tGiven a propulsion system with constant power, how long does it take to travel a given distance in space? (non-rel. and relativistic)

Question

Suppose a space ship of mass $m$ is travelling away from our solar system, starting with "starting speed" $v_{start} > v_{escape\ \odot}$, meaning it will escape sun and have some velocity left over, $v_0$.

On board of the ship, there is a power source from which a constant power $P$ is drawn for propulsion. The propulsion system works by exhausting relativistic particles at such high speeds (and thus gamma factory) that the kinetic energy drawn from the power source dominates and we do not loose significant rest mass. (the power source itself works by converting the rest mass of little parts of the ship into energy though).

Given a target, i.e. knowing the distance $s$ we have to travel, we want to know the time $t_s$ it takes to get there as well as the speed $v(t)$ and acceleration $a(t)$ over time.

Non-relativistic

First the simplified task: $m$ is considered constant and we assume a start in empty space, ignoring sun.

so to summarize, we have

• mass $m$
• starting time $t_0 = 0$
• initial speed $v_0$
• distance $s$
• power $P=const.$ of the propulsion system

and we are looking for

• travel time $t_s$
• velocity $v(t)$
• acceleration $a(t)$

In the non-relativistic case for $v_0 = 0$, I think I've got the solution, but not 100% sure it is correct: we start with $$P=F\cdot v = m\dot vv$$ $$\Rightarrow v(t)\dot v(t)=\frac{P}{m}$$ Integration: $$\int_0^t{v(t')\dot v(t') dt'}=\frac{v^2(t)}{2}=\frac{P}{m}t$$ which gives us $$v(t)=\sqrt{2\frac{P}{m}t}$$ and $$a(t)=\dot v(t)=\sqrt{\frac{P}{2mt}}$$ another integration gives $$s(t)=\int_0^t{\sqrt{2\frac{P}{m}t'} dt'} = \sqrt{\frac{8}{9}\frac{P}{m}t^3}$$ $$\Rightarrow t_s=\left(\frac{9s^2m}{8P}\right)^{\frac{1}{3}}$$

In all three results the units are correct, which gives me some confidence in the solutions. Intuitively, I can't grasp this though, as it leads to different changes in speed at different velocities relative to earth although we are in empty space, thrusting forward with constant power. But constant power does not lead to constant force, guess I'll just have to live with that xD

But that's the reason I'm missing $v_0$ here:

Since with constant power, the change in velocity is apparently dependent on the velocity we already have, I don't know hot correctly to integrate a non-zero $v_0$.

Can you help me with that?

Relativistic

Trying the same approach here leads me to some overwhelming equation I can't solve. Starting with the relativstic force $$F=\gamma^3(v(t))m_0\dot v(t) $$ with rest mass $m_0$ of the ship; I will not write indexes in the following formulas. using the same approach as above, we have $$P=F\cdot v = m_0\left(1-\frac{v^2}{c^2} \right)^{-\frac{3}{2}} v\dot v$$ $$\Rightarrow\left(\frac{P}{m_0}\right)^2=\frac{v^2\dot v^2}{\left(1-\frac{v^2}{c^2}\right)^3}$$ expanding the binomial $$\left(\frac{P}{m_0}\right)^2\left(1-3\frac{v^2}{c^2}+3\frac{v^4}{c^4}+\frac{v^6}{c^6}\right)=v^2\dot v^2$$

now I don't know how to solve this for $v$, $\dot v$, or $t$. One idea was that maybe a substitution $z:=v^2 \rightarrow \dot z = 2v\dot v$ would make things easier. Also, to save some space, let's use $k:=\left(\frac{P}{m_0}\right)^2$ $$\frac{\dot z^2}{2} = \frac{k}{c^4}Z^3 + \frac{3k}{c^2}z^2-\frac{3k}{c^2}z+k$$

but neither am I able to solve this for $z(t)$ to then re-substitute it to get v, nor do I know if I'm moving in the right direction at all. Or if the results are in the earth or the ship system, for that matter. And of course here, too, $v_0$ is missing. Happy to hear your solutions. If needed, I can provide more intermediate steps in the calculation.

Bonus

Given $s$ as a single-digit number of light years and the ship has a quite modest propulsion, so that $t_s$ is around $1 000$ years , do the following considerations / changes alter the result significantly?

1. $v_0$ is not actually the start velocity, but the excess speed above the escape velocity from the sun; we actually start with $v_{start} = v_{escape\ \odot} + v_0$ at $1$ AU from sol (but largely outside of Earth's gravity well).
2. the mass is not actually constant. Our power source slowly "eats" the ship by converting is's rest mass into the energy for propulsion. Will it become significant if we don't have 100% efficiency and can only use around $\frac{1}{10}th$ of the power for propulsion? (i.e. "eating" 10 times more rest mass of our ship than we get out in kinetic energy)

Generally, there is a lot of material to be found dealing with constant acceleration or constant force, but not for constant power, oddly enough.

Answer 1

The propulsion system works by exhausting relativistic particles at such high speeds ... that the kinetic energy drawn from the power source dominates and we do not loose significant rest mass.

I don't know what you mean by "the kinetic energy drawn dominates". The best you can do for limiting mass loss is to make the exhaust as low-mass as possible is for the exhaust to be photons.

The problem is that the lighter the exhaust and the more mass efficient your drive, the less energy efficient it becomes. Your engine might generate a constant power, but most of that energy goes into the exhaust, not into the ship (at least in a frame where the ship is not moving at high speed).

A constant power draw engine means constant thrust. Regular rocket engines are this way, just they have run times in the minutes, not years. And for even very short burns, the mass of the craft is not constant. For a photon drive, you could consider the mass to be constant for quite some time.

The P in the $P = Fv$ equation you have is the power delivered to the spacecraft. It is not equal to the power consumed by the engine (since some of the power has to go into accelerating the exhaust). In your constant-consumption engine, you do not have constant power delivery. Since you have constant force, you can use those other sources you have found.