# How much power would artificial gravity and inertial dampeners require?

Ignoring, for the sake of this question, how exactly the gravitational fields required for this to work are generated:

## The question

Our stalwart adventurers have a spaceship, perhaps a gateship, fitted with inertial dampeners (for preventing their organs from being liquefied) and artificial gravity.

The artificial gravity exerts a constant downward acceleration of 10 m/s2, and the inertial dampeners provide a variable multi-directional acceleration exactly counter to the acceleration of the ship itself.

Assuming that the artificial gravity generator is perfectly efficient (i.e. only the energy required for the gravity itself has to be accounted for) and that the artificial gravity only affects the things within the ship, not the ship itself: How much power is required to run the toys?

## Examples that need to be accounted for

For each example, the ship contains 1 000 kg of stuff that needs to be affected by the gravity fields created, referred to as the payload, and itself weighs 49 000 kg; for a total mass of 50 000 kg. For the purposes of cross sectional area, the ship can be modelled as a cylinder with a diameter of 3 meters and a height of 10 meters, where the front and back of the ship are the flat ends.

### Coasting

The ship is in a stable orbit around a planet. The ship experiences an apparent acceleration of 0 m/s2, and the payload experiences a uniform downward acceleration of 10 m/s2.

### Manoeuvring

The ship is transferring between two orbits, burning prograde with a uniform acceleration of 5 m/s2. The crew is experiencing no lateral acceleration and a uniform downward acceleration of 10 m/s2.

### Under attack!

The ship is in orbit, experiencing an apparent acceleration of 0 m/s2 when a photon torpedo explodes 10 meters off the port side of the ship, releasing one gigajoule of energy. The ship holds, and experiences some lateral acceleration as a result, but the crew experiences no lateral acceleration, and a uniform downward acceleration of 10 m/s2.

For each of the above examples, accounting for all forces, how much power is required by the artificial gravity generator?

• You might need to give a better explanation of artificial gravity, since it's not a real thing its going to be hard to make estimates for. – DaaaahWhoosh Feb 10 '15 at 18:54
• @DaaaahWhoosh How it works doesn't really matter, because what I need is the forces involved in making a net force. It could be magical fairies doing the work, so long as energy is conserved. – Williham Totland Feb 10 '15 at 18:56
• From a physical point of view, do you even need any energy? There is no energy change. Neither the kinetic nor the potential energy of the payload changes. It's just "stabilized" but no energy changes. Further one has to ask, where would the energy spent go to? Energy can not be destroyed and does not anihilate other energy. It's just transformed. So if the net energy (kin/pot) with and without the dampener is the same, where does the energy it "consumes" go to? – Hothie Jun 20 '16 at 17:37
• To clarify what I mean, If you are falling down, your potential energy is converted to kinetc. And if you hit the ground, your kinetic energy is transformed to heat or something and transfered to the ground and in the air. A Platform prevents you from falling down. But it does not consume any energy. It's just there. So the dampener is just like this platform, preventing that you potential energy is converted to kinetic. This in itself does not consume any energy. – Hothie Jun 20 '16 at 17:42
• @Hothie Yup, the only energy consumption would be from inefficiency - that is, waste heat (and other byproducts) of the artificial gravity machine itself. A 100% efficient machine might take energy to set up, but it would not have a constant drain over time. But if it's 90% efficient instead, how much of waste power are we talking about? :) – Luaan Nov 14 '16 at 11:42

Note I don't know how the artificial gravity is produced, but I am assuming you can selectively shape a potential gravity well so it affects only the things in the ship; the payload. There are problems with this kind of artificial gravity, and to make this answerable I am going to ignore them. Additionally, the individual forces exerted on the payload will be in various directions for various situations, and I assume the craft is capable of this.

# Coasting

It could just create a gravity well, with the Energy around that of Earth. The equation to figure this out would look like:

$$U=\frac{-Gm_E}{r_E}\sum{m_n}$$

where $G$ is the universal gravitational constant, $m_E$ is the mass of the earth, and $r_E$ is the radius of the earth, $m_n$ is the mass of an object you want to feel this force, and U is the energy of the gravity well. In your specific example $\sum{m_n} = 1000$. (The ship itself does not need to be under the effect of this gravity well, so it doesn't get added to $\sum{m_n}$ at all.) So your gravity well looks like it needs to be $6.249*10^{10}$(-ish) joules deep.

The craft itself contributes to the gravity well, but only a little. The more mass of the craft is centered on one side, the more it can can offset the gravity needed to keep people down. I have ignored the craft's gravity because it's negligible influence. If the ship were a ball that people stood on, with their center of masses 1m away, it would only contribute about $.003 J$. Not enough to substantially affect the gravity well.

Trying to determine the power requirements for bending space enough or producing this gravity well is tricky, because no one has every produced a gravity well like this before, nor have we seen gravity (of an appreciable amount) apply to bodies and then not. This is where sci-fi magic/handwaving comes in. To simply provide an answer, I am going to say that you need $6.249*10^{10} W$: you need to maintain that gravity well by giving it the required energy for the total depth every second.

