# Calculate acceleration from power [closed]

I am currently programming a simple computer space ship simulation. This is just for fun to train my programming, 3D graphics and a little bit of physics skills. What I want is the user to pick a space ship that has a mass and power. I could just use force, but somehow power sounds better, because you can compare it to cars if you have no knowledge in physics. Without any reason I would like that to have at least a small physics background that is not completely off. So here is my question:

What is the acceleration of a 75 kg space ship with a 1 hp rocket engine?

I know there is nothing like a 1 hp rocket engine, but let's define:

A 1 hp rocket engine is an engine that has the power to raise a 75 kg space ship against the earth's gravitational force of $9.80665 \, \text{m/s}^2$ over a distance of one metre in one second (see Wikipedia).

Now let's assume there is no air friction and the space ship somehow does not lose mass while accelerating. This does not have to be completely realistic. What is the acceleration of the space ship, if you turn off the earth's gravitational force?

I have

$$P = \frac{W}{t} = \frac{F\cdot s}{t} = m a v$$

so

$$a = \frac{P}{m v}$$

but I do not know the velocity $v$.

I hope you can somehow point me in the right direction or provide a better solution for my simulation.

• A rocket engine doesn't work the way a car engine works. A rocket engine produces a certain amount of thrust for a certain amount of time before the fuel runs out. Try to understand the concept of specific impulse: en.wikipedia.org/wiki/Specific_impulse – CuriousOne Jul 6 '15 at 21:09
• assume [that] the space ship somehow does not loose mass while accelerating. That's a very un-realistic assumption. The fuel for a launch-to-orbit burn may account for more than half the pre-launch weight/mass of the entire stack. The change in weight/mass as the rocket burns its fuel is a very significant part of the whole picture. – Solomon Slow Jul 6 '15 at 21:13
• That's true, but I think the user does not like the fuel to be empty all the time... – Toxiro Jul 6 '15 at 21:31
• If the user doesn't like that, then the user doesn't have the right stuff! – CuriousOne Jul 6 '15 at 21:47
• Ok, I am sorry, if I am in the wrong forum here, but this is for a small computer simulation/game, just for fun, so the fuel just should not be empty, because the user wants to keep flying through space. Probably I should read more about what power is. I think there is something that I just do not want to understand... – Toxiro Jul 6 '15 at 22:22

If you really had a "constant power" engine, and all that power was transferred to your rocket which does not lose mass, it would result in a linear increase in the kinetic energy.

And since the kinetic energy $E=\frac12 m v^2$, you can find the velocity at a given time from

$$P\cdot t = \frac12 m v^2\\ v = \sqrt{\frac{2 \cdot P \cdot t}{m}}$$

If you wanted the acceleration as a function of time, you would differentiate...

$$a = \frac{dv}{dt} = \sqrt{\frac{ P }{2m\cdot t}}$$

But note

• Rocket engines provide thrust not power
• Mass changes significantly as fuel is used up

So this is totally unrealistic.

• Wouldn't that be the equation from a static reference frame outside of the space ship? I thought a rocket engine provides a somehow constant acceleration or is that just because of the mass change? It also doesn't necessarily has to be a rocket engine, it could be an ion thruster or maybe a solar sail... – Toxiro Jul 6 '15 at 21:45
• Rocket engine providing constant thrust (not power) would indeed produce constant acceleration except for the fact that the mass of the rocket changes. But you asked about a rocket motor with constant power. Maybe that was not what you meant... – Floris Jul 6 '15 at 21:58
• Or maybe I meant the power from a reference frame within the space ship or is that not possible? Somehow like a battery that's constantly providing 1hp electricity and this is fully used with an ion-thruster to accelerate ions - no energy is lost. I guess I am getting something totally wrong here... – Toxiro Jul 6 '15 at 22:15

As @Floris said, rocket motors provide thrust (reaction force), not power. The thrust is just the mass times velocity of exhaust, per second. The power is the mass times velocity squared, per second (over two).

