25
$\begingroup$

The answer to What happens to an astronaut who's floating in a spaceship (in space) when it begins to move? is that we can create g-force just by accelerating. So then why go all the way to create rotating ring?

The ring makes sense for ISS, but why use it for spaceships like the one I see on The Martian. Is there any reason why they can't just accelerate to create gravity?

$\endgroup$
  • 10
    $\begingroup$ No rotating spacecraft with artificial gravity has actually been built. The ISS does not rotate. $\endgroup$ – Tor Klingberg Jul 15 '16 at 12:40
  • $\begingroup$ @TorKlingberg: I believe there was an experiment in doing this by binding two crafts, "bolo style", the cable broke though. It would take a while to find the references. $\endgroup$ – SF. Jul 15 '16 at 13:29
  • 2
    $\begingroup$ If you have torchships, then you can. Of course that is in the realm of science fiction, not real world science or space exploration. $\endgroup$ – hyde Jul 15 '16 at 16:48
  • 1
    $\begingroup$ @TorKlingberg Gemini 11 succeeded to do so to some degree. $\endgroup$ – called2voyage Jul 15 '16 at 17:47
  • 2
    $\begingroup$ Hi Kirie - you could do exactly, precisely what you say - you are 1000% correct. But, extremely simply, it would use a mind-boggling amount of fuel. It would be utterly impossible to sustain it for more than a few minutes. It's that simple. $\endgroup$ – Fattie Jul 15 '16 at 20:34
54
$\begingroup$

Constant acceleration requires energy. Our current rocket engines need to use propellant to provide that energy. And there just cannot be enough propellant to generate artificial gravity for any meaningful duration. We would need a new type of space drive to be able to use acceleration that way.

The concept is well known from (science-)fiction (sometimes named "Torchship") and the artifical gravity provided is actually sort of a side effect. The main benefit of a ship able to accelerate at $1G$ fo a long time would be the speed with which it can travel across the Solar System - Mars in two days, Jupiter under one week. But we are not sure if such propulsion system is even possible in reality. Often cited possibilities which might allow it in theory are fusion and antimatter drives.

$\endgroup$
  • 2
    $\begingroup$ -1 Acceleration does not require energy. This statement is misleading, armwavey and frankly just plain WRONG. $\endgroup$ – Aron Jul 18 '16 at 8:00
  • 1
    $\begingroup$ @Aron might you please explain a bit more? In my understanding acceleration is generated by force acting on the ship (but not its payload/passengers, otherwise no artificial gravity would be felt - so no acceleration from gravity), and for spaceships the force is mostly generated by their engines converting the energy stored in propellant. I skipped solar/laser sails for which the energy source is external. $\endgroup$ – jkavalik Jul 18 '16 at 8:08
  • 1
    $\begingroup$ @jkavalik It is simple physics. Conservation of Energy. As long as the acceleration is tangential to the direction of motion, no energy transfer occurs. This can occur in any form of force field, for example, but not limited to, gravity, electrical, magnetic fields. $\endgroup$ – Aron Jul 18 '16 at 8:11
  • 2
    $\begingroup$ @jkavalik Much more significant to energy is that LINEAR acceleration in a spacecraft requires reaction mass, which cannot be regenerated on a spacecraft. Coupled with the finite limit of the exhaust velocity, we have the "tyranny of the rocket equation". $\endgroup$ – Aron Jul 18 '16 at 8:13
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – called2voyage Jul 18 '16 at 12:13
45
$\begingroup$

jkavalik gives the right answer. But to put this in perspective, let me add some numbers.

Let's say, we use a battery of state-of-the-art ion drives to retain a semi-comfortable, Martian level of gravity, of 3.73 m/s2 for a period of 24 hours.

24 h is 86400 seconds. At 3.73 m/s2 this gives 322,272 m/s of delta-V.

Let's use a large number of VASIMIR thrusters, of 12,000 s of specific impulse. Substituting to the Rocket Equation:

$$ \begin{align} \Delta v &= I_\mathrm{sp} g_0 \ln { m_0 \over m_\mathrm f }\\ \ln { m_0 \over m_\mathrm f } &= { \Delta v_0 \over { I_\mathrm{sp} g_0 } }\\ { m_0 \over m_\mathrm f } &= \mathrm e^{ \Delta v_0 \over { I_\mathrm{sp} g_0 } }\\ &= \mathrm e^{ 322272 \over { 12000 \cdot 9.8 } }\\ &= \mathrm e^{2.74}\\ &= 15.48 \end{align}$$

That means, that to maintain gravity of Mars, about third of Earth's, for a period of 24 hours, using ones of the best ion drives we have currently, the craft would need to use 14.48 times its own weight in fuel.

