Solved Burning prograde and angle of semi-major axis?

[SPQRALAN]

Let's say you have an elliptical orbit, if you burn prograde at the periapsis, you extend the apoapsis by some factor(according to vis-viva equation). But if you burn prograde on somewhere during the elliptical orbit, the obit will be "tilted", so the semi-major axis would be angled. But by how much? How do you calculate the new location of the apoapsis?

 

[SupremeDorian]

Altaïr Horus Lupercal It's your time to shine!

 

[Altaïr]

Wow, a very technical question.
What's sure is that burning prograde will always increase your semi-major axis, as you raises your orbital energy:
V^2/2 - mu/r = - mu/(2D) for an ellipse (D being the semi great axis)
In practice, I think it will increase both periapsis and apoapsis, and it will tilt your semi-major axis indeed: it will move the periapsis closer to your ship.

Now to calculate by how much it will increase/tilt your periapsis you will need more elaborated formulas. I'll have a look to see what I can provide to help.

 

[Altaïr]

Ok, I've made a few calculations. But it's not simple, I can't give a general answer.
But it's possible to calculate this for a given situation with a few formulas entered in a spreadsheet.

For this, I consider I know the position of my ship and its speed (actually I must know its radial speed and its tangential speed). Then I can calculate all orbital characteristics: periapsis, apoapsis, and argument. If I change the speed, I can calculate the new characteristics.

Here is my work (sorry for my awful handwriting ):

Sorry, it's really raw, but I'll make a cleaner presentation if you're interested.

 

[SPQRALAN]

Ok, I've made a few calculations. But it's not simple, I can't give a general answer.
But it's possible to calculate this for a given situation with a few formulas entered in a spreadsheet.

For this, I consider I know the position of my ship and its speed (actually I must know its radial speed and its tangential speed). Then I can calculate all orbital characteristics: periapsis, apoapsis, and argument. If I change the speed, I can calculate the new characteristics.

Here is my work (sorry for my awful handwriting ):
View attachment 20641 View attachment 20642
Sorry, it's really raw, but I'll make a cleaner presentation if you're interested.

Thank you. This is great. I'm really glad that someone actually spent so much effort on a question instead of just listing irrelevant materials(I feel guilty about my Yahoo Answers responses now). I guess the philosophers are right, you can never stop learning(although I'm just sighing at the amount of new orbital mechanics I need to do/learn).

Just two more things, I know, me the annoying nerd:
1) Is the angular speed just v/r or do you have to physically measure the angle traveled and divide it by time? If it is just v/r, then v/r = a component of v? That doesn't make sense. If not, ignore this question.
2) What is responsible for the radial velocity? Because gravity pulls inward, not outward. Please elaborate.

Other than that, thank you, I'll have to do a lot of studying on my own. Just curious, are you studying a physics major in college or something?

 

[Altaïr]

Haha, I don't think that Yahoo will be able to answer that indeed.
You can get the main equations that way, but at some point you'll have to go into it.

1) Is the angular speed just v/r or do you have to physically measure the angle traveled and divide it by time? If it is just v/r, then v/r = a component of v? That doesn't make sense. If not, ignore this question.

You can calculate the angular speed that way, but you have to only account for the tangential speed (the one I called V_theta):
Angular speed = V_theta/R
The radial component doesn't contribute to the angular motion.
A more general way to know it is by using the fact that areal velocity is a constant:
R × V_theta = R^2 × angular velocity =A (A is a constant)
A can easily be calculated by considering the ship at the periapsis or the apoapsis. As the velocity is fully tangential at those points, you can calculate A easily:
A = Rp×Vp = Ra×Va
Then at any point, the angular velocity is A/R^2.

Trying to divide the angle travelled by time is very difficult, it can only be done numerically.

2) What is responsible for the radial velocity? Because gravity pulls inward, not outward

That's because of the centrifugal effect associated to the circular motion. If the orbit is perfectly circular, the centrifugal effect exactly compensates gravity, which is why the orbit radius never changes.
In an elliptical movement, they are slightly different, so there are 2 phasis:
- the lower part of the ellipse: around the periapsis, the tangential speed is important, and the centrifugal effect is greater than gravity. The result is that the radial speed increases: it's negative during the descending phasis, null at the periapsis, and positive after that.
- the upper part of the ellipse: Now the ship slows down and its the contrary: the gravity becomes higher than the centrifugal effect, and the radial speed diminishes: from positive (ascending phasis) it becomes null (apoapsis) and then negative.

Tell me if that's not clear, I'll try to be more explicit.

Just curious, are you studying a physics major in college or something?

No, I've studied mathematics and physics (I'm an engineer now), but it's been long I'm not a student anymore. I just love toying with equations and trying to understand things by myself.

 

[Lt. Snakestrike]

Would it make more sense to measure angular velocity in Radians/time? (Or degrees...). I feel like that'd make it a lot less complicated, as it'd apply to everywhere on the object.

 

[Altaïr]

Would it make more sense to measure angular velocity in Radians/time? (Or degrees...). I feel like that'd make it a lot less complicated, as it'd apply to everywhere on the object.

That's what it's supposed to be.
The angular speed is measured in radians/s, whereas the tangential speed is in m/s.
You can use whatever you like, but in terms of complexity it doesn't drastically change things. Otherwise I would gladly do it

 

[Danny Batten]

I imagine this is solved now?

 

[Altaïr]

Regarding the initial demand I think I did the best I could, so yes we can consider it solved.
But you can still ask on that thread if you need SPQRALAN

 

[SM84CE]

*mind blown*

 

[SPQRALAN]

You solved it. How do I mark it as complete?

 

[SupremeDorian]

You solved it. How do I mark it as complete?

I've already done that.