Entering SOI's Clockwise or Counter-clockwise

[700NitroXpress]

No. Just.... No. Draw a couple vectors, will you. Your chairs or whatever are falling in different directions, yes they are falling at the same speed in a particular frame of reference but on impact their horizontal surface speeds will not be identical because the surface itself is moving. Seriously. Draw a picture. Those often help.

Due to the conservation of energy, they will have the exact same amount of energy and thus, you need the same amount of counter energy to stop them. For every action their is an equal opposite reaction.

[700NitroXpress]

No, because at that altitude the curvature of the planet is no longer negligible. The craft in a ballistic trajectory will take longer to hit the planet because the planet is curving away from it, and thus it has farther to fall.

Path A-B is notably shorter than path A-C.

Note that if the horizontal velocity of the red dot is high enough, the trajectory path will miss the planet entirely - which we refer to as an "orbit." When this happens the time it takes to land on the surface becomes infinite, as it will never land at all.

It's not an interesting question if you understand calculus, and that acceleration is the time derivative of velocity, which itself is the time derivative of position... so when you integrate acceleration over time a "2" pops out all on its own.

In conclusion, objects that deorbit from the same height and orbiting at the same speed will hit the ground at the same time. But if one is in a ballistic trajectory and the other does not, then they don't have the same speed and won't hit the ground at the same time.

However, the velocity of the one deorbiting prograde will have a much lower velocity relative to the ground than the one orbiting retrograde, unless additional fuel is used to compensate. Again, this is without atmosphere effects.

Edit: Since I'm drawing pretty pictures, how about one more?

The velocity of the ground is shown as yellow arrows, and the horizontal velocity of the object is shown as black arrows. Magnitude isn't that important but direction is. Notice how the arrows on the left are pointing in the same general direction, while the arrows on the right are pointing opposite? That's the difference. To land safely you need to turn the black arrow on the right completely around to point in the same direction as the yellow arrow, otherwise you'll be traveling sideways when you hit the ground - that's called a crash.

Changing the direction of that arrow takes energy - energy which, if you don't have an atmosphere to suck it up, requires fuel.

=Smidge=

You are forgetting one very important detail that you drew that contradicts what you said about it. The velocity on the surface for B, is the same as the velocity in orbit for A. These two arrows cancel each other out because they are both moving at the same speed at the same time. This means that now you have three arrows for both halfs of the sphere. When A is stationary like that the projected point D and the projected point C are not moving faster or slower than each other. This is because the rotational speed of the surface is the same in one point as it is in another. When you are stationary above point B, you move at the same speed it moves. That point moves across the surface at the same speed that all the other points move across the surface. This is because the surface speed is constant. The surface does not accelerate ever. This is what's known as the Coriolis effect. The person on the ground gets the illusion that A is moving faster towards it than D, than the same person on the ground would see A moving toward C.

All objects of the same mass fall at the same speed of gravity rather or not they are launched or dropped. Object A gets launched in both directions from it's position above B. V = GT both objects arrive in the same position at the same time.

Or if you still don't believe that, Mythbusters says the same thing:

[Rusty6899]

A retrograde planetary orbit has the major disadvantage of meaning that orbiting one of the planet's moons (assuming it is retrograde like all moons in KSP) is almost impossible, with the exception of Laythe due to the option of aerobraking. As for landing on a planet, it is more efficient to land or take off prograde.

One thing I wonder about is orbiting, though. I haven't really looked into the numbers but I am aware that in real life most captured asteroids around the gas giants orbit retrograde, which would suggest to me that it could be more efficient to get into orbit retrograde, although I am aware there could be other reasons for this.

[jwilhite]

A retrograde planetary orbit has the major disadvantage of meaning that orbiting one of the planet's moons (assuming it is retrograde like all moons in KSP) is almost impossible, with the exception of Laythe due to the option of aerobraking.

Yeah, I've tried to get into orbit around Laythe (using aerobraking) coming from a retrograde Jool orbit. It's insanely hard. I think it took me about 10 tries to get the altitude exactly right. When you are coming in THAT fast there just isn't any margin for error at all.

[Specialist290]

All objects of the same mass fall at the same speed of gravity rather or not they are launched or dropped. Object A gets launched in both directions from it's position above B. V = GT both objects arrive in the same position at the same time.

Or if you still don't believe that, Mythbusters says the same thing:

Nitro, no one is contesting this.

What we are contesting is your assertion that both objects have the same velocity, which is patently wrong. Yes, a bullet dropped from a height will fall to the ground at the same time as a bullet fired out of a rifle from that same height, because they have the same vertical acceleration. However, the bullet fired from the rifle has a much higher total velocity on impact, because its horizontal velocity component, in the frame of reference of the planetary surface, is much higher than that of a bullet that is simply released to fall to the ground.

Remember, for velocity, you have to consider both distance and direction for your calculations to make sense.