Example, shooting a rifle bullet has low reaction force, but high energy. Shooting a bowling ball with the same reaction force takes a lot less energy.

So if a rocket engine could eject bowling balls, it would take a lot less energy than if it could eject something very light (at very high speed).

So bowling balls would make excellent exhaust mass. The trouble is, you'd run out of them very quickly.

The reason rockets have very high power is just so they can make the fuel last as long as possible. That's why ion-engines are being considered for long-term thrusting in space. They sip fuel by ejecting it at a velocity far higher than a chemical reaction could generate.

• The good thing is, I can program the rocket engine in my simulation so that it has an infinite amount of bowling balls available, maybe by beaming them into the space ship. Now, if the 75kg space ship shoots bowling balls to the ground so that it moves up 1m/s in the air above the earth and I would now switch off the gravity, how fast would the ship accelerate? – Toxiro Jul 6 '15 at 22:45
• I also understand from your answer that if the thrust is constant, the power is constant too, e.g. 2 kg exhaust per second and 5 m/s velocity compared to the current space ship's speed: 2 kg * 5 m/s per second = 10 kg m/s per second (thrust) and 2 kg * 5 m/s * 5 m/s per second = 50 kg (mm)/(ss) per second (power) – Toxiro Jul 6 '15 at 22:54
• Well, to be precise, rockets have high power to increase their thrust-to-weight ratio (rocket need to lift), not to make the fuel last as long as possible. Having a fast ejection speed will increase the specific impulse, that number gives you the ratio between consumption of fuel vs thrust. Ion engines do excel in ejection velocity (specific impulse), not power. – Laurent Grégoire Aug 18 '16 at 16:26
• @Laurent: You can get the same thrust by ejecting half the reaction mass per time at twice the velocity, requiring two times the power. So, for a given thrust, to minimize the use rate of reaction mass, maximize the power. – Mike Dunlavey Aug 19 '16 at 0:20

What is the acceleration of the space ship, if you turn off the earth's gravitational force?

The acceleration of the ship is

$$a = \frac{F + G}{m}$$

where $F$ is thrust of the engine (force), $m$ is mass of the rocket and $G=-mg$ is gravity force. If thrust is such as to exactly cancel gravity, it is $F=mg$ so the acceleration with no gravity would be $g$.

In your question, however, you mention power instead of thrust. From your formulation, I assume you mean part of energy liberated from the fuel per unit time that turns into kinetic energy of the rocket. Unfortunately, this number changes as the rocket accelerates, because it is equal to

$$P = F.v$$

where $v$ is velocity of the rocket. $F$ is constant, but as $v$ changes with time, $P$ changes with time as well. When the rocket starts, $v = 0$ so $P=0$. The faster the rocket moves, the higher the power $P$ is. Quantity $P$ is thus not good for characterizing the capability of the rocket engine (or jet engine) alone.

While the efficient power $P$ changes in time, total power liberated $Q$ (including the energy leaving with exhausts) may be constant. It can be written as

$$Q = B.\epsilon$$

where $\epsilon$ is energy liberated per unit mass of exhaust gas and $B$ is mass loss per unit time (rate of fuel consumption). If $B$ is constant, so is $Q$.

From the law of conservation of momentum, thrust can be expressed as $$F = B.u,$$

where $u$ is speed of the gas exhausts leaving the nozzle of the rocket in the frame of the engine. It is constant in time, a characteristic of the fuel mixture and the engine.

Combining the last two relations, we obtain

$$F = Q \frac{u}{\epsilon}$$

and the acceleration is thus given by

$$a = \frac{Qu}{m\epsilon}.$$

Thus, all other things being equal, acceleration would be proportional to the liberated power $Q$. In practice, changing $Q$ requires changing other things as well, which will affect $u$ and if mixture ratio changes, even $\epsilon$. Thus it is not necessarily true that bigger engine with higher $Q$ will have proportionally higher $a$.