Entering SOI's Clockwise or Counter-clockwise

[SirSnuggles]

What's the difference? Also, AFAIK, all the planets rotate clockwise, no? Why do you always see people saying they are entering the SOI the wrong direction & flip it so that it is rotating w/ the planet?

[700NitroXpress]

Honestly it doesn't matter at all. The difference would be how you plot your landing zone. Rotating with the planet and you overshoot the LZ a little because you're compensating for the rotation. Rotating against the planet means that you have to undershoot your LZ. It boils down to personal flight preference. This would only be different if you're landing on a planet with atmosphere because you would want to overshoot the LZ due to the aerobraking of re-entry. Trying to land back on Kerban rotating against the planet would be difficult if you're trying to aim at the KSC as being your LZ. Unless your lander has a lot of power and can come in straight down instead of at an angle.

[kurja]

you do need to spend more fuel on powered landing if your orbit is opposite to the planet's rotation

[astropapi1]

When ascending from a planet, you get a little boost from it's rotation. That's why we always launch to the right (counter-clockwise).

[700NitroXpress]

you do need to spend more fuel on powered landing if your orbit is opposite to the planet's rotation

No, you would be orbiting at the same speed rather it's clockwise or counter clockwise. Your orbital speed in relation to a point on the surface would be greater if you were going opposite of the planet's rotation. You would need to burn the same amount of fuel to land no matter which way you're orbiting.

[700NitroXpress]

When ascending from a planet, you get a little boost from it's rotation. That's why we always launch to the right (counter-clockwise).

Not by much, if you start the turn early enough, then the tiny boost you get from the rotation gets canceled out and you end up with the same result as shown below. About 4% difference in fuel consumption.

Edited November 11, 2013 by 700NitroXpress

[RexKramer]

One good reason to try to enter an orbit heading 90ish is if you are planning on docking with a vessel from the surface, when an atmosphere is present.

For example, on Laythe and Duna the easiest way to orbit from the surface is to circle 90 degrees. Even on bodies with no atmosphere it takes less fuel to reach orbit by heading 90 than 270. So if you bring your recovery ship in heading 270, you just made it harder for your surface ship to match orbits.

Also, if you enter a planet's SOI with the intention of orbiting a moon or landing on a moon, it's easier to capture a moon's orbit if you are orbiting the planet in the same direction as the moon.

[Spyritdragon]

Not by much, if you start the turn early enough, then the tiny boost you get from the rotation gets canceled out and you end up with the same result as shown below. About 4% difference in fuel consumption.

4% is a pretty neat bonus, when using large lifter stages. If you start off with a few dozen thousand units of fuel, you can power a NERVA for quite a while.

[kurja]

No, you would be orbiting at the same speed rather it's clockwise or counter clockwise. Your orbital speed in relation to a point on the surface would be greater if you were going opposite of the planet's rotation. You would need to burn the same amount of fuel to land no matter which way you're orbiting.

erm, I disagree. You are correct in that the orbital speed is the same at a given altitude regardless of which way you're going, but the surface speed is not, and that's what you need to negate when landing. Like you said yourself, "orbital speed in relation to a point on the surface would be greater if you were going opposite of the planet's rotation" - it is exactly that relative speed that needs to go to zero when landing, and to make that larger change in speed more fuel needs to be used than if you orbited along the planet's rotation.

[Kasuha]

It is completely irrelevant which direction you enter the SOI.

It is also irrelevant which direction you orbit a planet.

It is relevant which direction you lift off or land, but not always:

On planet with very low surface speed it is irrelevant. Mun is such example.

On planet with atmosphere, only liftoff direction is relevant. You can land in any direction as the atmosphere will slow you down before you reach surface. This is limited for Duna, though.

[700NitroXpress]

erm, I disagree. You are correct in that the orbital speed is the same at a given altitude regardless of which way you're going, but the surface speed is not, and that's what you need to negate when landing. Like you said yourself, "orbital speed in relation to a point on the surface would be greater if you were going opposite of the planet's rotation" - it is exactly that relative speed that needs to go to zero when landing, and to make that larger change in speed more fuel needs to be used than if you orbited along the planet's rotation.

