# How would a rail gun affect the trajectory of a spacecraft?

What sort of force would a railgun need to produce to have a measurable affect on the trajectory of a large spacecraft?

In the scifi book I am working on the primary craft of the story is a military ship classed as a heavy destroyer and equipped with 3 massive railguns that extend along the central spine of the ship. As none of them can be directly in the center, I am mainly wondering if firing them one at a time would have enough force to cause the trajectory of the ship under power to alter slightly, or if the ship not under power might start to rotate along its length (like a helicopter, not a bullet).

Obviously, that would depend on the mass and velocity of the projectile, but when trying to determine a feasible mass the math got a little beyond me. Therefore, I have settled for determining minimum force, as that should allow me to come up with possible weight and speed of the payload. Likewise, if it turns out that the difference would be negligible or impractical/impossible, then I can just move on with the story and completely ignore the possibility all together.

For a clearer analogy of my concern: If I shoot a gun and it pushes my right shoulder back so that I end up turning slightly to the right, how much force would I need to make a spaceship do the same thing?

I'm not sure what values might be necessary, but I have some rough numbers that I imagine might be helpful:

Ship Length: 330 m

Ship Width: 60 m

Mass: 35,000 t

Railgun offset from ship center axis: 8 m

If it matters, assume ship center of mass is equal to geometric center. Please let me know if there is anything else I'm not considering.

• Momentum is mass times velocity. What is the minimum velocity change the ship could have which you consider measurable? Divide that by the mass of the ship. You now have the momentum of the projectile. What are you missing from there? Jul 22, 2019 at 22:42
• Assume for sake of argument, that you put a BB gun so that it shot a BB out one millimeter off of the exact line of thrust needed to continue straight. You shoot that gun once on a multi-megaton spaceship. Unless some countervailing force is applied, that ship is now rotating, albeit incredibly slowly (perhaps one rotation per century?) There is no limit on mass or velocity to start the rotation, there is only a limit on the sensitivity of measuring devices used to detect the rotation. Jul 22, 2019 at 23:32
• This is why spaceships always use "pew-pew"-type of weapons rather than kinetic missiles :) Jul 22, 2019 at 23:35
• if you place the railguns on the points of an equilateral triangle around the center of mass, you can fire all of them without introducing rotation. (all of them must be fired at once for that). You will still slow down the ship by a bit (or introduce kickback), but you won't rotate Jul 23, 2019 at 7:01
• @LaughingVergil Technically, there is no "continue straight" when applying any amount of thrust in any direction while in a freefall environment, even if we're courteous enough to consider "ahead in the current orbit, as if no thrust was applied at that point" to be "straight" (which it is, depending on your frame of reference). No matter how you do it, thanks to our good ol' buddy Sir Isaac Newton and friends, there will be some delta-v incurred from that impulse, which will result in the orbit changing, possibly ever so slightly but still by a non-zero amount.
– user
Jul 23, 2019 at 14:29

Alright, this is my first answer on Stack Exchange, so let me know if I've made any grave mistakes in my analysis. These numbers are definitely back of the envelope calculations, but they do give a good picture of what's going on in your ship.

TL;DR Yes, it will rotate the ship, but it's nothing that the ship's thrusters can't compensate for if your ship is designed to even move at all. In the context of your story, I wouldn't worry about it.

What we need to find is the maximum impulse your ship can impart onto the projectile without exceeding the maximum corrective torque that can be produced by your thrusters. This is because any torque put on the ship is going to cause a change in trajectory, even if basically infinitesimal. We'll set up an equilibrium with the torque produced by the impulse of the railgun and the force provided by the thrusters. I'm also going to assume that your thrusters are mounted at the edges of your ship, but you can change the variables to match whatever configuration you want.

Just to get an idea of the magnitude of the forces you're talking about, using the definition of torque $$\tau = I\alpha$$ (I is your moment of inertia), the angular acceleration caused by your railgun is going to be 3.2 x 10^-11 times the reaction force of the railgun, based on the dimensions and mass of your ship. Here's the calculation:

$$\tau = I\alpha$$ $$I = \frac{m(a^2 + b^2)}{12}$$ Assuming that your ship is a rectanglular plate (thickness doesn't affect distrubution of mass [and moment of inertia] in this calculation) $$Fr = \frac{m(a^2 + b^2)}{12}\alpha$$ $$F = \frac{m(a^2 + b^2)}{12r}\alpha$$ $$F = \frac{(3.18 * 10^7 kg)((60m)^2 + (300m)^2)}{12(8m)}\alpha$$ $$F = 3.1 * 10^{10}\alpha$$ $$\alpha = 3.2 * 10^{-11}F$$ (Note: this constant of proportionality does have units, but they're not important if we're just using this as a ratio between force applied and angular acceleration)

A factor of 3.2 x 10^-11 is really small. To put a 0.1 degree per sec^2 angular acceleration on your ship (a pretty minor acceleration), you need 5.31 x 10^7 Newtons of force. I think the highest thrust rocket engine we've built, the F-1, produces 6.6 x 10^6 Newtons of thrust. You'd need the equivalent of 8 of those just to get your 0.1 degrees per sec^2 angular acceleration. And that's just the force the railgun would have to make.