[kurja]

The velocity on the surface for B, is the same as the velocity in orbit for A. These two arrows cancel each other out because they are both moving at the same speed at the same time. This means that now you have three arrows for both halfs of the sphere. When A is stationary like that the projected point D and the projected point C are not moving faster or slower than each other. This is because the rotational speed of the surface is the same in one point as it is in another. When you are stationary above point B, you move at the same speed it moves. That point moves across the surface at the same speed that all the other points move across the surface. This is because the surface speed is constant. The surface does not accelerate ever. This is what's known as the Coriolis effect. The person on the ground gets the illusion that A is moving faster towards it than D, than the same person on the ground would see A moving toward C.

ookay, so velocity of A when moving left is the same as B. Why not. So horizontal surface velocity of A in that scenario is nil. What is the horizontal surface velocity of A when it moves right to D? Hint: it's not nil.

All objects of the same mass fall at the same speed of gravity rather or not they are launched or dropped. Object A gets launched in both directions from it's position above B. V = GT both objects arrive in the same position at the same time.

Or if you still don't believe that, Mythbusters says the same thing:

They can fall at the same rate all they want, this whole insane ten page argument is about difference in horizontal speed that is due to rotation of a planet. We've been through this, speed at which an object falls matters squat for it's horizontal speed.

[jwilhite]

Kurja, Specialist, Smidge,

You have all valiantly defended the truth. Well done. But now there really aren't any other arguments to make that haven't already been aired many times in this thread, and I think there is zero chance that some casual observer of this thread is going to be led astray by the misinformation here. So, it's time to let it go.

http://xkcd.com/386/

Jeff

[700NitroXpress]

ookay, so velocity of A when moving left is the same as B. Why not. So horizontal surface velocity of A in that scenario is nil. What is the horizontal surface velocity of A when it moves right to D? Hint: it's not nil.

They can fall at the same rate all they want, this whole insane ten page argument is about difference in horizontal speed that is due to rotation of a planet. We've been through this, speed at which an object falls matters squat for it's horizontal speed.

Horizontal velocity In direction D is the same as the opposite horizontal velocity to B. If you have two craft in orbit and are 100 meters apart and you are in between them. You decide to you want to go to D at a speed of 5 m/s. You are also getting further away from the craft C at 5 m/s because that was your change in velocity. This means that you are traveling to D at +5 m/s and leaving C at -5 m/s. Thus you are using the same amount of fuel for the burn to change your velocity 5 m/s total. The velocity of the ground has nothing to do with this. By the time you reach D, time will have passed for 10 seconds. You will now be at D and you will be 150 meters away from C. The opposite is true if you decide to go to C instead of D.

[kurja]

Horizontal velocity In direction D is the same as the opposite horizontal velocity to B yadda yadda yadda

So you have the same velocity in both directions... in you very particular frame of reference. Awesome! Now take the direction in which the surface moves into account. ... no, I'm taking xkcd's and jwilhite's word for it and going to sleep, fare well.

[Smidge204]

Back to the OP, will a retrograde planetary orbit cause you to need more delta-v for encounters with said planets moons? I think it should but I'm not certain.

Absolutely, yes. Potentially by a LOT more. You need to (roughly) match the moon's velocity to be captured by it, and if you're in a retrograde orbit you're coming completely the wrong way... meaning you'll effectively have to slow down and reverse direction.

Due to the conservation of energy, they will have the exact same amount of energy and thus, you need the same amount of counter energy to stop them. For every action their is an equal opposite reaction.

For someone who likes to quote Newton a lot, you seriously lack any depth of understanding. :/

The problem with the video you posted is it's simply not applicable. I'll explain why in a bit, because the video you posted with Adam Savage is not applicable for exactly the same reason.

The velocity on the surface for B, is the same as the velocity in orbit for A. These two arrows cancel each other out because they are both moving at the same speed at the same time. ... When A is stationary like that the projected point D and the projected point C are not moving faster or slower than each other.

Um. They're moving in opposite directions, though. You DO understand that velocity is a vector, right? That velocity has a magnitude AND a direction, and that direction matters? The MAGNITUDE of the velocity for paths AC and AD are the same, but the DIRECTION is not, and therefore the VELOCITIES are not equal either. If we arbitrarily choose right-to-left as the positive direction, then the ball at point C has velocity +V and the ball at point D has velocity -V.

This is because the rotational speed of the surface is the same in one point as it is in another. When you are stationary above point B, you move at the same speed it moves.

Wrong. If you are moving at the same speed it [the ground] moves at point B, then you are NOT STATIONARY with respect to the image's frame of reference. You are literally saying "When you are stationary, you are moving" - doesn't that hurt your head just a little bit?

The crux of your difficulty stems from the inability to understand reference frames. You fritter back and forth between an orbital reference in which the planet moves to a planetary reference in which the planet does not move. Because of this, you seem unable to get past any of the explanations being offered. Please fix this.

This is what's known as the Coriolis effect.

There is no Coriolis effect because the frame of reference in my diagram is not rotating. See above.

