Yes, it’s associative. But if you thrust at 90 degrees to the rocket’s direction of motion, you aren’t thrusting in a constant direction, but in a changing direction as the trajectory changes. This set of vectors in different directions will add up to a different combined vector than a single vector of the same total length pointing at 90 degrees to the direction of motion that the rocket had at the start of the thrusting.
...Are you just trying to point out that thrusting in opposite directions will cancel out? That seems obvious, and irrelevant. My post and all the subsequent discussion are assuming burns of epsilon duration.
...Are you just trying to point out that thrusting in opposite directions will cancel out?
No.
I’m pointing out that continuous thrust that’s (continuously during the burn) perpendicular to the trajectory doesn’t change the speed.
This also means that (going to your epsilon duration case) if the burn is small enough not to change the direction very much, the burn that doesn’t change the speed will be close to perpendicular to the trajectory (and in the low mass change (high exhaust velocity) limit it will be close to halfway between the perpendiculars to the trajectory before and after the burn, even if it does change the direction a lot). That’s independent of the exhaust velocity, as long as that velocity is high, and when it’s high it will also tend not to match the ship’s speed since it’s much faster, which maybe calls into question your statement in the post, quoted above, which I’ll requote:
One interesting questions is at what angle of thrust does the effect on the propellant go from negative to positive? I didn’t do the math to check, but I’m pretty sure it’s just the angle at which the speed of the propellant in the planet’s reference frame is the exact same as the rocket’s speed.
Ok now I’m confused about something. How can it be the case that an instantaneous perpendicular burn adds to the craft’s speed, but a constant burn just makes it go in a circle with no change in speed?
The trajectory is changing during the continuous burn, so the average direction of the continuous burn is between perpendicular to where the trajectory was at the start of the burn and where it was at the end. The instantaneous burn, by contrast, is assumed to be perpendicular to where the trajectory was at the start only. If you instead made it in between perpendicular to where it was at the start and where it was at the end, as in the continuous burn, you could make it also not add to the craft’s speed.
Going back to the original discussion, yes this means that an instantaneous burn that doesn’t change the speed is pointing slightly forward relative to where the rocket was going at the start of the burn, pushing the rocket slightly backward. But, this holds true even if you have a very tiny exhaust mass sent out at a very high velocity, where it obviously isn’t going at the same speed as the rocket in the planet’s reference frame.
Yes, it’s associative. But if you thrust at 90 degrees to the rocket’s direction of motion, you aren’t thrusting in a constant direction, but in a changing direction as the trajectory changes. This set of vectors in different directions will add up to a different combined vector than a single vector of the same total length pointing at 90 degrees to the direction of motion that the rocket had at the start of the thrusting.
...Are you just trying to point out that thrusting in opposite directions will cancel out? That seems obvious, and irrelevant. My post and all the subsequent discussion are assuming burns of epsilon duration.
No.
I’m pointing out that continuous thrust that’s (continuously during the burn) perpendicular to the trajectory doesn’t change the speed.
This also means that (going to your epsilon duration case) if the burn is small enough not to change the direction very much, the burn that doesn’t change the speed will be close to perpendicular to the trajectory (and in the low mass change (high exhaust velocity) limit it will be close to halfway between the perpendiculars to the trajectory before and after the burn, even if it does change the direction a lot). That’s independent of the exhaust velocity, as long as that velocity is high, and when it’s high it will also tend not to match the ship’s speed since it’s much faster, which maybe calls into question your statement in the post, quoted above, which I’ll requote:
Ok now I’m confused about something. How can it be the case that an instantaneous perpendicular burn adds to the craft’s speed, but a constant burn just makes it go in a circle with no change in speed?
The trajectory is changing during the continuous burn, so the average direction of the continuous burn is between perpendicular to where the trajectory was at the start of the burn and where it was at the end. The instantaneous burn, by contrast, is assumed to be perpendicular to where the trajectory was at the start only. If you instead made it in between perpendicular to where it was at the start and where it was at the end, as in the continuous burn, you could make it also not add to the craft’s speed.
Going back to the original discussion, yes this means that an instantaneous burn that doesn’t change the speed is pointing slightly forward relative to where the rocket was going at the start of the burn, pushing the rocket slightly backward. But, this holds true even if you have a very tiny exhaust mass sent out at a very high velocity, where it obviously isn’t going at the same speed as the rocket in the planet’s reference frame.
I don’t understand what “at the start” is supposed to mean for an event that lasts zero time.
In the case where it’s instantaneous, “at the start” would effectively mean right before (e.g. a one-sided limit).