As a mental exercise, I created the following specific scenario:
Your spaceship, which is out in space traveling at some unspecified speed, has a dry weight mass of 12 tons (12000Kg). It is loaded with a million American baseballs (fixed to 145g each). Your ship has a rail gun that can shoot a baseball out of the ship at a speed of 1000m/s.
You aim the rail gun and start shooting baseballs in the same direction (say, straight out the back). The first baseball adds a very small amount of speed; the second adds an ever so slightly less-small amount of speed, each causing a tiny bit more acceleration because of the tiny reduction in overall mass of your ship.
As you look out the back window, you see a long line of baseballs flying away from your ship, each one moving away from the one that is chasing it. Someone on another ship sees your ship being followed by a long, spreading line of baseballs (assuming you had significant relative speed to begin with).
Now, I tried to do a spreadsheet to illustrate this with numbers, but I did something wrong. The ΔV between each line should be consistent (increasing slightly as you go down the chart), but for different initial V numbers, I get wildly different acceleration patterns.
The resulting V is used to calculate ship.Ek on the next line. The progression grows in a nominally consistent way for a given initial V, but the delta from line to line varies widely depending on the iniial V. For example, if the initial V is 5000 m/s, I get a believable initial acceleration of 2.4mm/s/s but if I use 100,000, I get 4.6cm/s/s, which is about what I get an initial V of 10.
I think the problem is in assuming that the baseball kenetic energy is a constant. Since they are being shot with constant velocity out of the back of an accelerating ship, their velocity with respect to a stationary observer is going to be 1000m/s minus the velocity of the ship. So each successive baseball will be moving slower than the one before it as its shot out the back until the velocity of the ship reaches 1000m/s at which point the baseball shot out the back has zero velocity and so zero energy.
The best way to solve this is probably to work out the specific amount of acceleration generated by the Nth ball, and then use calculus to integrate that from one to a million.
Zero energy from an external point of reference perhaps but still 1000m/s from the ship’s frame of reference. If not, the baseball accelerator would do its thing (I am imagining a pinball plunger here) and nothing happens; it just hangs there at the tip of the plunger. Obviously this is absurd – the ball is going to recede away from the ship at 1km per second, same as the ones before it. Mass thrown one way, acceleration the other, even beyond the velocity of the mass thrown.
I’m talking about less than relativistic speeds. Taking that into consideration makes my brain hurt.
Also glossed over is the source of the energy for the baseball slingshot.
To accelerate one baseball to 1000m/s relative to the ship requires a certain amount of energy. You will need the magical energy source that produces that acceleration a million times.
Just treat it as continuous. With a million baseballs, it’ll be very close to the same answer.
You haven’t said how many baseballs/second you’re shooting out. I’ll assume 1 baseball/sec.
Ship mass m(t) = (157000 - 0.145t). Force is 145 N (0.145 kg * 1000 m/s / 1 s). Therefore, a(t) = (157000 - 0.145t) / 145.
To get the total delta V, just use the Tsiolkovsky rocket equation. Fairly easy to derive (I did so here), but just using an online calculator is more convenient. Plug in the numbers and get 2571.3 m/s. Note that this is independent of thrust (i.e., balls/sec).
You could maintain constant acceleration (or any other function) by varying the rate at which you shoot out balls, but it won’t change the delta V.
I should note that it’s really annoying to do these kinds of calculations with energy. Just too easy to make mistakes by forgetting which reference frame you’re in, missing the KE of some objects, etc. Easier with momentum. Each baseball shot out gives the same impulse to the ship. The amount of delta V will vary solely based on mass. And since mass decreases linearly with time, assuming a constant rate of shooting out baseballs, it’s easy to integrate thrust.
I’ll note that real rockets work the same way. They tend to have roughly constant thrust, which means their acceleration goes way up as they get to the last drops of propellant. They’ll limit thrust at times to reduce aerodynamic or gee loads, but for the most part they just run at the maximum available thrust.
Yes, 1000m/s from the ship’s frame of reference, but if you do the calculation for the ship’s frame of reference reference frame the rocket always has zero velocity and so zero energy so you wouldn’t be able to do any calculations. Not that you it would be right to do so anyway since the whole point of the problem is that the rocket is accelerating and there is not one fixed inertial reference frame that represents . In order for the energy calculation to work, you need to pick one inertial reference frame and stick with it.
I agree that the ship will continue to accelerate even when the balls shot out the back are effectively stationary. Its just that the kenetic energy calcuation will involve the energy being removed from the ball as its velocity drops to zero being added to the ship.
Example: Lets assume a baseball weighs 1kg ship that weighs 1000 kg, so that our gun so when shoots the ball backwards at 1000m/s accelerates the ship forward at 1m/s to conserve momentum.
If the ship starts at at rest, before the shot the equation total kenetic energy is zero,
after the shot the total energy is
E=1/2 ((1000m/s)^21kg+(1m/s)^21000kg) = 500500 kg m/sec*2
Which represents the energy output of the gun.
Now suppose the ship along with its ball is moving at 1000m/s. So the intial
E=1/2 ((1000m/s)^2*1001kg) = 500500000 kg m/sec^2
After I shoot the ball out the back, the velocity of the ball is zero while the velocity of the ship is now 1001 m/sec
E=1/2 ((0m/s)^21kg+(1001m/s)^21000kg) = 50100500 kg m/sec*2
The difference being 50100500 -500500000=500500 kg m/sec*2 the same energy as before.
Ok, suppose you are in a reference frame such that you can measure a zero velocity (in space, there really is no such thing as literally motionless) for the last ball that was shot out. That would mean that all the previous balls that were shot out had a negative velocity (velocity is a vector quantity, which has a sign). In your reference frame, the next ball shot out will have a positive velocity: it will be following the ship, which will be moving away from it at 1000m/s.
Only 1 m/s. The ship is then going at 1001 m/s, and the ball at -1000 m/s relative to that. In any one ball’s reference frame, there is a series of balls forward and back at -1 m/s, 1 m/s, -2 m/s, 2 m/s, etc.
Buck_Godot is right that you can’t use the rocket’s reference frame, since it’s accelerating. You have to pick one inertial reference frame and carry that through the whole calculation.
Your math is all messed up. From where you are, you observe my ship moving from left to right at 1000m/s; then you observe me shoot a ball out the back at 1000m/s; at that point, from your perspective, the ball is now motionless and my ship is continuing away from it at slightly more than 1000m/s. If I appear to have a velocity of 1000m/s, I would have to shoot the ball at 2000m/s for it to have a velocity of -1000m/s from your point of view.
I said relative to “that”, meaning the ship. If the ship is at +1001 m/s relative to me, and the next ball at -1000 m/s relative to the ship, then it is at 1 m/s relative to me. And the next one at 2 m/s, and so on.
(see, this is why I said doing the physics in energy terms is error prone)
There are lots of problems where solving it from an energy standpoint is the way to go (e.g. determining the velocity of a pendulum at a given point in its swing), But this is definitely not one of them.
Definitely. Physics offers all kinds of ways to solve a problem–conservation of momentum, conservation of energy, principle of least action, etc. For any given problem usually some are better than others.
