So You Want To/Write a Hard Science Fiction Story With Space Travel: Difference between revisions

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:

 

Simplifying:<br />

 

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 :

 

Simplifying:<br />

Taking the square root of both sides:<br />

 

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:

 

 

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.

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'':

 

Simplifying:<br />

Simplifying:<br />

Taking the square root of both sides:<br />

 

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:

 

 

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 1''g'' (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:

 

Subtracting 3,500,000 m/sec from both sides:<br />

Dividing both sides by -9.8 m/sec²:<br />

 

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, let's say you want your inertial-dampener-equipped space ship to accelerate at 1,000''g'', or 9800 m/sec², and you want to follow the same course you did with your piddling little 1''g'' 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:

 

 

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:

 

 

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:

 

 

... 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 [[Useful Notes/Local Stars|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:

 

 

=== Orbits ===

Plugging those two numbers into the formula, we get:

 

 

... 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:

 

 

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.

As an example, the Earth's radius is 6,371,000 meters, and Earth's average density is 5515 kg/m³. Plugging those into the formula above, we get:

 

 

... which is, in fact, how fast things accelerate downward near the surface of the Earth when you drop them. Mars, by contrast, only has a radius of 3,380,000 meters and an average density of 3930 kg/m³, so its surface gravity is only 3.71 m/s², about 38% of Earth's.