- #51

A.T.

Science Advisor

- 10,869

- 2,413

Where did you get that from?But the forces along and perpendicular to the original velocity would still vary in magnitude from √2F (i.e. greater than F) to zero.

- Thread starter thinkagain
- Start date

- #51

A.T.

Science Advisor

- 10,869

- 2,413

Where did you get that from?But the forces along and perpendicular to the original velocity would still vary in magnitude from √2F (i.e. greater than F) to zero.

- #52

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,373

- 4,947

Sorry I didn't complete the sentence. I am referring to the centrally directed force (circular path) and, although it will take longer, the argument given by haruspex seems to imply that the two forces will always be less than F.Where did you get that from?

- #53

A.T.

Science Advisor

- 10,869

- 2,413

No matter what strategy you were talking about, perpendicular components can never be greater than the total magnitude. Isn't that obvious?Sorry I didn't complete the sentence. I am referring to the centrally directed force (circular path)

- #54

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,373

- 4,947

What I need is an explanation of how that two lines of reasoning actually applies when comparing the two cases. The actual wording seems to ignore the fact that, at some places on the path, the momentum change (or at least the force) will be greater in the circular case.

Look, I have accepted (the sums are trivial) that the 45° system does it faster. What I was questioning was

- #55

A.T.

Science Advisor

- 10,869

- 2,413

I see no problem with the argument in post #43. It's a valid proof, as far I can see. And no, I can't make any sense of your objections to it.I still doubt that the argument is actually valid, as it stands.

What are you talking about? The force magnitude is F, all the time, in both cases.The actual wording seems to ignore the fact that, at some places on the path, the momentum change (or at least the force) will be greater in the circular case.

- #56

- 35,135

- 6,271

No, |Fx|=|Fy|=|F/√2|. In all paths, if the direction of the thrust at time t is ##\theta(t)## then ##F_x(t)=F\sin(\theta(t))## etc.|Fy| = |Fx| = |√2F|, in the 45°case, where F is the available thrust.

- #57

Merlin3189

Homework Helper

Gold Member

- 1,653

- 769

If this is some sort of rocket thruster, do you mean energy or do you mean impulse?... the total amount of energy available needs to be specified -....

If the path were circular, I'd have thought no energy is required, but to apply force does imply impulse.

Of course for a rocket to function at all requires energy input even if no useful energy is output.

Since this is a fixed thrust engine, both impulse and energy input are proportional to time.

So we come back to the original question, what is the shortest time we can operate the rocket to make the required change.

- #58

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,373

- 4,947

It's interesting that haruspex, him,self is making no comment here. My difficulty is with this:I see no problem with the argument in post #43. It's a valid proof, as far I can see. And no, I can't make any sense of your objections to it.

What are you talking about? The force magnitude is F, all the time, in both cases.

where he talks about the net change in momentum in theTo see that the 45 degree must be the fastest way of finishing with the same speed, consider that the net change in momentum required is in that direction. With a thrust limited to F in magnitude, it must be quickest to keep that F in the desired direction throughout. If we direct the thrust in any other direction for a time, it reduces the momentum change in the desired direction.

- #59

- 35,135

- 6,271

P√2 = F tIt's interesting that haruspex, him,self is making no comment here. My difficulty is with this:

where he talks about the net change in momentum in the45° direction.The change in momentum in the original direction needs to be -P and the change in momentum at right angles needs to be P and the change in the 45° direction will have magnitude P√2, irrespective of the path taken or the direction of the thrusters at any time. How is it 'obvious' that the way that change is delivered is quickest if the thrust is always delivered along the 45° line? For other paths, the rate of momentum along the two axes will vary with time, being sometimes more in magnitude and sometimes less than P. Can you see at least something in my problem. I am not just arguing; I need some expansion of those few words in haruspex's assertion (not proof).

- #60

A.T.

Science Advisor

- 10,869

- 2,413

I think haruspex's explanation qualifies as proof. If you need expansion on it, you should pinpoint exactly which point of it you want expanded.I need some expansion of those few words in haruspex's assertion (not proof).

As an alternative consider the reference frame where the astronaut was initially at rest. Here he has to accelerate from 0 to some speed along the 45° line as quickly as possible. Still not obvious?

- #61

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,373

- 4,947

P√2 = F t_{45}. Suppose we try some other strategy for time t_{45}. Let the 45 degree direction be the y direction. If for any part of the time we direct some of the available F in the x direction then during that time F_{y}< F. Thus the integral of F_{y}over time t_{45}will be less than P√2, and we will not have achieved the desired result.

Thanks for persevering, chaps. I think I have it now.I think haruspex's explanation qualifies as proof. If you need expansion on it, you should pinpoint exactly which point of it you want expanded.

As an alternative consider the reference frame where the astronaut was initially at rest. Here he has to accelerate from 0 to some speed along the 45° line as quickly as possible. Still not obvious?

- #62

- 19

- 0

So is the 45 degree angle the optimum direction of thrust then? Thanks!

- #63

A.T.

Science Advisor

- 10,869

- 2,413

For the equal speed constraint, not for the fixed turn space constraint .So is the 45 degree angle the optimum direction of thrust then?

- #64

- 19

- 0

I see, so we still don't have an answer for that one then?

- #65

- 35,135

- 6,271

As I posted, that would make quite a hard problem. In all likelihood, there is no analytical solution. Certainly out of the scope of a homework forum.I see, so we still don't have an answer for that one then?

- #66

- 1,241

- 189

- Last Post

- Replies
- 4

- Views
- 5K

- Last Post

- Replies
- 2

- Views
- 4K

- Replies
- 3

- Views
- 1K

- Replies
- 18

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 5K

- Last Post

- Replies
- 0

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 5K