No, because you are going at a specific velocity to the ground in relation to your altitude. A craft orbiting a moon at 1000 m/s would need to slow down 1000 m/s to stop rather or not it's going clockwise or counter clockwise. Also, that same craft would need to descend to the ground. So if you have an orbital altitude of 100km and you kill the horizontal velocity, that velocity is the direction you are traveling and has nothing to do with the ground because you are not in contact with the ground. So the amount of fuel it takes to slow down going 1000 m/s and land for a powered landing from 100 km is the same in both cases because your are going to descend at a set velocity and that velocity is dependent on the objects gravity and the gravity alone because there is no other force acting on the object in orbit. If wouldn't matter if the moon was rotating 1000 times per second or 1 time per day, the gravity of the moon is dependent on mass and the moon's mass is the same. Therefore the gravity is the same, and thus you're downward velocity is the same no matter which way it rotates or you rotate to begin with.

And that is Newtons laws of physics which are fact and the physics that this game uses, so you sir are wrong.

If you still disagree, then I suggest you read this: http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

Edited November 11, 2013 by 700NitroXpress

[wasmic]

It is always easiest to launch heading 90 degrees, but on planets/moons with slow rotation, it doesn't matter a lot. Also, since all moons orbit counter-clockwise in KSP, there is no reason to do a clockwise orbit.

[700NitroXpress]

4% is a pretty neat bonus, when using large lifter stages. If you start off with a few dozen thousand units of fuel, you can power a NERVA for quite a while.

True, but as Kasuha stated below, this bonus is only really relevant for takeoff of Kerban and Eve. Otherwise it doesn't matter. But since my statement was directed towards takeoffs from Kerban, you're right, the bonus is there and we all want to save fuel.

[kurja]

No, because you are going at a specific velocity to the ground in relation to your altitude. A craft orbiting a moon at 1000 m/s would need to slow down 1000 m/s to stop rather or not it's going clockwise or counter clockwise. Also, that same craft would need to descend to the ground. So if you have an orbital altitude of 100km and you kill the horizontal velocity, that velocity is the direction you are traveling and has nothing to do with the ground because you are not in contact with the ground. So the amount of fuel it takes to slow down going 1000 m/s and land for a powered landing from 100 km is the same in both cases because your are going to descend at a set velocity and that velocity is dependent on the objects gravity and the gravity alone because there is no other force acting on the object in orbit. If wouldn't matter if the moon was rotating 1000 times per second or 1 time per day, the gravity of the moon is dependent on mass and the moon's mass is the same. Therefore the gravity is the same, and thus you're downward velocity is the same no matter which way it rotates or you rotate to begin with.

And that is Newtons laws of physics which are fact and the physics that this game uses, so you sir are wrong.

If you still disagree, then I suggest you read this: http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

See bolded part of quote. That velocity to the ground is different depending on your direction. In your example the orbital speed of 1km/s at a given altitude is the same regardless of direction, but if you nullify it, that does not make your horizontal surface speed zero which has everything to do with landing. No point in arguing I guess, I'll do a test this evening when I get home...

[700NitroXpress]

Now when we're talking about SOI, it is highly important which way you exit an SOI, exiting prograde to a planets rotation around the sun will eject you outward from the planet's orbital path and ejecting retrograde will put you inside the planets orbital path. This is because you have to think about the planet's position in it's orbit around the sun as a periapsis. To get to Moho and Eve from Kerban, you go retrograde and to go everywhere else, you go prograde.

[Dunbaratu]

No, because you are going at a specific velocity to the ground in relation to your altitude.

No, sideways relative velocity with the ground matters a heck of a lot. On Kerbin, for example, at sea level the ground is moving at 174 m/s from west to east. It's moving slightly faster at higher altitudes.

When a hill hits you at 389 miles per hour, the fact that it attacked you from the SIDE instead of you attacking it from above will be of little consequence. The hill will still win.

The only reason you're not noticing it much on the Mun is because the Mun is tide-locked and so it rotates very slowly compared to most of the bodies in the system.

Your claim that you don't need to kill the relative difference between your horizontal velocity and the ground's horizontal velocity to get a safe landing would only be true of the ground was a perfectly smooth sphere, and it was frictionless.