Or if you still don't believe that, Mythbusters says the same thing:

Okay, and now back to this completely separate misunderstanding of yours. In both of the videos you referenced, there is no curvature of the ground taken into consideration. We, discussing orbital mechanics, have no such luxury of assuming the ground is actually flat. If Adam has a gun capable of firing a bullet at a whopping ~7,900 m/s (and there was no atmosphere or buildings/mountains/etc in the way), he'd shoot himself in the back, and the bullet, unobstructed, would NEVER hit the ground ever again.

When you're talking about distances and velocities relevant to planetary scales, everything you've demonstrated that you know is woefully inadequate. This is why Newton had to invent calculus to explain it.

Since you like Newton so much, here's a diagram that he himself drew:

It bears more than a striking similarity to my own diagram, because that's how physics works.

=Smidge=

/Bedtime isn't for another hour!

[numerobis]

Back to the OP, will a retrograde planetary orbit cause you to need more delta-v for encounters with said planets moons? I think it should but I'm not certain.

For an encounter, no; you just need to have your orbital planes to cross. But if you want to stop at the moon, you'll need a lot more delta-V. If that moon has an atmosphere, no problem, the atmosphere can provide the deltaV to slow you down. If it doesn't, good luck.

[700NitroXpress]

So you have the same velocity in both directions... in you very particular frame of reference. Awesome! Now take the direction in which the surface moves into account. ... no, I'm taking xkcd's and jwilhite's word for it and going to sleep, fare well.

Right so if you slow down a set speed in either direction that speed is the same. This means that you will hit the surface at the same time. The angle at which you are going in the parabolic arc is equal opposite on both sides and it can't be any other way. This means that you would angle your craft in the equal opposite direction to your velocity to the surface. Which is... guess what, shown on the nav ball. This means that by coming into an object rotating, you would angle your craft and burn to zero the velocity which is the same in both cases because V = gt The rotation speed of the planet is it's angular velocity, which is the same on all points along the equator of a planet. Angular velocities are either equal at the poles or equal opposites. That means that it takes the equal opposite amount of fuel to decelerate and land given the rotational direction and the angle of approach. This is because angular velocity is a pseudovector. As described here: http://en.wikipedia.org/wiki/Pseudovector.

Equal magnitude, but opposite direction.

[boomerdog2000]

For an encounter, no; you just need to have your orbital planes to cross. But if you want to stop at the moon, you'll need a lot more delta-V. If that moon has an atmosphere, no problem, the atmosphere can provide the deltaV to slow you down. If it doesn't, good luck.

That's what I thought after the predicted 4000 dv needed to circularize a Vall orbit, good news is the mission was a success after a quickload.

Also Nitro think of it this way. The planet is orbiting 50m/s we'll call its direction positive for this.

First Example:

You are traveling in the same direction 100 m/s.

To slow down to the planet's 50m/s from your 100 m/s you have to slow down 50 m/s.

Second Example:

If you are traveling retrograde you are traveling -100m/s.

Now you have to speed up to the planet's 50m/s.

To speed up to that you have to cancel out your 100m/s then speed up to the planet's 50m/s

First example:

50-100=-50 50m/s against the planet's rotation

Second exmple:

50--100=150 150m/s with the planet's rotation

[700NitroXpress]

That's what I thought after the predicted 4000 dv needed to circularize a Vall orbit, good news is the mission was a success after a quickload.

Also Nitro think of it this way. The planet is orbiting 50m/s we'll call its direction positive for this.

First Example:

You are traveling in the same direction 100 m/s.

To slow down to the planet's 50m/s from your 100 m/s you have to slow down 50 m/s.

Second Example:

If you are traveling retrograde you are traveling -100m/s.

Now you have to speed up to the planet's 50m/s.

To speed up to that you have to cancel out your 100m/s then speed up to the planet's 50m/s

First example:

50-100=-50 50m/s against the planet's rotation

Second exmple:

50--100=150 150m/s with the planet's rotation

This is only in relevance in a universe without mass and gravity. We do not live in such a universe.

If something is traveling at a speed in one direction over something else, is this object in orbit?

Also, does an object that rotates produce a different gravity than one that does not?

Does gravity act differently on two objects in orbit?

What forces are at work on an object in orbit as opposed to an object on the ground?

Do objects of the same mass that are released fall at the same speed?

You aim two guns on the equator opposite each other pointed East and West and fire them, also you drop the same bullet from a stationary elevated height at the exact micro second that the two guns are fired and at the same height that the two guns fire from. Which bullet hits the ground first?

Which bullet travels the greatest distance?

Which bullet of the two fired has the greatest range?

Answer all of these questions correctly if you can.

[numerobis]

Answer all of these questions correctly if you can.

Either your understanding of physics is wrong, or the understanding of physics of absolutely every other commenter here is wrong. Which is more likely?

[sjwt]

This thread has run about its longest course I believe, independent of which way you are orbiting it.