To maintain it over another (earlier) 24 h, it would need 14.48 times the weight of craft plus fuel needed for the second day – about 240 times its own weight. Third day? 3700 times.

So, for 3 days of travel at mild Martian gravity, in a capsule of 10 tons, we'd need to deliver 37 thousand tons of propellant to the orbit.

$\endgroup$
  • 3
    $\begingroup$ You forgot to mention how you'd need more energy than the entire planet provides to power those thrusters. At 24 hours left of thrust, you'd need more than a terrawatt of power. $\endgroup$ – BWG Jul 15 '16 at 19:06
  • 1
    $\begingroup$ @BWG: For such a battery of ion thrusters, probably. Of course we do have engines that would run on less power, but they don't have nearly as good ISp. An NTR of 900s, surely capable of such acceleration, would use 3.5 craft's weight in fuel in an hour. Our best hope would be a huuuuuuuge solar sail. At 1AU we have 4.8N per km^2 of the sail, so getting enough thrust to accelerate a 10 ton craft to 1g would require a sail about the size of Israel. $\endgroup$ – SF. Jul 15 '16 at 20:49
  • $\begingroup$ The best option IMO is to push off a common background, such as the earth. Making an ungodly strong magnetic field is much easier than finding a terrawatt power source. $\endgroup$ – BWG Jul 15 '16 at 22:11
  • $\begingroup$ @BWG: I don't quite see sustaining acceleration for any considerable period of time that way. I mean, you'll be escaping the magnetosphere in no time. $\endgroup$ – SF. Jul 16 '16 at 14:31
  • 2
    $\begingroup$ @Luaan: Actually, electromagnetic propulsion is being researched - but instead of "ungodly" powers, we're using a very modest current, and the thrust is barely enough to keep the satellite orbiting, never mind accelerating out of the system. Unfortunately any electric or magnetic loop produces net zero thrust. So it needs to be "open circuit" - emit positive ions (or grab free electrons from the void) on one end, emit electrons on the other - and these emitters have a rather poor efficiency. $\endgroup$ – SF. Jul 16 '16 at 22:31
9
$\begingroup$

Basically, we just don't have engines that can accelerate at 1G, or anywhere near that, for more than a few minutes. Not only do we not currently have such engines, we aren't even sure when we will. Nuclear-thermal designs can get at least twice the $I_{SP}$ of chemical rockets, meaning the same amount of thrust can be maintained for twice as long for the same starting mass. (This doesn't mean 1G for twice as long, exactly, but it's fairly close.) Of course, twice as long as "a few minutes" still isn't very good. Project Orion and similar nuclear pulse propulsion designs can achieve much higher $I_{SP}$, but require setting off 1kt nuclear charges repeatedly. Both of these rely on fairly well-proven principles, but we don't have mature designs for them yet, and the current climate toward nuclear reactors and bombs in space is unfavorable, so there's not much serious research.

In particular, the Medusa variant of Orion (using a sail to intercept the nuclear blast along with long tethers reeled in and out to spread its acceleration bursts over time) can likely achieve $I_{SP}$ of close to 100000 s. Consider a mission to Mars at a fairly close approach (<75 million km, with the closest approach being about 40 million km) and a constant 0.3g acceleration/deceleration (to match Mars gravity ahead of time). Per Atomic Rockets, the equation for total time is $t = 2 \sqrt{D/a}$ (which comes out to be about 3.7 days), and the equation for required $\Delta V$ is $\Delta V = 2 \sqrt{D \cdot a}$. Finally, the Tsiolkovsky rocket equation is of course $$\Delta V = v_e \ln {m_0 \over m_1}$$ Assuming a 100 ton dry mass (probably rather low) we can substitute: $$2 \sqrt{D \cdot a} = v_e \ln {m_0 \over m_1}$$ $$m_0 = m_1 \cdot e^{2 / v_e \cdot \sqrt{D \cdot a}}$$ The result is a starting mass of around 270 tons, which means 170 tons of fuel pellets in the form of 1kt nuclear shaped charges, so probably about half highly-enriched plutonium or U-238 and half bomb casing. Needing to acquire close to 100 tons of weapons-grade plutonium for a short trip to Mars is difficult, but not outright impossible. Accepting a lower gravity of, say, 0.1g the whole time (which may or may not be enough for comfort, but over a short duration, in this case about 6.3 days, is not going to affect bone density much) would cut the required propellant mass in half.