Let's say your railgun has a muzzle velocity of 5 km/s (over Mach 14!) and accelerates your projectiles over the whole 300m length of your ship. From basic kinematic equations, this means your projectile takes 0.12s to get from the back to the front of your ship when starting at rest. The projectile's acceleration is 41,700 m/s^2, and now to find the reaction force due to this, just multiply the mass times the acceleration. We'll assume 1000kg (and yes, a 1000kg metal rod going at Mach 14 will do a lot of damage). This gives a force of 4.17x10^7 Newtons, which is pretty close to the force required to make a 0.1 degree per sec^2 angular acceleration.

Now, the torque produced by your corrective engines will have to be the exact same torque as the torque produced by your railgun to keep the ship from gaining angular velocity. This is where the width of your ship comes in: the radius from the center of mass to the edge of your ship is 30m, and your railgun is at 8m. The ratio between the two is 3.75. This means that your engines can be 3.75 times as weak (or 0.26 times as strong) as your railgun. In this case, the engines will have to put out 1.11 x 10^7 Newtons of thrust, or about two F-1 engines at full power. On a ship your size, this isn't unreasonable.

Anyways, all that calculation was just to show you that although the railgun puts out a (much more than a literal) ton of reaction force, the ship's thrusters have to be at least one or two orders of magnitude greater than this just to get the ship moving at any reasonable speed. Just to get your massive 35,000-ton ship accelerating at 10 m/s^2, you need 3.18x10^8 Newtons of thrust, so correcting for railgun blasts should be well within the tolerance of what your maneuvering engines can provide. I hope this helps.

• could you give units for the 10^-11 number you cite as reaction force in the beginning? Jul 23, 2019 at 5:26
• It's a ratio: just clarified it. Thanks for catching that. Jul 23, 2019 at 5:38
• Could you expound a little on how you arrived at that number? OP gave a length, width and a mass of the ship, and the offset of the railgun. You probably modelled the ship as a cylinder of these measures, with the mass equally distributed? Jul 23, 2019 at 7:30
• "The angular acceleration caused by your railgun is going to be 3.2 x 10^-11 times the reaction force of the railgun": this is quite very obviously impossible from dimensional analysis. Angular acceleration has units of T^-2, rad/s² in SI. Force has units of M×L×T^-2, kg·m/s² in SI. An angular acceleration cannot be equal to a number, any number, times a force. Jul 23, 2019 at 14:13
• @AlexP I'm using the magnitude of that relationship as a constant of proportionality to illustrate the forces going on here. It does technically have units, but they're not really important in this analysis since the relationship between force and angular acceleration is known. Jul 23, 2019 at 16:26

Simplest case: fire the projectile in radial direction.

The sum momentum of the whole system has to stay equal.

$$\bar I_\text{ship} = m_\text{ship} \bar v_\text{projectile} \\ \bar I_\text{projectile} = m_\text{projectile} \bar v_\text{projectile} \\ \Sigma \bar I = \bar I_\text{ship} + \bar I_\text{projectile} = \text{const.} \\ m_\text{ship,t0} \bar v_\text{ship,t0} + m_\text{projectile,t0} \bar v_\text{projectile,t0} = m_\text{ship,t1} \bar v_\text{ship,t1} + m_\text{projectile,t1} \bar v_\text{projectile,t1}$$

So if you fire a projectile perpendicular to the ship axis, the ship gets a momentum in the other direction, equal to the projectile. So if the masses do not change, you can solve it this way:

$$\bar v_\text{ship,t1} = \cfrac{\left(m_\text{ship} + m_\text{projectile}\right) \bar v_\text{ship,t0} - m_\text{projectile} \bar v_\text{projectile,t1}}{m_\text{ship}}$$

So if your ship is not moving and you fire a 1t projectile with 35000m/s velocity, your ship starts to move in the opposite direction with 1m/s velocity.

More complex case, fire the railgun in tangential direction

If you fire the railgun tangential to the ship, your ship starts to rotate. In this case I would use the energy conservation, it's simpler.

$$E_\text{rotational} = \cfrac{1}{2} I \omega^2 = E_\text{kinetic,projectile} = \cfrac{1}{2} m_\text{projectile} v_\text{projectile}^2$$

$$I$$ is the inertial momentum of the ship, it depends mostly on its form. $$\omega$$ is the angular momentum: $$2 \times \pi \times \frac{1}{T}$$, where $$T$$ is the time it takes for the ship to rotate once.

Even more complex, do both

In this case you have to decompose the speed of the projectile in tangential and radial direction. Use the radial speed in the first formula and use the tangential speed in the second formula.

• I reformatted your formulas using Mathjax instead of inline images. I'm pretty sure I got it right, but you may want to double-check, just in case. Here's a handy-dandy guide to Mathjax syntax.
– user
Jul 23, 2019 at 14:22
• Thx, I did not found anything in the help... Jul 23, 2019 at 15:12
• Only momentum and angular momentum conservation apply here: the energy conservation equation is not correct (energy need not be equally shared, and the gun firing adds mechanical energy) Jul 24, 2019 at 20:16