Edited November 11, 2013 by Steven Mading

[700NitroXpress]

See bolded part of quote. That velocity to the ground is different depending on your direction. In your example the orbital speed of 1km/s at a given altitude is the same regardless of direction, but if you nullify it, that does not make your horizontal surface speed zero which has everything to do with landing. No point in arguing I guess, I'll do a test this evening when I get home...

NO, your descent velocity is from you to the surface, that velocity is caused by one force and one force only GRAVITY and GRAVITY IS CONSTANT therefore your decent velocity to the ground is the same.

There are two forces at work here: Force A = orbital velocity, Force B = Gravitational velocity

Force B is constant and cannot be changed

Force A gets changed to 0 from a deceleration burn.

Therefore only Gravity is acting on the craft. Your speed is only going in two directions horizontal and vertical. The rotational speed of an object doesn't affect it's gravity or the distance between you and the ground. There is absolutely no force of an objects rotation that impacts distance or gravity. If it take 100 seconds to descend 10km then it takes 100 seconds to descend 10km. If it takes 100 units of fuel to stop and land because you're countering your vertical velocity, then it takes 100 units of fuel to stop. Rotation speed means nothing, it's Newtons laws!

[Dunbaratu]

A rider sitting atop a motorcycle is typically less that 1 meter up above the ground. If the motorcycle is stationary and the rider falls off, in the time it takes to fall from that height, the vertical component of velocity is still quite small and only results in a minor owie. In the alternate universe you're talking about, 700NitroXpress, a person going 100 miles per hour on a motorcycle who falls off the motorcycle should be just fine, because its no worse than falling off it when it was at rest.

Friction with the moving ground MATTERS.

[700NitroXpress]

No, sideways relative velocity with the ground matters a heck of a lot. On Kerbin, for example, at sea level the ground is moving at 174 m/s from west to east. It's moving slightly faster at higher altitudes.

When a hill hits you at 389 miles per hour, the fact that it attacked you from the SIDE instead of you attacking it from above will be of little consequence. The hill will still win.

The only reason you're not noticing it much on the Mun is because the Mun is tide-locked and so it rotates very slowly compared to most of the bodies in the system.

If I'm in orbit around Kerban at 75,000 meters traveling at 2300 m/s and I decide to come to a stop and land on the surface. I kill the 2300 m/s horizontal velocity. Right now the only force acting on the craft is GRAVITY. Gravity pulls me to the surface of Kerban at a rate of 9.8m/s/s. During the descent I deploy my parachutes and softly land perfectly straight up right on the ground. This is because the rotation of the planet doesn't affect the craft's velocity at all. Newton's Laws.

[Dunbaratu]

If I'm in orbit around Kerban at 75,000 meters traveling at 2300 m/s and I decide to come to a stop and land on the surface. I kill the 2300 m/s horizontal velocity. Right now the only force acting on the craft is GRAVITY. Gravity pulls me to the surface of Kerban at a rate of 9.8m/s/s. During the descent I deploy my parachutes and softly land perfectly straight up right on the ground. This is because the rotation of the planet doesn't affect the craft's velocity at all. Newton's Laws.

Bull****. It's because Kerbin has an atmosphere and air drag is doing the work of matching speeds with the ground for you. Try that trick on a body that spins fast (not the Mun) and doesn't have an atmosphere. Kill your orbital velocity until the navball says (in orbit mode, not surface mode) that you are falling absolutely vertically , right toward the ground. As you approach the ground, notice how fast it's moving sideways. Try to land without killing that sideways motion.

Edited November 11, 2013 by Steven Mading

[Smidge204]

See bolded part of quote. That velocity to the ground is different depending on your direction. In your example the orbital speed of 1km/s at a given altitude is the same regardless of direction, but if you nullify it, that does not make your horizontal surface speed zero which has everything to do with landing. No point in arguing I guess, I'll do a test this evening when I get home...

A picture is worth a thousand words:

In order to land, you must match the speed of the ground. This requires more change in velocity if you're traveling the wrong direction. The same is true for takeoffs in the wrong direction.

So, if you plan to land on a body, entering the SOI in the right direction is important.

