Constant acceleration Rockets

One recurring idea is to use continuous thrust in a spacecraft to create artificial gravity. Of course this would need a whole lot of fuel and is thus very impractical. But how impractical exactly? I tried to solve this; my question is whether this is correct:

First of all, we need the travel time if we accelerate constantly, for the first half of the trip, and break for the second half: $$t_t = \sqrt{\frac{x}{a{}}}$$ with acceleration a and distance x

For the travel itself we have the following relationship: $$\dot{m} v_e = m(t)a$$ with $\dot{m}$ being the mass stream of the exhaust and $v_e$ its velocity.

I solve this - but I'm not good at differential equations - to: $$m(t)=m_0 e^{-\frac{a}{v_e}t}$$ and $$\dot m(t)=-m_0 \frac{a}{v_e} e^{-\frac{a}{v_e}t}$$

I have a certain payload, $m_t$ I want to get to the target, so knowing my time to target $t_t$ I get $$m_0=m_t e^{2\sqrt{\frac{x}{a}}\frac{a}{v_e}}$$ or, simpler, $$m_0=m_t e^{t_t\frac{a}{v_e}}$$

My fuel need, then, is the difference between $m_t$ and $m_0$.

• You might find this of interest: williamhaloupek.hubpages.com/hub/Calculations-for-science-fiction-writers-Space-travel-with-constant-acceleration-nonrelativistic – Jerard Puckett Mar 15 '14 at 14:28

You are missing a factor of two in your exponent. The half-way time is $t_h = \sqrt{x/a}$, so the total time is twice that.

However, you apparently incorporated that factor of two here:

$m_0 = m_t \exp\left(2\sqrt{\frac x a}\frac a {v_e}\right)$

Note that this simplifies to $m_0 = m_t \exp\left(\frac {2\sqrt{a x}} {v_e}\right)$

Suppose you want to go a mere 384,500 km (the mean distance between the Earth and the Moon) at 1g and are using a chemical rocket with an effective exhaust velocity of 4400 m/s. This means your initial mass is 1.32×1012 times the final mass. Another way to look at it: Over 99.9999999999% of the initial mass of the vehicle is fuel.

We don't know how to build such a vehicle. We will never know how to build such a vehicle. A realistic rocket has the fuel comprising at most 90 to 95 percent of the initial mass. Your $m_t$ isn't just payload. It also includes the rocket, structure, and empty fuel tanks.

One way around the nastiness of the ideal rocket equation is to use multistage rockets. The Apollo vehicles, for example, were essentially a six stage rocket. It launched with the three stage Saturn rocket, and the payload itself was another three stage rocket. This helps, but not near enough so that you can accelerate / decelerate the whole way. The Apollo vehicles coasted most of the way to the Moon and then coasted most of the way back to Earth. The vehicle that is en route to Pluto is coasting and has been for years.

Another way around this is to use a much lower acceleration. Why do you need to go at 1g? Yet another way around this is to use a better rocket. 4400 m/s is barely enough to get us off the surface of the Earth. Ion thrusters have a much higher specific impulse. However, ion thrusters have no oomph.

Back to your original question,

One recurring idea is to use continuous thrust in a spacecraft to create artificial gravity. Of course this would need a whole lot of fuel and is thus very impractical. But how impractical exactly?

The answer is that it is extremely impractical with the rockets we now know how to make. Maybe it would be possible if we had nuclear rockets, or even better, antimatter rockets. We don't. For now, those are the stuff of science fiction.

High specific impulse engines do fire all the time, but not to create artificial gravity. Ion thrusters have to fire almost all the time because that's the only way to make the rocket go anywhere. For example, it took over a year for SMART-1 to reach the Moon.

• I believe you meant an effective exhaust velocity of 4400 m/s, not an escape velocity of 4400 m/s. – a CVn Jan 12 '17 at 15:58
• Another Way to do it is to use some as-yet-nonexistent high-thrust/high-ISP engine. That said I think Zurbin thrusters belong in Worldbuilding, not sx. – UIDAlexD Jan 12 '17 at 16:38