Other designs either don't have enough thrust to create decent gravity, or don't have enough $I_{SP}$ for continuous running during a voyage, or are very speculative. (Antimatter rockets, say, are a very long ways from being practical.)

$\endgroup$
  • $\begingroup$ This answer isn't wrong, but I get the feeling that it's far more technical than what the OP is looking for. $\endgroup$ – a CVn Jul 15 '16 at 12:07
  • $\begingroup$ "...we aren't sure when we will."[have engines capable of such thrust and duration]". While true, that statement is less than honest. There is no known technology which can come even close to sustained 1-g acceleration. Including VASIMR. period. Until we get to the geopolitical point where we can use nuclear reactors in space, our options consist of low speed long duration spaceflight. $\endgroup$ – user16253 Jul 15 '16 at 18:03
  • $\begingroup$ @LiZhi That is exactly what I said and meant. The answer mentions the unfavorable attitudes toward nuclear reactors, and alludes to the other designs that don't have enough thrust. $\endgroup$ – Nathan Tuggy Jul 15 '16 at 18:47
  • $\begingroup$ I assume that even Orion cannot sustain 1g for a trip through the solar system, with reasonable designs and assumptions about available nuclear material etc $\endgroup$ – Peter A. Schneider Jul 15 '16 at 19:45
  • $\begingroup$ @PeterA.Schneider I'd have to do a more detailed workup, but I think either Orion or its more efficient tractor variant Medusa could reasonably maintain at least Mars gravity if not full 1G for the trip to Mars. It would only be a few days. $\endgroup$ – Nathan Tuggy Jul 15 '16 at 20:04
5
$\begingroup$

Acceleration costs the energy. Let's ignore everything else - suppose we are in free space, there is no gravity, no drag, no other motions. then energy $E$ needed to gain velocity $v$ is $E=1/2 mv^2$.

If you want to maintain given acceleration $g$ you will ned a power source that can provide output of $P(t)=mgv(t)$ (I've just derived the energy by time) all the time.

If we use rotation to get feeling of gravity we just need to rotate the ring. The "gravity" will be $g=\omega^2 r$ and the energy needed to accomplish that is $E=1/2J\omega^2=\frac{J\cdot g}{2r}$.

No matter how the engines would work, how effective they would be; the gravity by rotation will be more effective that gravity by propulsion.

$\endgroup$
  • 2
    $\begingroup$ Also important: acceleration costs mass, as you have to sacrifice reaction mass to conserve momentum. Sooner or later you run out. $\endgroup$ – Hohmannfan Jul 15 '16 at 14:09
  • $\begingroup$ "a power source that can provide output of P(t)= ... all the time" - Important to note: To maintain constant acceleration, you'll need more and more power (~t²) because v is a multiplicand in the equation. $\endgroup$ – JimmyB Jul 15 '16 at 16:20
  • 1
    $\begingroup$ @Hohmannfan - even for a hypothetical reactionless drive, acceleration still costs energy continuously, and rotation in principle only requires one spin-up's worth of energy. $\endgroup$ – Joffan Jul 15 '16 at 16:34
  • $\begingroup$ @JimmyB: it's actually even worse than quadratic. ∝ 𝑡² is only when you have something to push off from, like a car or jetplane has. For any spacecraft, it gets exponential instead, because you can only gain momentum by expelling propellant that you need to bring yourself and first accelerate to you own speed. $\endgroup$ – leftaroundabout Jul 16 '16 at 13:35
  • $\begingroup$ @Hohmannfan a photon rocket needs no sacrificial mass. It is well within the laws of physics to eject pure momentum and no mass other than the needed kenetic energy. $\endgroup$ – JDługosz Jul 18 '16 at 4:54

protected by Community Jul 15 '16 at 18:48

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.