If you plan to rendezvous with a moon, it's especially important: Imagine that the moon you want to encounter is the surface of a really large, imaginary planet... if you're orbiting the opposite direction, you have to stop and turn around to match the velocity... which could be a delta-V of 10,000+ m/s. That's not including getting yourself captured into orbit around said moon, or landing!

=Smidge=

Edited November 11, 2013 by Smidge204

[700NitroXpress]

A rider sitting atop a motorcycle is typically less that 1 meter up above the ground. If the motorcycle is stationary and the rider falls off, in the time it takes to fall from that height, the vertical component of velocity is still quite small and only results in a minor owie. In the alternate universe you're talking about, 700NitroXpress, a person going 100 miles per hour on a motorcycle who falls off the motorcycle should be just fine, because its no worse than falling off it when it was at rest.

Friction with the moving ground MATTERS.

A man moving on a motorcycle falls at the same velocity of 1g on Earth rather he's moving at 100 m/s or not at all because GRAVITY IS CONSTANT!

He hit's the ground at the same time because the same force is pulling him to the ground in both cases.

Friction has absolutely no effect on an object that isn't in contact with another object.

They both fall at the same rate. When you fall to a planet, you fall at the same rate. If this man on a motorcycle was in space and orbiting Minmus at 100 m/s he would not touch the ground because he is in orbit. Objects in orbit are falling towards the surface still because they are being acted on by gravity. When you stop going 100 m/s, you are no longer countering the force of gravity. You then fall at the rate of gravity of the object. Again Newton's laws.

[Dunbaratu]

A picture is worth a thousand words:

In order to land, you must match the speed of the ground. This requires more change in velocity if you're traveling the wrong direction. The same is true for takeoffs in the wrong direction.

=Smidge=

I think the problem is that 7000Nitro is forgetting that atmosphere moves with the body, and therefore it's pushing you sideways. With atmosphere present, you end up eventually being pushed sideways until you asymptotically approach matching speeds with the ground anyway even when you don't try to, and 7000Nitro was not paying attention to the fact that this is happening, instead thinking "I'm still falling straight down in orbital terms" when that's not really true. Thus my suggestion to try it somewhere with good rotation and not atmosphere. (Kerbin has atmo, and the Mun only rotates once a month, so if that's all the experience he has with the game he won't realize he's wrong) That experience will correct the misconception pretty quickly.

[Dunbaratu]

A man moving on a motorcycle falls at the same velocity of 1g on Earth rather he's moving at 100 m/s or not at all because GRAVITY IS CONSTANT!

He hit's the ground at the same time because the same force is pulling him to the ground in both cases.

Friction has absolutely no effect on an object that isn't in contact with another object.

They both fall at the same rate. When you fall to a planet, you fall at the same rate. If this man on a motorcycle was in space and orbiting Minmus at 100 m/s he would not touch the ground because he is in orbit. Objects in orbit are falling towards the surface still because they are being acted on by gravity. When you stop going 100 m/s, you are no longer countering the force of gravity. You then fall at the rate of gravity of the object. Again Newton's laws.

Gravity isn't constant. It's stronger the lower your altitude. Not that this error in what you're saying has anything to do with the enormous foot you insist on stuffing into your mouth with this bogus argument, but it does highlight the fact that you should calm down and take a LOOK at what you're saying and think about it harder.

Your claim was not merely that you fall at the same rate either way, but that this means there's no need to match velocity with the ground in order to land. That is enormously false, as "landing" is not "crashing". And "crashing" is exactly what you'll do if you try to land while the ground is whizzing past horizontally. You keep repeating the same thing without responding to what was said. Please explain why it's perfectly safe for a hill to hit you at 389 miles per hour sideways, or for a motorcycle rider to fall off while the ground is whizzing past at 100 miles per hour just because it's a short fall.

[kurja]

As a simplified analogy, let's say you're a bird flying in calm weather at air speed of 2m/s. Below you, there are two conveyors both moving 0.5m/s (relative to the ground), one in same direction as you and the other in the opposite direction. So your relative speed to one of the conveyors is 1.5m/s and 2.5m/s to the other. Do you need to slow your horizontal speed more or less to land on one of the conveyors? Or if you make your air speed (analogous to orbital speed, here) zero, what is your speed relative to the conveyors, can you land softly on either?

I hear linking to wiki articles makes me more right so here's one