Obviously, this value can change depending on your views of how artificial gravity production works, but it's the answer I'm running with. If you think that, once a gravity well is made, you needn't give it more energy to maintain it, your power requirements go to 0 after it's made. If you think gravity wells must apply their energy over plank time, then you'll get a very large power requirement (Around $32*10^{54}$ Watts!)

# Maneuvering

For the people inside the ship to notice no acceleration, the ship's inertial dampener must compensate for the movement of the ship. The well will have to get deeper or shallower by the amount of work needed to keep the people on the floor. This depends greatly on which way the ship accelerates. The equation for this work, though presents a problem:

$$W=\int_{x_1}^{x_2}{Fdx}$$

No, the problem is not the calculus, it is the fact the work required depends on how long the acceleration is experienced. If you attempt to solve this for our specific situation, you get $$W=\int_{x_1}^{x_2}{5dx}=5*(x_2-x_1)=5*\Delta x$$

That's not hard. Your well changes by 5 times the distance (in meters) traveled. That being said, you don't need to produce this energy all at once; your power requirements may not change all that much. The total fuel you need and the total energy you need to produce will!

The power demand with fluctuate depending on how much you travel per second. It would look like: $$P = \frac{5*\Delta dx}{dt} = -5*v$$ That $v$ is velocity. This changes, of course, with how fast you're going. Since you're not jumping from one speed to the next every second, the above equation gives you the instantaneous power you need. I suggest you look at your top speed, as that is the maximum amount of power you need to reach that top speed.

# Under Attack

1 Gigajoule $(10^9 J)$ just got added to the mix! I need to assume that this energy is radiated in a sphere, not just directly impacting the ship. This makes the impact profile of the ship really important. The surface area of a sphere of radius $10m$ is about $1256.4 m^2$ Our serendipitously cylindrical space ship has a side surface area of $10m*\pi*3m=94.24...m^2$. This means the ship gets a dose of ($94.24/1256.4=.075...$) about 8% of the gigajoule, or $8*10^{7} J$. Once again, the gravity well needs to increase by this amount to prevent our payload from getting harmed.

The power needed for this? Well, it depends on how long the explosion lasts. If the explosion lasts a small fraction of a second, you multiply that $8*10^{7}$ by the reciprocal of that fraction. An explosion which transfers its energy over 1/100 of a second, for example, will require $8*10^{9}$ Watts to totally negate.

• What does this work out to in watts? – Williham Totland Feb 10 '15 at 19:23
• @WillihamTotland You asked for the energy needed; watts is a measure of power. Power is energy per unit of time. So, the wattage needed depends on how quickly you need to do these things. If you just chose 1s of time, the numbers would stay the same. I would say come into chat to talk about this, but you've not the rep required. – PipperChip Feb 10 '15 at 19:26
• Then I was unclear, correcting question. – Williham Totland Feb 10 '15 at 19:28
• @WillihamTotland There you go! Asking for Power makes the answers more messy, less "here-is-a-number," but the honest-to-goodness truth. – PipperChip Feb 10 '15 at 20:06
• And if the operation of the artificial gravity requires, say, 6.249E10 watts, where does that energy go? – Williham Totland Feb 11 '15 at 13:27

In order to counteract the acceleration of the craft, you need to bend spacetime to produce gravity. The simple presence of mass/energy does that. It means you can't switch it on and off because your battery or fuel source, being a store of energy, will produce this gravity!

Note that any other answer will violate Einstein's formula for general relativity, so postulating a way to do that means you can make up whatever answer you like (and make a perpetual motion machine in the process).

The best way to store energy in such a compact form is with mass.

So, imagine a superdense plate—far denser than normal atom-based matter—that can be moved forward and back. Actually, move the people's beds forward and back. As the ship accelerates beyond the gs that a human can withstand, move them closer to the dense plate in front of them, so its gravity counters the acceleration.

If you could somehow apply an acceleration to all the particles of the person's body individually rather than just pushing on one surface and letting the compression transmit the force throughout the body, he would feel no accelleration. As suggested by Abulafia's answer, the energy would be the usual amount to accelerate that mass,

But, you would have no need to counter the ship's acceleration using that technique. That would be how you accelerate the ship (or portions of it) in the direction you want to go, and the occupents will feel nothing. So, it's not energy in addition to the thrust, but simply the manner of applying the energy you need according to Newton's second law, anyway.

You might still need extra energy though: if the device is carried on the ship, it might be recoiling from accelerating the people etc. which is the opposite of where you want it to go. So you need another thrust to push the device (and everything it's carrying), if you can't divert the effect directly. So, for the sake of plot, a very understandable approach is that the energy needed to thrust doubles to stay put and then applies a third time to really move as intended. Thrust with inertial dampening triples the normal energy of acceleration (less the 1g residual you want, plus inefficiency in the mechanism).

• So your rocket has a planets worth of mass to create gravity for the crew. Even if you use osmium, the most dense element under reasonable conditions, The mass used has to be greater than 1/20 th of the earths mass. – Donald Hobson Jun 20 '16 at 13:48
• I was thinking of superdense matter as described in a few short stories by Robert L Forward, which is collapsed matter not normal atoms. In one an asteroid is compressed to a hand-sized disk. In another tiles are used to increase gravity to be hospitible to aliens, or to cancel Earth's gravity by holding it overhead. He wrote hard SF. – JDługosz Jun 20 '16 at 13:52