View Full Version : Math of an accelerating spaceship
I'm considering doing a sci-fi comic or novel in an Einsteinian universe, no wormholes or faster-than-light travel. I haven't seen this done very often. It's been a while since I've worked with math, though, so I'm having some trouble.
First, I just want you to check my math. The relativistic mass of an object is
m(v) = m_o / \sqrt{1-v^2/c^2}
From that I've arrived at
v = c - m_o c / m(v)
How's my algebra? Now, just to be sure, m(v) is defined as the kinetic energy of my spaceship converted into mass plus the rest mass of the spaceship, correct? So, supposing the ship is powered by anti-matter or otherwise can perfectly convert its fuel into kinetic energy, than the amount of fuel needed would then be:
m(f) = m(v) - m_o
Is that right?
Now here's where I'm running into trouble. I can't figure out how much fuel is needed for the spaceship to accelerate up to speed, and then decelerate back to zero velocity. My instincts are saying to square the mass of the fuel, but I can't prove it.
And here's where it's starting to get really complicated. I'm probably going to want to have the spaceship constantly accelerating at 1G up until the midpoint in the trip, and then decelerating at 1G until it reaches its destination. I don't remember anything about differential equations, so I don't trust myself enough to figure out the length a trip would take either from the reference frame of the ship or the reference frame of the planets. Furthermore, would that even work, or would I need to incorporate general relativity as well? I'm hoping that would be negligible, as 1G is basically Newtonian.
From:
http://www.ksc.nasa.gov/facts/faq04.html
Relativistic rocket equation (constant acceleration)
t (unaccelerated) = c/a * sinh(a*t/c)
d = c**2/a * (cosh(a*t/c) - 1)
v = c * tanh(a*t/c)
Relativistic rocket with exhaust velocity Ve and mass ratio MR:
at/c = Ve/c * ln(MR), or
t (unaccelerated) = c/a * sinh(Ve/c * ln(MR))
d = c**2/a * (cosh(Ve/C * ln(MR)) - 1)
v = c * tanh(Ve/C * ln(MR))
Wow thank you, that answers most of my questions. Not sure how trig got in there, but I have the answer I'm looking for either way. I do have a few questions though:
d = c**2/a * (cosh(a*t/c) - 1)
d = c**2/a * (cosh(Ve/C * ln(MR)) - 1)
What is the meaning of the double asterisk?
Also, I'm still not sure on what total fuel is required because, for one, the exhaust velocity equation doesn't really apply in my case and, two, it still doesn't answer the question of deceleration. Does it?
The double asterick means "to the power of", I don't know why they didn't use "^".
Unless you are going for some type of "Reactionless drive"( in which case, if you are going to break one law of physics, why not others?), you are going to have to throw something out the back to get your ship moving forward.
The faster what you throw out moves, the more efficient your drive.
Assuming you convert your antimatter into photons, and use the photons as your exhaust, then the exhaust velocity will be c.
In that case use the second equation to solve for the Mass Ratio (using c for Ve), to find out how much fuel it would take to accelerate up to speed. Add this mass to the empty mass of the ship to get your new empty mass, and with this and the mass ratio determine the total fuel.
Example, assume you come up with a mass ratio of 2, (ship + fuel)/ship =2
This puts your fuel mass at 1 ship mass. (this will actually be the fuel needed to decelerate).
This puts your "starting ship" mass for acceleration at 2 (empty ship mass + fuel needed for later deceleration).
The fuel needed to accelerate up to speed will be one " starting ship" mass, meaning 2 "empty ship" masses.
This gives a total fuel of 3 "empty ship" masses.
Meaning three quarters of your total mass of the ship will be fuel in this example.
Wow thank you, that answers most of my questions. Not sure how trig got in there, but I have the answer I'm looking for either way. I do have a few questions though:
d = c**2/a * (cosh(a*t/c) - 1)
d = c**2/a * (cosh(Ve/C * ln(MR)) - 1)
What is the meaning of the double asterisk?
It's just a weird way of writing c^2.
Also, I'm still not sure on what total fuel is required because, for one, the exhaust velocity equation doesn't really apply in my case and, two, it still doesn't answer the question of deceleration. Does it?
You can find big tables of effective exhaust velocities for different propulsion methods on the wikipedia article on spacecraft propulsion (http://en.wikipedia.org/wiki/Spacecraft_propulsion) as well as this page (http://www.projectrho.com/rocket/rocket3c2.html), they say an antimatter rocket would have an effective exhaust velocity of 10 million - 100 million m/s. The speed of light is a little under 300 million m/s, so that would be the theoretical upper limit on exhaust velocity for any conceivable rocket.
