While it's theoretically possible that a planet out there might harbor life as we know it, it would have to fit a long, narrow list of parameters, and even then, the kind of life that might have actually evolved there will most likely be very different from the multicellular-eukaryote-rich biome inhabiting Mother Terra. In order for a planet to be able to support life as we know it on its surface at all, it will have to lie in a very narrow range of distances from its parent star. Too close, and any water would evaporate. Too far, and any water would freeze. Liquid water — and life as we know it requires liquid water — can only exist if the planet lies within that narrow zone where it's receiving just the right amount of energy from its star for the surface temperature to allow it. This is called the star's "comfort zone," or "Goldilocks Zone" (as in: not too close, not too far, but juuuuuuuust right). The exact width of a star's Golilocks zone is a matter of some debate, due to the fact that some atmospheres can trap heat (*cough* Venus *cough*) and some can't, and a number of other factors that astrogeologists can make whole careers out of. All we can say for sure is that, for a star as bright and hot as the sun, Venus is too close, Earth is clearly within the Goldilocks zone, and Mars is probably close to the tail end of it. How far away from the star the Goldilocks zone is depends on the star's energy output. A very dim red dwarf star, like Wolf 359, would require a planet to be only about 1.5 million kilometers away from it to receive as much energy as Earth does from our sun — that's only 0.01 A.U., 1% of the Earth-sun distance. A bright and powerful star like Sirius A, on the other hand, would require a planet to be 5 A.U. away from it to receive as much energy as the Earth does from the sun. Interestingly, both of those distances have potentially disastrous consequences. If a planet is only 0.01 A.U. away from its star, the star's tidal influence is going to be enormous. The strength of tidal forces varies directly with the larger object's (i.e. the star's) mass, but inversely with distance cubed. The tidal forces on a planet only 0.01 A.U. from a star 1/10 the mass of the sun are, therefore, going to be 0.1 / 0.013 = 100,000 times as strong as the tidal forces the Earth experiences from the sun. This all but guarantees that the planet will be locked in synchronous rotation with its star — that is, its rotational period must match its orbital period, so the same side is always facing the star. One side of such a planet would be in perpetual daylight, while the other would be in perpetual night. The climate on such a world would be much different than the climate on Earth. Dim stars also have the disadvantage that their Goldilocks Zones are going to be narrower. There is disagreement as to exactly how wide

Traveling - The First Thing on Your To-Do List

There is nothing quite like traveling, like seeing a new place for the first time or returning to a favorite place. People of all ages, from all countries, travel to foreign places for many different reasons – namely work, family and leisure. Whether by plane, train, ship or automobile, travel is generally a pleasurable experience, at least for the people who can financially afford comfortable and safe methods of travel. But it has more benefits than satisfying one’s need to make money, as well as to see loved ones and enjoy one’s self on vacation. There are other benefits of traveling that many people often overlook.


HOW DIETING AND EXERCISING CAN CHANGE YOUR LIFE?


One of the traveling benefits is finding and keeping humility. Too often, people get wrapped up in their lives, their daily routine of working, sleeping, eating and living. They become self-absorbed to the point it affects their health, their happiness, and their perspective. 

It’s a great, big world out there with billions and billions of people, who each day live their life and have their own unique experiences. 

Travel reminds those paying attention that they are not the only ship in the sea, that this is a huge world and that they are only a small, insignificant pea in it. This is quite a humbling experience – to go to another country and see large numbers of peoples living differently, and coming to understand how large the crazy world actually is. When people who learn return home, they keep with them this perspective for the rest of their life and they benefit from this is knowledge and perspective.

Another benefit to traveling is coming to see one’s native country in a different light, in a different way. This is done through being able to compare and contrast home from a foreign location, done most always through traveling. A new perspective may be formed.

Away from home, one comes to understand what “home” actually is and what it means. 

Perhaps their native country is not as free as they had been told or originally thought it to be, for example. One does not understand what it means to be a citizen of their native country until they have seen it from a distance, from another, completely different country. When traveling elsewhere and having to live according to a foreign place’s laws and social norms, one immediately thinks of how things are done in their own country and culture and begins to favor one way or another. This changes how one feels about their native country, whether in a better or worse light. This notion can be applied to various characteristics, such women’s rights, human rights, customs and traditions, beliefs, a trust of government, etc. Traveling is always beneficial for the individual experiencing it.

Another great benefit to traveling is the life experience. Many people in the world do not have the luxury of going to another country for pleasure, or even to another city in their native country for that matter. 

Traveling gets a person out of their comfort zone, away from all their normal pleasures and comforts and way of doing things. 

This forces them to be adventurous, to live life to the fullest, to take the most of this precious gift of life and use the time they have to discover new things, meet new people and experience a completely different life – much like people experience when reading fictional stories: They get to become whoever they are reading about, just like in travel they get to become the citizens of the country they are visiting, even if for just a short time. They live outside themselves.

To conclude, traveling is good for a person of any age. It not only helps people to form a better understanding of themselves, their beliefs and their lives, it also provides people with a better understanding of the world in which they live, even if it's beyond their immediate environment. And it may even help a person to feel connected to the many people living in the world, even if their lives never meet, even if their lives are so completely different that they may as well be from different planets.

There are no hesitations. Go. Privatewriting.com will take care of your academic success. Just place an order and get ready for the trip.

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