I’m not sure my perspective is significantly different than yours, but:
Using conservation of energy: imagine we have a given amount of mechanical (i.e. kinetic+potential) energy produced by expelling exhaust in the rocket’s reference frame. The total mechanical energy change will be the same in any reference frame. But in another reference frame we have:
the faster the rocket is going, the more kinetic energy the exhaust loses (or less it gains, depending on relative speeds) when it is dumped the other way, which means more energy for the rocket.
the further down a gravity well you dump the exhaust, the less potential energy it has, which means more energy for the rocket.
Both are important from this perspective, but related since kinetic+potential energy is constant when not thrusting, so it’s moving faster when it’s down in the gravity well. Yeah, it also works with it using a gun or whatever instead of exhaust, but it’s more intuitive IMO to imagine it with exhaust.
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.
I am not quite sure I understand the question, but when the thrust is at 90 degrees to the trajectory, the rocket’s speed is unaffected by the thrusting, and it comes out of the gravity well at the same speed as it came in. That would apply equally if there were no gravity well.
when the thrust is at 90 degrees to the trajectory, the rocket’s speed is unaffected by the thrusting, and it comes out of the gravity well at the same speed as it came in.
That’s not accurate; when you add two vectors at 90 degrees, the resulting vector has a higher magnitude than either. The rocket will be accelerated to a faster speed.
In the limit where the perpendicular side vector is infinitesimally small, it does not increase the length of the main vector it is added to.
If you keep thrusting over time, as long as you keep the thrust continuously at 90 degrees as the direction changes, the speed will still not change. I implicitly meant, but did not explicit say, that the thrust is continuously perpendicular in this way. (Whereas, if you keep the direction of thrust fixed when the direction of motion changes so it’s no longer at 90 degrees, or add a whole bunch of impulse at one time like shooting a bullet out at 90 degrees, then it will start to add speed).
How is that relevant? In the limit where the retrograde thrust is infinitesimally small, it also does not increase the length of the main vector it is added to. Negligibly small thrust results in negligibly small change in velocity, regardless of its direction.
In the limit where the retrograde thrust is infinitesimally small, it also does not increase the length of the main vector it is added to.
I implicitly meant, but again did not say explicitly, that the ratio of the contribution to the length of the vector from adding an infinitesimal sideways vector, as compared to the length of that infinitesimal vector, goes to zero of as the length of the sideways addition goes to zero (because it scales as the square of the sideways vector).
So adding a large number of tiny instantaneously sideways vectors, in the limit that the size of each goes to zero and holding to the total amount of thrust added constant, in that limit results in a non-zero change in direction but zero change in speed.
Whereas, if you add a large number of tiny instantaneous aligned vectors, the ratio of the contribution to the length of the vector to the length of each added tiny vector is 1, and if you add up a whole bunch of such additions, it changes the length and not the direction, regardless of how large or small each addition is.
I don’t understand how that can be true? Vector addition is associative; it can’t be the case that adding many small vectors behaves differently from adding a single large vector equal to the small vectors’ sum. Throwing one rock off the side of the ship followed by another rock has to do the same thing to the ship’s trajectory as throwing both rocks at the same time.
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.
Right—looking at energy change of the exhaust explains the initial question in the post: why energy is preserved when a rocket accelerates, despite apparently expending the same amount of fuel for every unit of acceleration (assuming small fuel mass compared to rocket). Note that this doesn’t depend on a gravity well—this question is well posed, and well answered (by looking at the rocket + exhaust system) in classical physics without gravity. The Oberth phenomenon is related but different I think
The Oberth phenomenon is related but different I think
Yes, I think that if you (in addition to the speed thing) also take into account the potential energy of the exhaust, that accounts for the full Oberth effect.
I’m not sure my perspective is significantly different than yours, but:
Using conservation of energy: imagine we have a given amount of mechanical (i.e. kinetic+potential) energy produced by expelling exhaust in the rocket’s reference frame. The total mechanical energy change will be the same in any reference frame. But in another reference frame we have:
the faster the rocket is going, the more kinetic energy the exhaust loses (or less it gains, depending on relative speeds) when it is dumped the other way, which means more energy for the rocket.
the further down a gravity well you dump the exhaust, the less potential energy it has, which means more energy for the rocket.
Both are important from this perspective, but related since kinetic+potential energy is constant when not thrusting, so it’s moving faster when it’s down in the gravity well. Yeah, it also works with it using a gun or whatever instead of exhaust, but it’s more intuitive IMO to imagine it with exhaust.
I am not quite sure I understand the question, but when the thrust is at 90 degrees to the trajectory, the rocket’s speed is unaffected by the thrusting, and it comes out of the gravity well at the same speed as it came in. That would apply equally if there were no gravity well.
That’s not accurate; when you add two vectors at 90 degrees, the resulting vector has a higher magnitude than either. The rocket will be accelerated to a faster speed.
In the limit where the perpendicular side vector is infinitesimally small, it does not increase the length of the main vector it is added to.
If you keep thrusting over time, as long as you keep the thrust continuously at 90 degrees as the direction changes, the speed will still not change. I implicitly meant, but did not explicit say, that the thrust is continuously perpendicular in this way. (Whereas, if you keep the direction of thrust fixed when the direction of motion changes so it’s no longer at 90 degrees, or add a whole bunch of impulse at one time like shooting a bullet out at 90 degrees, then it will start to add speed).
How is that relevant? In the limit where the retrograde thrust is infinitesimally small, it also does not increase the length of the main vector it is added to. Negligibly small thrust results in negligibly small change in velocity, regardless of its direction.
I implicitly meant, but again did not say explicitly, that the ratio of the contribution to the length of the vector from adding an infinitesimal sideways vector, as compared to the length of that infinitesimal vector, goes to zero of as the length of the sideways addition goes to zero (because it scales as the square of the sideways vector).
So adding a large number of tiny instantaneously sideways vectors, in the limit that the size of each goes to zero and holding to the total amount of thrust added constant, in that limit results in a non-zero change in direction but zero change in speed.
Whereas, if you add a large number of tiny instantaneous aligned vectors, the ratio of the contribution to the length of the vector to the length of each added tiny vector is 1, and if you add up a whole bunch of such additions, it changes the length and not the direction, regardless of how large or small each addition is.
I don’t understand how that can be true? Vector addition is associative; it can’t be the case that adding many small vectors behaves differently from adding a single large vector equal to the small vectors’ sum. Throwing one rock off the side of the ship followed by another rock has to do the same thing to the ship’s trajectory as throwing both rocks at the same time.
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).
Right—looking at energy change of the exhaust explains the initial question in the post: why energy is preserved when a rocket accelerates, despite apparently expending the same amount of fuel for every unit of acceleration (assuming small fuel mass compared to rocket). Note that this doesn’t depend on a gravity well—this question is well posed, and well answered (by looking at the rocket + exhaust system) in classical physics without gravity. The Oberth phenomenon is related but different I think
Yes, I think that if you (in addition to the speed thing) also take into account the potential energy of the exhaust, that accounts for the full Oberth effect.