As for deceleration, well, if you want to accelerate at a certain rate for half the trip and decelerate for the second half, just find the ratio between payload mass and initial total mass that would be needed for the ship to go half the distance, which tells you the mass that the rocket must have at the midpoint of the journey before decelerating; then treat this total mass as the "payload mass" for the first half of the trip. This will mean that the ratio between the total mass at the beginning of the trip and the final payload mass will just be MR^2.
By the way, this relativistic rocket page (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html) considers a perfectly efficient photon rocket with an effective exhaust velocity of exactly c, and finds that the mass ratio MR works out to e^(aT/c) - 1. T is the shipboard time to get to the destination at constant acceleration a, and the page earlier gave equations that give T if you know a and d; so if you want acceleration followed by deceleration, you could just find the T for a d that's halfway to the destination, plug it into that equation to get the mass ratio for half the trip, then square it to get the mass ratio for the total trip. For example, they show on a table that to get to Vega at 1G, 27 l.y. away, you'd need 57 kg of initial mass for every kg of final mass at the destination, so MR = 57, meaning that if you wanted to accelerate for 27 l.y. and then decelerate for another 27 l.y. the mass ratio would have to be 57^2 = 3249. And to get to the center of the galaxy 30,000 l.y. away, you'd need 62 tonnes (=62,000 kg) of total initial mass for each kg of final mass, which means that if you wanted to accelerate at 1G for 30,000 l.y. and then decelerate for another 30,000 l.y. you'd need a mass ratio of 62,000^2 = 3.844 billion. These sorts of numbers show why no one thinks that long interstellar journeys could be carried out by a rocket that carries its fuel with it; other options that don't require taking all your fuel with you would be things like the Bussard ramjet (http://en.wikipedia.org/wiki/Bussard_ramjet) which scoops interstellar hydrogen for fuel (though unfortunately recent calculations seem to show the idea wouldn't work for acceleration since the drag created by scooping the hydrogen would outweigh the forward acceleration; but it has been suggested that the drag could actually be exploited so that the scoop would work for decelerating an interstellar ship relative to the interstellar medium) or some kind of beamed propulsion (http://en.wikipedia.org/wiki/Beam-powered_propulsion) like the starwisp (http://en.wikipedia.org/wiki/Starwisp) where an ultrathin sail is pushed along by a giant maser (microwave version of a laser) constructed in our solar system.
Incidentally, if you're interested in reading a summary of recent thinking on plausible interstellar travel, you might check out the paper The Interstellar Conundrum (http://www.blackwell-synergy.com/doi/abs/10.1196/annals.1370.004) by Paul Gilster; it's not available for free online, but if you want I could email you a copy if you send me a PM with your email address.
In that case use the second equation to solve for the Mass Ratio (using c for Ve), to find out how much fuel it would take to accelerate up to speed. Add this mass to the empty mass of the ship to get your new empty mass, and with this and the mass ratio determine the total fuel.
Example, assume you come up with a mass ratio of 2, (ship + fuel)/ship =2
This puts your fuel mass at 1 ship mass. (this will actually be the fuel needed to decelerate).
This puts your "starting ship" mass for acceleration at 2 (empty ship mass + fuel needed for later deceleration).
The fuel needed to accelerate up to speed will be one " starting ship" mass, meaning 2 "empty ship" masses.
This gives a total fuel of 3 "empty ship" masses.
Meaning three quarters of your total mass of the ship will be fuel in this example.
Are you talking about accelerating and then decelerating for the same amount of time? As I said in my post, I thought in this case you'd find the mass ratio needed for half the trip and then square it; when you calculate the mass ratio for half the trip that tells you the total mass (payload + fuel) the ship will need to have at the midpoint of the journey in order to be left with nothing but payload after the second half is finished, and then that total mass can be treated as the "payload" that must be accelerated for the first half of the trip.
Are you talking about accelerating and then decelerating for the same amount of time? As I said in my post, I thought in this case you'd find the mass ratio needed for half the trip and then square it; when you calculate the mass ratio for half the trip that tells you the total mass (payload + fuel) the ship will need to have at the midpoint of the journey in order to be left with nothing but payload after the second half is finished, and then that total mass can be treated as the "payload" that must be accelerated for the first half of the trip.
Actually, on second thought I'm not sure this argument works. Although the deceleration and acceleration both cover the same distance in the frame of the Earth and the destination, if we go to the rocket's own instantaneous inertial rest frame at the midpoint and then look at the second leg of the trip from this frame, here the rocket is also accelerating at 1G from an initial rest state just like it did when leaving the Earth in the Earth's rest frame, but the distance between the midpoint and the destination in this frame is signficantly shorter than the distance between the Earth and the midpoint in the Earth's frame, which suggests the second leg of the trip would not require as large a mass ratio (edit: although in this frame the ship is actually accelerating away from the destination while the destination approaches the ship at a high constant velocity, which makes things more complicated).
Fortunately the relativistic rocket page (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html) has already worked out the math for a rocket that accelerates for the first half and decelerates for the second half:
What if we prefer to stop at the destination? We accelerate to the half way point at 1g and then immediately switch the direction of our rocket so that we now decelerate at 1g for the rest of second half of the trip. The calculations here are just a little more involved since the trip is now in two distinct halves (and the equations at the top assume a positive acceleration only). Even so, the answer turns out to have exactly the same form: M/m = exp(aT/c) - 1, except that the proper time T is now almost twice as large as for the non-stop case, since the slowing-down rocket is losing the ageing benefits of relativistic speed. This dramatically increases the amount of fuel needed:
d..................Stopping at:...............M
4.3 ly............Nearest star...............38 kg
27 ly.............Vega.........................886 kg
30,000 ly.......Center of our galaxy....955,000 tonnes
2,000,000 ly...Andromeda galaxy......4.2 thousand million tonnes
Compare these numbers to the previous case: they are hugely different! Why should that be? Let's take the case of Laurel and Hardy, two astronauts travelling to Vega. Laurel speeds past without stopping, and so only needs 57 kg of fuel for every 1 kg of payload. Hardy wishes to stop at Vega, and so needs 886 kg of fuel for every 1 kg of payload. Laurel takes almost 28 Earth years for the trip, while Hardy takes 29 Earth years. (They both take roughly the same amount of Earth time because they are both travelling close to speed c for most of the journey.) They travel neck-and-neck until they've both gone half way to Vega, at which point Hardy begins to decelerate.
It's useful to think of the problem in terms of relativistic mass, since this is what each rocket motor "feels" as it strives to maintain a 1g acceleration or deceleration. The relativistic mass of each traveller's rocket is continually decreasing throughout their trip (since it's being converted to exhaust energy). It turns out that at the half way point, Laurel's total relativistic mass (for fuel plus payload) is about 28m, and from here until the trip's end, this relativistic mass only decreases by a tiny amount, so that Laurel's rocket needs to do very little work. So at the halfway point his fuel:payload ratio turns out to be about 1.
For Hardy, things are different. He needs to decrease his relativistic mass to m at the end where he is to stop. If his rocket's total relativistic mass at the halfway point were the same as Laurel's (28m), with a fuel:payload ratio of 1, Hardy would need to decrease the relativistic mass all the way down to m at the end, which would require more fuel than Laurel had needed. But Hardy wouldn't have this much fuel on board--unless he ensures that he takes it with him initially. This extra fuel that he must carry from the start becomes more payload (a lot more), which needs yet more fuel again to carry that. So suddenly his fuel requirement has increased enormously. It turns out that at the half way point, all this extra fuel gives Hardy's rocket a total relativistic mass of about 442m, and his fuel:payload ratio turns out to be about 29.
Another way of looking at this odd situation is that both travellers know that they must take fuel on board initially to push them at 1g for the total trip time. They don't care about what's happening outside. In that case, Laurel travels for 28 Earth years but ages just 3.9 years, while Hardy travels for 29 Earth years but ages 6.6 years. So Hardy has had to sit at his controls and burn his rocket for almost twice as long as Laurel, and that has required more fuel, with even more fuel required because of the fuel-becomes-payload situation that we mentioned above.
This fuel-becomes-payload problem is well known in the space programme: part of the reason the Saturn V moon rocket was so big was because it needed yet more fuel just to carry the fuel it was already carrying.
Thank you both, you've been very helpful.
Janus:
Unless you are going for some type of "Reactionless drive"( in which case, if you are going to break one law of physics, why not others?), you are going to have to throw something out the back to get your ship moving forward.
No, of course it is going to in some way involve action and reaction. I guess I just didn't think it would be as simple as using the mass equivalent of the energy of the fuel combined with using the speed of light as the exhaust velocity. That just seemed too simple...
Hmm, quite a few different answers on how much fuel is needed. I guess I'll look into those links JesseM. It's surprising how complicated this has become.
The problem of the massive quantities of fuel needed is a tough one. I've been considering using stars as the fuel source. The sun produces enough energy in one second to accelerate a carnival cruise ship to 97% the speed of light. Pretty cool! I'm just having trouble thinking of just how to harness that. I guess you could encase the star in a dyson sphere and use the energy to power particle accelerators to produce antimatter, but then I've still got to think of a way to store it on the ship. And that's not very good for a round trip either! I'm considering just cheating a little and making up some kind of weird quantum field that drains the energy of a star into the spaceship at a distance, but it sounds a little out there. Still more plausible IMO than warp drive though. I'll definitely look into that link you gave for methods of space travel.
Thanks again.
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